Just trying to be economical, but I understand if it's not allowed. Can we use both sides of each engineering paper sheet (one per problem, or as an extension to a long problem)?

I'm willing to bet that it's okay. I think the Homework submission just states that you have to submit one pdf form and that new problems shouldn't be started on the same page as the previous.

I think it would be fine as some professors have stated that engineering paper is not specifically needed. Only graphing paper at the minimum to encourage neat work.

I don't think you can use the front and back of the engineering paper. The reason for this is because I believe there is only one certain side we are supposed to use.

According to my professor we are just supposed to use the lighter side, since the other side makes it difficult to read. Also using the back side could bleed into the other side making it difficult to read.

I know at this point you probably already know, but the syllabus states that we aren't allowed to do that. Only to use the side with the boxes at the top.

I think it depends on how it will scan in. We are not supposed to use the darker side of the paper since the gridlines are dark and will most likely show up when scanning in your work. If you are trying to be economical I recommend getting regular graph paper and drawing the heading boxes onto the paper. Then you could use the front and back of the paper.

In the announcement section, the prof. said this "Submitting HWs: No special paper is required. Use a proper app to scan it instead of a raw photo (app suggestions: CamScanner, Scannable). " Does this mean the I don't have to use engineering paper right?

If I run out of room on my paper for a problem, I'm assuming I can continue it on another sheet of paper, right? Or do does the whole problem have to fit on one paper?

I don't realistically think any professor would penalize you for using multiple papers; in my experience trying to force your work to fit in minimal space/on one sheet bothers them more than using the space you need for organized work that you and your professor/grader can both follow with ease.

No I believe you can use as many sheets as you want as long as you clarify that it is a continuation of the problem (i.e. page numbers or titling 'Homework X continued').

I think it is more professional to scan the homework, with a scanner app them save it as a pdf. This ensures that the homework is clear and legible to the professor can read it.

Personally I like to handwrite and scan it. However if you can write it onto an iPad and create a PDF I don't see the problem with that as long as you have the correct header for the formatting of the homework.

I use the app Adobe Scan which is free to use. Would definitely recommend. It appears also the best thing to do if a problem requires more than one sheet is to use another engineering sheet of paper rather than the "back side" of the first page.

I have only done physics problems like the ones on this homework with Metric Units, not British gravitational units. Does anyone know of any good resources to get a better understanding of how to work with lbs, slugs, and ft/s/s? I think the idea of slugs being a derived unit is slightly confusing.

I too have only ever worked with SI units and the concept of slugs being derived is also very confusing to me. If it helps, since slugs and kilograms are both units of mass, there is a direct conversion rate between kilos and slugs. Additionally, the conversion rate between ft/s/s and m/s/s is the same as simply converting meters to feet. Hope this helps!

I also found it helpful when I realized that slugs and kilograms were both measurements of mass. Using the acceleration due to gravity in imperial units with the mass helped me make force calculations analogous to SI units.

I always use Scannable to scan any hws as pdfs. The app allows me to directly save as a pdf on my iPhone, and even share it directly to my MacBook via AirDrop. It makes turning in assignments a breeze.

I seem to be having trouble visualizing some of the homework problems in my head, which make trying to make specific assumptions about the angles/cables difficult. Does anyone have any suggestions to try and better visualize problems?

It's a little hard to get used to but what helped me is to redraw what they have given in different colors to portray depth or difference in length. It also helps to lightly draw spotted lines between the drawings to show their relationship in position.

Diego, this is a great question! Certainly visualization can help solve problems. First thing you can try to make the best use of the things at hand. For example, you can use your book and water bottle to visualize H4B. Open your book to get an wedge with to side and put the bottle on the book. Than try to look at the front view. Second thing you can try is to make full use of the mathematical tools to help you! express the coordinate of the points, from which you can calculate the vectors (unit vectors), and use the concept of projection to find the components of the force vector to each axis. Then you can use the components to solve the equations. By doing this, you can simply your visualization effort to only locating the coordinates of the points of interest!
Hope this helps.

Sometimes I struggle with the steps of a problem. Like I first solve for all the needed components and then I normally get stuck when they ask for things vector form, because I cannot visualize the picture properly. Is the any advice on my problem?

I can relate to this as well. For me, once things become three-dimensional, picturing the situation becomes difficult enough to where its really not a viable strategy. The way I go about things is just finding the unit vector of a force and multiplying by its magnitude to create the vector.

Does anyone know if/when there will be video solutions of the HW problems uploaded? I just want to see if there are more efficient ways of doing the HW problems.

For 4.B, can we expect the forces coming off the inclined planes to be perpendicular to the planes? I feel like the planes cant exert force at another angle unless it has friction. Is that right?

Hi David. In this problem, you can neglect friction and assume that the sphere is contacting the two surfaces at a single point respectively. A single point contact implies a reaction perpendicular to the surface.

I think this would be nice. It is nice to have the lectures released so early, so I feel like it would help me stay on top of things if the homework was also released in a similar manner.

For the homework in each section, I find it best to use the most recent example problems. It has been very helpful in finding the correct steps in solution.

I've been rounding for every calculation, but I generally keep it at 4 decimal places so I have a little better accuracy. Once I get to the final answer I round a little more appropriately.

Hi Samuel, the 'correct' formula for stretched springs should be F = -k * delta_d. If the spring is stretched, delta_d is positive and force at the ends of the spring is negative (meaning that the spring would like to restore to relaxed position). But for the object linked to the spring, that means spring is pulling the objects towards the spring.

In homework 5, is it purposeful or a typo that problem A lists W as a mass, yet problem B lists W as a weight force? I ask because I do not want to be counted off for having the weight force of block A be Wg if W was meant to be the weight initially.

For hw5.a, they say the MASS OF THE block is W. This means that we would have to include g in some way shape or form in finding the force of gravity from the block, but it doesn't say put yourr answer in terms of W, k, d, AND g. Am I missing something?

When using F=-k*d for HW 5.A "d" is the string stretch or compression from the original length. In order to find the original length my thought is to add "d" to the current stretched length. How do we find the current stretched length?

The current stretched length would be the magnitude of the original spring vectors. Finding d would be the difference of the stretched length and the unstretched length.

I'm working on HW 6 and problem A asks you to find the moment arm d of F around point A. From what I can tell from the lecture 6 examples, the moment arm would be AC, which is directly parallel to the force vector F, meaning any moment would be 0. The next two parts of the question ask us to find the moment in different ways, though it should still be 0. Is this a trick, or am I thinking about it wrong?

In terms of the calculation itself, they are the same, so yes. However I believe "torque" is generally used to refer to a movement while a moment is a static force.

Yes. Moment about a point can, in fact, be thought of as torque. Since torque can be defined as a twisting motion about a point and Moment is the "tendency of a force to turn a body to which the force is applied" we can come to the conclusion that we can determine the terms Moment and Torque can be used interchangeably.

Was there a typo for H6.B part c? It says to compare to answer found in part a. However part a is just the force DE and part c is the moment about point A due to force DE. Was it meant to say compare answer to part b, because part b is also the moment about point A due to force DE but different method from part c?

Is there more than one answer for part d on H6.B? I think there should be because if you use vector rCE you get one answer but if you use rCD you get another. Help please ðŸ™‚

I know for Moment problems, you are not required to create Free Body Diagrams, but I am just wondering if they are helpful enough to create them in a Moment problem.

Personally I think it's always helpful to create a FBD in most if not all force-related questions. Just make sure that you are appropriately labeling all the information you need and that you don't confuse yourself with your drawing

For homework problem H7.A, is part a solved by basically solving for P so that the sum of all of the forces is 0, since that would mean that the 3F and the 5F (now 3F since P would be 2F in the other direction) would be a couple M? Also, once that answer is found, how are we supposed to solve for part b? Are we supposed to start at point D and find the sum of the moments about that point made by the other three forces? I don't believe that this is the case, since P wouldn't be factored in, meaning that it is no longer a couple, but I'm not sure.

Hi Glenn, I just checked and both tabs work for me. Can you check again and let me know? In the meantime, you can find the problem statement in the Assignment description in Gradescope (as a backup). We will solve this problem as soon as possible.

For 7b, I am having trouble trying to find position vector Rob. I understand the magnitude is just the radius R, but don't we need it as a vector in order to do the cross product?

In the lecture video nine with the example of the forked beam, I am confused how Professor looks at the rolling pin at point B and instantly can assume the direction of the force it will apply on the forked beam, as well as the direction of its moment. I know clockwise is negative and CCW is positive, but I am confused how he so easily draws this information as looking at the diagram does not seem very intuitive to me as far as direction of force applied is concerned.

Are there any scenarios in which it would be easier for us to solve if we flipped the axis' (x,y,z)? I am just curious if flipping the axis would change anything in the scenario except for visualization things.

Hi Diego. Changing the axis would not change the process for solving the question or make it easier to solve. It would merely change the magnitudes corresponding to their respective directions.

For Homework 11.B and similar problems, why in our moment equation do we not include the reaction moments around point B that revolve around the y and z axes?

Since there are two hinges, the translational x/y/z forces from the other hinge negate any net moment. This is also done to prevent overconstraining the system i.e. there would be too many variables.

Hi Benjamin, for magnitude of force, Yes. Add them together with the signs. For position of the load, use moment equation. The single force equivalent load creates the same moment about a point as that created by the distributed load.

For HW 12 A, to find the equivalent force, I'm assuming to find the magnitude and location of the single-force equivalent load, we have to integrate the OA and BC portions of the line load separately. If we do, would the limits of the BC portion be relative to point O, with the upper and lower limits being 4d and 2d, respectively?

I believe that you only have to do that for when you are finding the distance the equivalent force is from O. When you are looking for only the magnitude of the equivalent force, you do not have to do this, allowing for simplification. Just make sure that if you are not using limits relative to O that you are changing your W(x) equations accordingly.

For centroid problems, in a practice problem, the reference axis was in the middle of the structure so one of the shapes had a negative centroid. Are there cases in which it would be easier to shift the axis and then take that shift into account later?

Hi Diego, This is a very good question. Remember that centroid is defined for a geometry (object). It doesn't necessarily involves a coordinate system. So, you can always choose a proper coordinate system X'Y'Z' and calculate the centroid based on that coordinate system (X'Y'Z'). Once you know the position of the centroid w.r.t. the chosen coordinate system X'Y'Z' and the geometry, you can express that position in the coordinate system XYZ that's given.

For Homework 13.A, do we just search up the centroid for a quarter circle and a triangle or are there formulae we are expected to know to calculate these coordinates?

The centroid for a quarter circle is 4R/3pi, a triangle is 1/3 from the fat end and 2/3 from the pointy end. Professor Jones has given these to use, however I am not sure about other professors.

If you kept your origin at (0,0), then it would be 4R/3pi + R. This is because you have to add the space between the x axis and where the small semi circle (R) starts to get to the centroid.

Is there a specific x-bar equation for trapezoids, because I can see myself making a calculation error when breaking up the trapezoid into its composite parts just to put them back together to for the average x-bar for the force equivalent like we did with Fluid Statics today.

For HW 15.A, how would we go about solving for the volume submerged? Would we have to do a ratio from the big cone to the small cone, or is there another step that needs to be completed?

I'm pretty sure it's just a ratio. It's something we did in calc, and while I can't remember it very well, I believe that, since we know the ratio of height to radius, we can get the radius in terms of height (r = 1/5 h). If we sub that into the equation pi*r^2*(h/3), we get pi*(h/5)^2*(h/3), and simplify it if we feel like it. Then we do all the buoyancy and such calculations, using d as the height of the cone for the submerged volume.

Yes exactly. I have used similar triangles to find the submerged radius however i'm getting the answer in terms of d. As far as i know, there isn't a way to convert between them

For HW 15.A, I'm pretty sure I understand most of it, although the pulleys are confusing me. I notice they aren't all of the same size, and are connected kind of strangely (or maybe not, I don't know how pulleys work). Do the pulleys change the ratio of forces at all, or are they just there to redirect them so having two downward forces still makes sense?

We can neglect the size of the pulleys since we know the system is in static equilibrium. You should, however, take note that there appear to be two cables in the system holding up the weight of block A.

Since there are two cables in the system holding up block A, in a free body diagram concerning the block, would there be equivalent upward force vectors on each side of the pully acting on the block, or would there only be one force vector? For example, if the weight of block A was 60N, would there be upward tensions on the left and right of the pully at 30N each, or only one because there is only one pully?

Given that the system is in static equilibrium, I believe we can assume that the weight of the block is being distributed evenly across the two parts of the pulley holding up the block.

I believe the first step in order to solving homework 15.A is to identify the forces acting on the system in the x and y. Would the weight just have to be greater than the tension force and the buoyancy force?

I know I'm late, but you were on the correct track. Since the pulleys had no mass and the weight was to be a minimum, the buoyancy and the tension were all you had to work with.

For HW 15.a, I got the answer in terms of 'd' as that's the submerged part of the cone but the problem statement says that the answer should be in terms of R. I can't find a way to convert between R and d. How are we supposed to go about this?

It might be a bit late now, but in the problem, you can put d as 5R because it is asking for the minimum weight of Wa. This will happen when all the cone is submerged.

Yes that is what I did. Since it is asking for the moment when the cone is just at the surface of the water you can assume that the dimension of 5R starts at the surface and goes down and is completely in the water.

Are hydrostatic loads dependent on the horizontal extent of the water at all? The formula does not include it, but it feels wrong that it doesn't influence the force at all.

It is not dependent on the horizontal extent of the water at all, you are correct. With consistent density, hydrostatic force is only a function of depth.

For Homework 16.A do we not need to consider the part of the water that doesn't touch the beam itself? Or does finding finding the force of the triangular part of the water account for all the water to the right of it too?

You were looking at the definition of friction force, which expresses the friction force as the magnitude of the normal force times the static coefficient. However, since we do not know the static coefficient yet, we need to equate the friction force with something else. The only other force that is also acting in the x-direction is the normal force at C. You need to find the magnitude of the normal force at C, and equate that with the friction force at B using a static equilibrium equation.

If I am working with a 2D problem and I have a force acting in either x or y, do I need to include an over-bar for that force in the free body diagram and in my solution?

For example: F acts in the positive x direction and B acts in the negative x direction. Sum of the forces in x = F-B = 0. Should I include over-bars for both F and B in this equation?

I am just trying to understand the concept of impending motion regarding "Slipping and Tipping" a little better. So what it boils down to if the Force that we are dealing with is a "Causation Force", once we calculate the slipping and tipping forces, the smaller of those two values will be the correct condition, but if the force we are dealing with is "Preventing" the motion, then the bigger of the calculated forces will be the correct condition. This is the proper way to understand this concept, corret?

You do that in gradescope. When you go into gradescope you will select the individual assignment, and then the actual problem from that assignment. there should be regrade request buttons on the bottom right of the page. I got a point back on the test because of a regrade!

You should figure out the direction of the net force on the plane of the object (just figure if it is positive or negative). The friction force will be in the opposite direction to the net force on the plane the object is located.

Is there some sort of formula to determine the distance d that the normal force moves, or are we only given situations where we assume that it is at the very edge?

The normal force is not on the edge unless the object is in a state of impending tipping. When the block is in a state of impending slipping, then the normal force will move away from the edge a distance d. There isn't really a formula to find the distance d, but you can use the moment equations to find the distance if it is required.

So here is one way to think about this. Lets say that you are pushing a block at it's top left corner with a horizontal force F to the right. If we take the moments about the bottom left corner of the block, we do not know where d will be. So we set up the equations so that the sum of the moments equal zero. We get the force applied F times H, the block height, as one moment. The equalizing moment will be the the force of friction times d. Since d would be the only unknown, we solve for d. Now, if d is greater than the width of the block, we know that we will tip the block since that is not possible for d to be farther out than the block width. If d is less than the block width, and F is larger the the force of friction, then the block slips.

Does anyone have any tips/tricks trying to understand the specific values that you get for (Normal Force, Force Assuming Tipping, Force Assuming Slipping, etc), for example, whenever I get the Preventative Force of Tipping to be negative, what does that mean in terms of my diagram?

I'm pretty sure that a preventative force of tipping (assuming that means a restorative pushing force) is meant to prevent slipping and tipping. In the first example of lecture 18, Professor Jones mentioned at the end that if we had gotten a negative pushing force, it would mean we would have to pull the box down the slope with that amount of force to get it to slip before it tips.

I've been working on the homework due tonight, and I was rewatching the lectures for the assignment at the same time. I have been having trouble watching lecture videos because Kaltura is globally down. Is there anywhere else that I can view lectures videos tonight?

We don't for only pulley C, since the tensions on both sides of the pulley will be equal regardless of friction. However, you still need to assume friction on the pulley D.

For Homework 20A , do we account for the reaction forces at point O? On my FBD, I included the O_x and O_y forces, but I'm not sure how to account for this. It gives two additional unknowns that do not allow me to relate F and the normal force at A directly.

You are correct, there are normal forces at O in the x and y direction. However, you need to use a moment equation about O to be able to get rid of these forces. When you set up a moment equation around O, the Ox and Oy forces will disappear, and you'll be left with the Force, the Normal Force, and the friction with the wedge.

do we have to show all the forces acting on a body on the FBD or only the ones that are going to actually have an effect? for example in h20A, we never use the reaction forces at O in any of our calculations so do they need to be shown on the FBD?

I believe your FBD should always show all the forces acting on the body. Even if they are not used to solve the problem, they are still relevant to why the body is in static equilibrium.

I agree with Julianna. A FBD is meant to show the forces acting on a system/object, and one should always show all the forces to accurately represent what forces are acting on the body and where they are acting.

I was rewatching the lecture where H21.A 's set up is explained and the member AD was ignored for the sum of the forces and the free body diagram. Was that a mistake?

On H21.B, I have found it very difficult to know the orientation of DH relative to the horizontal. Can we assume that triangle CDH is a 3-4-5 triangle?

Concerning homework 22.A and B, in the diagrams it says horizontal members of length 3d and then again horizontal members of length 4d
which one should we take as the vertical length?? 4d or 3d??

For the reaction at K, there is a fixed support. Additionally, two truss members act on K, combined in both the x and the y direction. Therefore, there must be a reaction force at K in both the x and y direction. Two member trusses also act in the x and y direction at A as well, however A is not a fixed support. Since A is a smooth horizontal roller support, the reaction force at A can only act in the vertical direction.

I think the method of sections comes in handy when you are able to split the truss into different section. If there's no easy way to split it up, then I would just look at each individual joint.

For Homework 23B, how can we determine which members of the B-E-D triangle are zero force members. I've assumed that HE is a zero force member, however, from point E, I cannot cancel out other members since they are in a triangle. Any ideas on how to work through this ?

I believe that from joint E, only HE is a zero-force member. The reason you cannot cancel out the other members is because the entire BED triangle carries a load. The way I solved question b) does not require solving the specific loads for the members in that triangle though.

I have not solved this yet but I also believe that the BED triangle does not have any zero-force members. However, I do not believe that it is necessary to solve the forces for these members to find the answers for the question. If I'm not mistaken, HE and HI are zero force members and that assumption allows you to solve for the required answers.

For Frames, do we always need to draw the overall FBD and or the FBD of a two force member? If not, would we have to state the certain assumptions that we are making?

You should only have to make FBDs for members with more than two parallel forces, but drawing more won't hurt. You never need to explicitly state these assumptions, however that won't hurt to put down either. How you choose to tackle this problem relies entirely on how you feel about the material.

Since the members in HW24 are not two-force members, you cannot simply solve for the reactions at A or B using global equilibrium. You will need to look at individual members as well to solve.

I believe pin joints inherently allow rotation but not translation. That being said, since they are not actively resisting rotation, they do not supply a reaction moment.

Hey I solved for M being a negative value as opposed to that in the diagram but it asks for the answer as a vector and the diagram shows a negative K moment so I was wondering if these would cancel and the answer should be a positive value in the K direction.

Although the diagram shows that M is in the negative direction, I would just solve using M in the positive direction. When you do this you get a positive value in the K direction. If you solve it how you did you get a negative value, but if you look at the diagram, that means that the moment is counter clockwise. Thus, your vector will be in the positive K direction.

In 25.A do we need to account for the reaction force By when taking the sum of moments around point A or is By's affect on the moment around point A already included in moment M?

Moment M is a force couple pure moment, meaning it can be applied at any location on the beam and still retain its value. Therefore, I do not believe this couple includes the moment created by By

I know we have talked about creating shear force and moment diagrams at an internal point in the system, but what do these graphics look like for an overall system? On H27.A we are asked to find the overall shear force and bending moment diagram

In the lecture we found shear force and bending moment diagrams at internal points inside the system in terms of a distance "x" from the origin. We found equations instead of raw values and these equations can be applied for each section of the member. By graphing out these equations using the x axis and the x variable in each equation, we can create a diagram of shear force and bending moments along the member, with each section of the member having its own set of equations that should connect to make an overall diagram that is consistent.

Since H26 was dropped, are we okay to skip the lecture 26 altogether? Or does watching lecture 26 aid in the work we will need to do in H27, along with lecture 27?

I think it is recommended to watch all the lectures. The homeworks do not always cover just that single lecture, but rather multiple lectures. After watching both lectures, you may be able to solve the homework more effectively.

Lecture 26 still provides important information and helped my understanding of Homework 27 so I recommend watching it. Also, if you are doing the honors extension for this class it provides information needed to complete your project.

In lecture, professor mentioned that V'(x) should equal p(x). However, according to both his lecture example and the answer I got for H28.A, V'(x) = -p(x). What is the reason for the flipped sign?

In H29.A, upon first glance it looks like the distributed load equates to the downward force P and cancels it out. How should I approach this? When I work out the segments, should I account for portions of the distributed load or just use the equivalent load in the middle?

The way I went about this problem is I drew the overall FBD and found the moment about point B. I then broke it up into two segments from 0 <x<2 and 2<x<4. For the first segment, you should account for the portion of the distributed load, and for the second segment account for the equivalent load. By doing this you can determine the moment and shear equations which makes it easier to graph the shear and moment.

You can find V(x) by solving V'(x)=-p(x) where p(x) is the magnitude of the distributed load. The negative sign comes from the fact that the load is usually pointing in the -y-direction.
This will make the V(x) graph linear if the distributed load is constant, it will just not be parallel to the x-axis.

On stress and strain problems, are there any scenarios where it would help to section off a certain segment to analyze the axial load, or does it not matter which side you analyze after making a section.

On H30.B, there isn't a diameter value given for the cable even though we need it's cross-sectional area to solve for stress. How should we go about finding it?

If you watch the online-Krousgrill lecture, He states the value of d is .5 inches. Therefore, I solved in terms of d first than at the end just inserted that value of d in. I did this because the find doesn't state to leave your answer in terms of d.

If it is not too late I also just left my answer in terms of d. I saw that the online lecture that the value for d was given, however I did not use this value as he also had different numbers for other values that we do have. I assume that it was just a mistake for them to either not give us the value of d or to not ask for the answer in terms of d.

Hi John, I was confused about that as well. Since the value of d was not shared with everyone I believe it is safe to assume leaving your answer in terms of d will not result in a deduction of points. Hope that helps ðŸ™‚

Hey Moyosoreolu,
I believe that is the correct method for this problem. Professor Krousgrill advised us to do a similar thing in problem 30A during his lecture.

Hi Ryan! Yes, I attended Tutorial Room Office Hours and the TA said we must look up the tensile yield strength for aluminum 6061-T6. Here's the source I referenced, although apparently this tensile yield strength value can also be found in our lecturebook. http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA6061T6 Hope my comment helps!

I didn't use it for the specific way I approached the problem, but I know some people said they did. Basically, it is up to you to choose the way you want to do it, and that will dictate what values you need to solve.

Hello,
In 32b fo my shear calculation I multiplied the yield strength in shear by 2 since there are two shear planes. I think this is right but if anyone can confirm I would appreciate it.

This multiplication is technically correct. Since there are two areas of pure shear, the area should be multiplied by two, but, you can move this multiplication to the other side and instead multiply your yield strength by two. The most important thing is that this factor of two is only multiplied into either the yield strength or the area, and not both.

I believe either will work to reach the correct answer. I computed the reactions at A and B for my work, however that is only because it looked like the easiest way to solve the problem at the time.

Hi Andrew! rho is 3d/2 only for segment 1. This is because rho is equal to the radius of the shaft and since the diameter is equal to 3d, the radius would be 3d/2

I know it's past due but does anyone know why I kept getting the wrong answer for 33B for the shear stress when I was using all of the length measurements in feet?

I believe you needed to convert everything to inches since ksi is measured in square inches. I could be wrong though its been awhile since I looked at it.

In general for the maximum shear stress of a system, does the sign matter? Is it right to believe that the answer should be merely the largest absolute value of calculated stresses?

Hi Jadah, to the best of my knowledge the sign is only important in determining whether the component is in tensile stress or compression. Secondly, it is correct to use the absolutel value of the shear stress to determine the maximum shear stress of the system.

Hi Sera, I am not sure if you are refering to 34B instead, it doesn't seem as though we need to calcualte torque load in 34A. I would suggest going to the main homework discussion page and reading thorugh the comments of the specific homework assaignment you have questions about. There are very good answer and tips there. If you do mean 34B then I would suggest cutting the system into sections, with each cut exposing an internal torque load of one of the components. For example if you make a cut between C and D you will expose the torque load of component three. Then you can do a summation fo moments to solve for the torque load of component 3 (the only other moment in that case would be the applied moment 2T at D). Then you would repeat this process with cuts in the other sections to solve for the remaining torque loads. I hope this helps.

Just trying to be economical, but I understand if it's not allowed. Can we use both sides of each engineering paper sheet (one per problem, or as an extension to a long problem)?

I think I read that we can't

I'm willing to bet that it's okay. I think the Homework submission just states that you have to submit one pdf form and that new problems shouldn't be started on the same page as the previous.

I think it would be fine as some professors have stated that engineering paper is not specifically needed. Only graphing paper at the minimum to encourage neat work.

I think I read in the syllabus or brightspace to only use the light side of the engineering paper, if you are using engineering paper.

I don't think you can use the front and back of the engineering paper. The reason for this is because I believe there is only one certain side we are supposed to use.

My professor also commented on the importance of the boxes at the top of the engineering paper, which are not present on the back of each page.

I would assume you can only use one side because of the heading boxes.

According to my professor we are just supposed to use the lighter side, since the other side makes it difficult to read. Also using the back side could bleed into the other side making it difficult to read.

I use both sides since theres no bleeding thru and all we gotta do is scan it.

I don't think we can do it both sides of the paper

I think i saw on the syllabus that it wasnt allowed, i would re-check, but my other ME courses don't allow it

I know at this point you probably already know, but the syllabus states that we aren't allowed to do that. Only to use the side with the boxes at the top.

I googled it and the websites I read suggested against using the backside of the engineering paper. Sorry to break your bank.

No, because it is hard to read ur answers

I think it depends on how it will scan in. We are not supposed to use the darker side of the paper since the gridlines are dark and will most likely show up when scanning in your work. If you are trying to be economical I recommend getting regular graph paper and drawing the heading boxes onto the paper. Then you could use the front and back of the paper.

I'm assuming not, as my experience in the past with classes that use engineering paper has strictly upheld the one-side rule.

In the announcement section, the prof. said this "Submitting HWs: No special paper is required. Use a proper app to scan it instead of a raw photo (app suggestions: CamScanner, Scannable). " Does this mean the I don't have to use engineering paper right?

Yes, you don't have to use engineering paper.

If I run out of room on my paper for a problem, I'm assuming I can continue it on another sheet of paper, right? Or do does the whole problem have to fit on one paper?

I don't realistically think any professor would penalize you for using multiple papers; in my experience trying to force your work to fit in minimal space/on one sheet bothers them more than using the space you need for organized work that you and your professor/grader can both follow with ease.

Hey Matthew, you can continue to use new papers as long as it is included in the final pdf you submit!

No I believe you can use as many sheets as you want as long as you clarify that it is a continuation of the problem (i.e. page numbers or titling 'Homework X continued').

Is there any specific way we need to name the PDF of our homework when turning it into gradescope?

Nope

Does the homework have to be handwritten and scanned? Or are we allowed to do it on an iPad?

I think it is more professional to scan the homework, with a scanner app them save it as a pdf. This ensures that the homework is clear and legible to the professor can read it.

As long as its legible, is turned in as a PDF, and follows the homework format, I don't think it matters whether you do it digitally or on paper.

Personally I like to handwrite and scan it. However if you can write it onto an iPad and create a PDF I don't see the problem with that as long as you have the correct header for the formatting of the homework.

I use the scanner app Genius Scan and it works pretty well, but i always make sure it's legible before i turn it in.

I use the app Adobe Scan which is free to use. Would definitely recommend. It appears also the best thing to do if a problem requires more than one sheet is to use another engineering sheet of paper rather than the "back side" of the first page.

I use the Apple App "Scanner". I find it to be very effective and the pictures can be exported directly to your email in combined PDF form.

If you have an iPhone there is a built in scanner in the notes app.

I have only done physics problems like the ones on this homework with Metric Units, not British gravitational units. Does anyone know of any good resources to get a better understanding of how to work with lbs, slugs, and ft/s/s? I think the idea of slugs being a derived unit is slightly confusing.

The first chapter of the textbook has a brief overview of the imperial units.

I too have only ever worked with SI units and the concept of slugs being derived is also very confusing to me. If it helps, since slugs and kilograms are both units of mass, there is a direct conversion rate between kilos and slugs. Additionally, the conversion rate between ft/s/s and m/s/s is the same as simply converting meters to feet. Hope this helps!

I also found it helpful when I realized that slugs and kilograms were both measurements of mass. Using the acceleration due to gravity in imperial units with the mass helped me make force calculations analogous to SI units.

I always use Scannable to scan any hws as pdfs. The app allows me to directly save as a pdf on my iPhone, and even share it directly to my MacBook via AirDrop. It makes turning in assignments a breeze.

I seem to be having trouble visualizing some of the homework problems in my head, which make trying to make specific assumptions about the angles/cables difficult. Does anyone have any suggestions to try and better visualize problems?

It's a little hard to get used to but what helped me is to redraw what they have given in different colors to portray depth or difference in length. It also helps to lightly draw spotted lines between the drawings to show their relationship in position.

Hope this helps

Thank you so much! I have actually started to implement this and it has been really helping!

Diego, this is a great question! Certainly visualization can help solve problems. First thing you can try to make the best use of the things at hand. For example, you can use your book and water bottle to visualize H4B. Open your book to get an wedge with to side and put the bottle on the book. Than try to look at the front view. Second thing you can try is to make full use of the mathematical tools to help you! express the coordinate of the points, from which you can calculate the vectors (unit vectors), and use the concept of projection to find the components of the force vector to each axis. Then you can use the components to solve the equations. By doing this, you can simply your visualization effort to only locating the coordinates of the points of interest!

Hope this helps.

Thank you! I have not tried to physically try and visualize it with everyday items, but I think that would be extremely helpful!

For me the Free body Diagrams help a lot because you can clearly redraw the angles and components of the problem.

Hello,

When are they going to post the solutions to the homework? From what I've seen they haven't posted any for the ones that have already been graded.

Thanks

I think they normally post it the day after a lecture

Sometimes I struggle with the steps of a problem. Like I first solve for all the needed components and then I normally get stuck when they ask for things vector form, because I cannot visualize the picture properly. Is the any advice on my problem?

I can relate to this as well. For me, once things become three-dimensional, picturing the situation becomes difficult enough to where its really not a viable strategy. The way I go about things is just finding the unit vector of a force and multiplying by its magnitude to create the vector.

Does anyone know if/when there will be video solutions of the HW problems uploaded? I just want to see if there are more efficient ways of doing the HW problems.

For 4.B, can we expect the forces coming off the inclined planes to be perpendicular to the planes? I feel like the planes cant exert force at another angle unless it has friction. Is that right?

Hi David. In this problem, you can neglect friction and assume that the sphere is contacting the two surfaces at a single point respectively. A single point contact implies a reaction perpendicular to the surface.

Would it be possible to have the homework released slightly earlier so we don't have to wait until two days before it's due to start working on it?

I think this would be nice. It is nice to have the lectures released so early, so I feel like it would help me stay on top of things if the homework was also released in a similar manner.

For the homework in each section, I find it best to use the most recent example problems. It has been very helpful in finding the correct steps in solution.

Have you guys been rounding for each calculation, or only the final answer?

I've been rounding for every calculation, but I generally keep it at 4 decimal places so I have a little better accuracy. Once I get to the final answer I round a little more appropriately.

I'm confused on when we're expected to draw FBD. Should it become a habit to draw them on every problem?

Do you need to write force equilibrium equations like: summation of forces in x direction = 0 ?

If yes, then you are strongly encouraged to draw FBD.

For HW5.A, what is the force equation we are supposed to use for the springs. Is it F=-k*d ?

Hi Samuel, the 'correct' formula for stretched springs should be F = -k * delta_d. If the spring is stretched, delta_d is positive and force at the ends of the spring is negative (meaning that the spring would like to restore to relaxed position). But for the object linked to the spring, that means spring is pulling the objects towards the spring.

In homework 5, is it purposeful or a typo that problem A lists W as a mass, yet problem B lists W as a weight force? I ask because I do not want to be counted off for having the weight force of block A be Wg if W was meant to be the weight initially.

It doesn't matter which one because the TAs have been instructed to grade it both ways, so you won't lose points.

For hw5.a, they say the MASS OF THE block is W. This means that we would have to include g in some way shape or form in finding the force of gravity from the block, but it doesn't say put yourr answer in terms of W, k, d, AND g. Am I missing something?

Previously, @Radhika Kulkarni said TA's have been instructed to grade with or without g.

For H5.B, do we leave the tension answer as a vector or a scalar? It doesn't say

When using F=-k*d for HW 5.A "d" is the string stretch or compression from the original length. In order to find the original length my thought is to add "d" to the current stretched length. How do we find the current stretched length?

The current stretched length would be the magnitude of the original spring vectors. Finding d would be the difference of the stretched length and the unstretched length.

I'm working on HW 6 and problem A asks you to find the moment arm d of F around point A. From what I can tell from the lecture 6 examples, the moment arm would be AC, which is directly parallel to the force vector F, meaning any moment would be 0. The next two parts of the question ask us to find the moment in different ways, though it should still be 0. Is this a trick, or am I thinking about it wrong?

Can we think of Moment about a point as torque?

In terms of the calculation itself, they are the same, so yes. However I believe "torque" is generally used to refer to a movement while a moment is a static force.

Yes. Moment about a point can, in fact, be thought of as torque. Since torque can be defined as a twisting motion about a point and Moment is the "tendency of a force to turn a body to which the force is applied" we can come to the conclusion that we can determine the terms Moment and Torque can be used interchangeably.

For problem 6B, do we compare C and A? or answer C and answer B? Because I'm getting the same for B and C, not the same for A and C

Yes, I'm pretty sure it's a typo and is supposed to be c) to b).

Was there a typo for H6.B part c? It says to compare to answer found in part a. However part a is just the force DE and part c is the moment about point A due to force DE. Was it meant to say compare answer to part b, because part b is also the moment about point A due to force DE but different method from part c?

Is there more than one answer for part d on H6.B? I think there should be because if you use vector rCE you get one answer but if you use rCD you get another. Help please ðŸ™‚

I know for Moment problems, you are not required to create Free Body Diagrams, but I am just wondering if they are helpful enough to create them in a Moment problem.

Personally I think it's always helpful to create a FBD in most if not all force-related questions. Just make sure that you are appropriately labeling all the information you need and that you don't confuse yourself with your drawing

For homework problem H7.A, is part a solved by basically solving for P so that the sum of all of the forces is 0, since that would mean that the 3F and the 5F (now 3F since P would be 2F in the other direction) would be a couple M? Also, once that answer is found, how are we supposed to solve for part b? Are we supposed to start at point D and find the sum of the moments about that point made by the other three forces? I don't believe that this is the case, since P wouldn't be factored in, meaning that it is no longer a couple, but I'm not sure.

Pretty late to ask about 6A but whats the difference between questions b and c? Aren't rAC and moment arm d and the same line?

For 7b is it alright to assume the top most arrow is perpindicular to the circle (90 deg angle)?

The problem statement tab is not working on the blog

Hi Glenn, I just checked and both tabs work for me. Can you check again and let me know? In the meantime, you can find the problem statement in the Assignment description in Gradescope (as a backup). We will solve this problem as soon as possible.

Clarifying question: A mass-less rod wont have mass/weight, but it can still exert a force on a surface that it is placed against, correct?

Depends on the overall picture and where are all the other forces applied.

For 7b, I am having trouble trying to find position vector Rob. I understand the magnitude is just the radius R, but don't we need it as a vector in order to do the cross product?

Because the force is applied on the perimeter of the disk and B is on the -y-axis you have that Rob is simply .

In 7.A, it asks for the equivalent couple, but there are four forces so wouldn't it be 2 couples?

They're all vectors so any two couples can be added together.

In the lecture video nine with the example of the forked beam, I am confused how Professor looks at the rolling pin at point B and instantly can assume the direction of the force it will apply on the forked beam, as well as the direction of its moment. I know clockwise is negative and CCW is positive, but I am confused how he so easily draws this information as looking at the diagram does not seem very intuitive to me as far as direction of force applied is concerned.

Are there any scenarios in which it would be easier for us to solve if we flipped the axis' (x,y,z)? I am just curious if flipping the axis would change anything in the scenario except for visualization things.

Hi Diego. Changing the axis would not change the process for solving the question or make it easier to solve. It would merely change the magnitudes corresponding to their respective directions.

For Homework 11.B and similar problems, why in our moment equation do we not include the reaction moments around point B that revolve around the y and z axes?

Since there are two hinges, the translational x/y/z forces from the other hinge negate any net moment. This is also done to prevent overconstraining the system i.e. there would be too many variables.

I have to assume for HW 12 A and B, it's just a matter of finding the individual F_eq and xbar values and then adding them together, yes?

Hi Benjamin, for magnitude of force, Yes. Add them together with the signs. For position of the load, use moment equation. The single force equivalent load creates the same moment about a point as that created by the distributed load.

For HW 12 A, to find the equivalent force, I'm assuming to find the magnitude and location of the single-force equivalent load, we have to integrate the OA and BC portions of the line load separately. If we do, would the limits of the BC portion be relative to point O, with the upper and lower limits being 4d and 2d, respectively?

I believe that you only have to do that for when you are finding the distance the equivalent force is from O. When you are looking for only the magnitude of the equivalent force, you do not have to do this, allowing for simplification. Just make sure that if you are not using limits relative to O that you are changing your W(x) equations accordingly.

For centroid problems, in a practice problem, the reference axis was in the middle of the structure so one of the shapes had a negative centroid. Are there cases in which it would be easier to shift the axis and then take that shift into account later?

Hi Diego, This is a very good question. Remember that centroid is defined for a geometry (object). It doesn't necessarily involves a coordinate system. So, you can always choose a proper coordinate system X'Y'Z' and calculate the centroid based on that coordinate system (X'Y'Z'). Once you know the position of the centroid w.r.t. the chosen coordinate system X'Y'Z' and the geometry, you can express that position in the coordinate system XYZ that's given.

For Homework 13.A, do we just search up the centroid for a quarter circle and a triangle or are there formulae we are expected to know to calculate these coordinates?

The centroid for a quarter circle is 4R/3pi, a triangle is 1/3 from the fat end and 2/3 from the pointy end. Professor Jones has given these to use, however I am not sure about other professors.

What is the centroid of a semi-circle?

For quarter circle:

Xc = 4R/3pi

Yc = 4R/3pi

For half circle centered on origin:

Xc = 0

Yc = 2R/3pi

In the lecture book (pg 155) it states that for a filled in semi circle with a radius of R, the centroid would be:

Xc = 0

Yc = 4R/3pi

I think you were referring to a semi circle that was not filled in when you said "Yc = 2R/3pi"

Try to work on each quarter-circle and sum all of them together.

For 13.B, will the y-centroid for a small semi-circle be equaled to 4r/3pi + R? or just 4r/3pi?

If you kept your origin at (0,0), then it would be 4R/3pi + R. This is because you have to add the space between the x axis and where the small semi circle (R) starts to get to the centroid.

It's 4r/3pi + R because of the off-set

For 13.B can you multiply the entire bar by 2 in order to account for there being a half circle?

Is there a specific x-bar equation for trapezoids, because I can see myself making a calculation error when breaking up the trapezoid into its composite parts just to put them back together to for the average x-bar for the force equivalent like we did with Fluid Statics today.

This is a website that has the formulas and some practice problems for calculate the centroid of a trapezoid.

https://www.vedantu.com/formula/centroid-of-a-trapezoid-formula#:~:text=Centroid%20of%20a%20Trapezoid%20Formulas%20%2D%20Find%20the%20Centroid%20of%20a%20Trapezium&text=G%20is%20its%20centroid.,given%20by%20the%20following%20formula.

For HW 15.A, how would we go about solving for the volume submerged? Would we have to do a ratio from the big cone to the small cone, or is there another step that needs to be completed?

I'm pretty sure it's just a ratio. It's something we did in calc, and while I can't remember it very well, I believe that, since we know the ratio of height to radius, we can get the radius in terms of height (r = 1/5 h). If we sub that into the equation pi*r^2*(h/3), we get pi*(h/5)^2*(h/3), and simplify it if we feel like it. Then we do all the buoyancy and such calculations, using d as the height of the cone for the submerged volume.

This makes sense, but we are not allowed to have our answer in terms of d. But I do not see any way to express d in terms of R

Yes exactly. I have used similar triangles to find the submerged radius however i'm getting the answer in terms of d. As far as i know, there isn't a way to convert between them

For HW 15.A, I'm pretty sure I understand most of it, although the pulleys are confusing me. I notice they aren't all of the same size, and are connected kind of strangely (or maybe not, I don't know how pulleys work). Do the pulleys change the ratio of forces at all, or are they just there to redirect them so having two downward forces still makes sense?

We can neglect the size of the pulleys since we know the system is in static equilibrium. You should, however, take note that there appear to be two cables in the system holding up the weight of block A.

Since there are two cables in the system holding up block A, in a free body diagram concerning the block, would there be equivalent upward force vectors on each side of the pully acting on the block, or would there only be one force vector? For example, if the weight of block A was 60N, would there be upward tensions on the left and right of the pully at 30N each, or only one because there is only one pully?

There has to be two equal forces. If this were not true, it would cause rotation on the pulley, bringing it out of static equilibrium.

Given that the system is in static equilibrium, I believe we can assume that the weight of the block is being distributed evenly across the two parts of the pulley holding up the block.

I believe the first step in order to solving homework 15.A is to identify the forces acting on the system in the x and y. Would the weight just have to be greater than the tension force and the buoyancy force?

I know I'm late, but you were on the correct track. Since the pulleys had no mass and the weight was to be a minimum, the buoyancy and the tension were all you had to work with.

For HW 15.a, I got the answer in terms of 'd' as that's the submerged part of the cone but the problem statement says that the answer should be in terms of R. I can't find a way to convert between R and d. How are we supposed to go about this?

It might be a bit late now, but in the problem, you can put d as 5R because it is asking for the minimum weight of Wa. This will happen when all the cone is submerged.

Yes that is what I did. Since it is asking for the moment when the cone is just at the surface of the water you can assume that the dimension of 5R starts at the surface and goes down and is completely in the water.

Are hydrostatic loads dependent on the horizontal extent of the water at all? The formula does not include it, but it feels wrong that it doesn't influence the force at all.

It is not dependent on the horizontal extent of the water at all, you are correct. With consistent density, hydrostatic force is only a function of depth.

For Homework 16.A do we not need to consider the part of the water that doesn't touch the beam itself? Or does finding finding the force of the triangular part of the water account for all the water to the right of it too?

Hi. For HW 17.A I am having trouble solving for f is terms of W only. Is it alright to have Âµ in our answer as well? Or, am I messing something?

Hi Paul. If you solve the normal forces in B and C as a function of W only you would also get the friction force dependant on W only.

You were looking at the definition of friction force, which expresses the friction force as the magnitude of the normal force times the static coefficient. However, since we do not know the static coefficient yet, we need to equate the friction force with something else. The only other force that is also acting in the x-direction is the normal force at C. You need to find the magnitude of the normal force at C, and equate that with the friction force at B using a static equilibrium equation.

If I am working with a 2D problem and I have a force acting in either x or y, do I need to include an over-bar for that force in the free body diagram and in my solution?

For example: F acts in the positive x direction and B acts in the negative x direction. Sum of the forces in x = F-B = 0. Should I include over-bars for both F and B in this equation?

I am just trying to understand the concept of impending motion regarding "Slipping and Tipping" a little better. So what it boils down to if the Force that we are dealing with is a "Causation Force", once we calculate the slipping and tipping forces, the smaller of those two values will be the correct condition, but if the force we are dealing with is "Preventing" the motion, then the bigger of the calculated forces will be the correct condition. This is the proper way to understand this concept, corret?

Are we able to submit a regrade request? If so, how do I do that?

You do that in gradescope. When you go into gradescope you will select the individual assignment, and then the actual problem from that assignment. there should be regrade request buttons on the bottom right of the page. I got a point back on the test because of a regrade!

Generally, how can you tell what direction the frictional force is going if there are multiple forces acting upon an object? Is it just common sense?

You should figure out the direction of the net force on the plane of the object (just figure if it is positive or negative). The friction force will be in the opposite direction to the net force on the plane the object is located.

Is there some sort of formula to determine the distance d that the normal force moves, or are we only given situations where we assume that it is at the very edge?

The normal force is not on the edge unless the object is in a state of impending tipping. When the block is in a state of impending slipping, then the normal force will move away from the edge a distance d. There isn't really a formula to find the distance d, but you can use the moment equations to find the distance if it is required.

So here is one way to think about this. Lets say that you are pushing a block at it's top left corner with a horizontal force F to the right. If we take the moments about the bottom left corner of the block, we do not know where d will be. So we set up the equations so that the sum of the moments equal zero. We get the force applied F times H, the block height, as one moment. The equalizing moment will be the the force of friction times d. Since d would be the only unknown, we solve for d. Now, if d is greater than the width of the block, we know that we will tip the block since that is not possible for d to be farther out than the block width. If d is less than the block width, and F is larger the the force of friction, then the block slips.

Does anyone have any tips/tricks trying to understand the specific values that you get for (Normal Force, Force Assuming Tipping, Force Assuming Slipping, etc), for example, whenever I get the Preventative Force of Tipping to be negative, what does that mean in terms of my diagram?

I'm pretty sure that a preventative force of tipping (assuming that means a restorative pushing force) is meant to prevent slipping and tipping. In the first example of lecture 18, Professor Jones mentioned at the end that if we had gotten a negative pushing force, it would mean we would have to pull the box down the slope with that amount of force to get it to slip before it tips.

I've been working on the homework due tonight, and I was rewatching the lectures for the assignment at the same time. I have been having trouble watching lecture videos because Kaltura is globally down. Is there anywhere else that I can view lectures videos tonight?

For 19A why do we not need to account for friction in the pulleys?

We don't for only pulley C, since the tensions on both sides of the pulley will be equal regardless of friction. However, you still need to assume friction on the pulley D.

For Homework 20A , do we account for the reaction forces at point O? On my FBD, I included the O_x and O_y forces, but I'm not sure how to account for this. It gives two additional unknowns that do not allow me to relate F and the normal force at A directly.

You are correct, there are normal forces at O in the x and y direction. However, you need to use a moment equation about O to be able to get rid of these forces. When you set up a moment equation around O, the Ox and Oy forces will disappear, and you'll be left with the Force, the Normal Force, and the friction with the wedge.

do we have to show all the forces acting on a body on the FBD or only the ones that are going to actually have an effect? for example in h20A, we never use the reaction forces at O in any of our calculations so do they need to be shown on the FBD?

I believe your FBD should always show all the forces acting on the body. Even if they are not used to solve the problem, they are still relevant to why the body is in static equilibrium.

I agree with Julianna. A FBD is meant to show the forces acting on a system/object, and one should always show all the forces to accurately represent what forces are acting on the body and where they are acting.

Should we consider the bars that make up each truss as massless?

Yes - for modelling trusses we assume the weight of the members is negligible compared to external loads.

Because none of the problems refer mass you should consider each truss massless. You only need to consider the 2 forces given in each problem.

I was rewatching the lecture where H21.A 's set up is explained and the member AD was ignored for the sum of the forces and the free body diagram. Was that a mistake?

On H21.B, I have found it very difficult to know the orientation of DH relative to the horizontal. Can we assume that triangle CDH is a 3-4-5 triangle?

Thanks in advance for any help

Yes. DH is the hypotenuse of a triangle with height 3d and length 4d, so the angle with the horizontal would then be arctan(3/4)

Concerning homework 22.A and B, in the diagrams it says horizontal members of length 3d and then again horizontal members of length 4d

which one should we take as the vertical length?? 4d or 3d??

There was a correction that said all verticle lengths are 3d.

For 22.A is the reaction force at K and A and is it just in the y direction?

For the reaction at K, there is a fixed support. Additionally, two truss members act on K, combined in both the x and the y direction. Therefore, there must be a reaction force at K in both the x and y direction. Two member trusses also act in the x and y direction at A as well, however A is not a fixed support. Since A is a smooth horizontal roller support, the reaction force at A can only act in the vertical direction.

Would method of sections be helpful in any case, or only in cases where only a few members are asked for?

I think the method of sections comes in handy when you are able to split the truss into different section. If there's no easy way to split it up, then I would just look at each individual joint.

For HW 22.A should which one should we use for horizontal and vertical length? Both say horizontal

I believe that the horizontal members are of length 4d and the vertical members are of length 3d.

For Homework 23B, how can we determine which members of the B-E-D triangle are zero force members. I've assumed that HE is a zero force member, however, from point E, I cannot cancel out other members since they are in a triangle. Any ideas on how to work through this ?

I believe that from joint E, only HE is a zero-force member. The reason you cannot cancel out the other members is because the entire BED triangle carries a load. The way I solved question b) does not require solving the specific loads for the members in that triangle though.

I have not solved this yet but I also believe that the BED triangle does not have any zero-force members. However, I do not believe that it is necessary to solve the forces for these members to find the answers for the question. If I'm not mistaken, HE and HI are zero force members and that assumption allows you to solve for the required answers.

For Frames, do we always need to draw the overall FBD and or the FBD of a two force member? If not, would we have to state the certain assumptions that we are making?

You should only have to make FBDs for members with more than two parallel forces, but drawing more won't hurt. You never need to explicitly state these assumptions, however that won't hurt to put down either. How you choose to tackle this problem relies entirely on how you feel about the material.

For HW 24.B, is theta supposed to be given in the parameters? I can't think of another way to find Ax without it.

It is posted on the blog that theta = 30 degrees.

For HW 24, do we just solve the overall system as a static equilibrium problem? Do we even need to consider internal reactions?

Since the members in HW24 are not two-force members, you cannot simply solve for the reactions at A or B using global equilibrium. You will need to look at individual members as well to solve.

Hi guys, another question - this one on HW 25.A. Do both A and B, being pin joints, have reaction moments? If so, how do we solve for M?

Thanks in advance

I believe pin joints inherently allow rotation but not translation. That being said, since they are not actively resisting rotation, they do not supply a reaction moment.

Hey I solved for M being a negative value as opposed to that in the diagram but it asks for the answer as a vector and the diagram shows a negative K moment so I was wondering if these would cancel and the answer should be a positive value in the K direction.

Although the diagram shows that M is in the negative direction, I would just solve using M in the positive direction. When you do this you get a positive value in the K direction. If you solve it how you did you get a negative value, but if you look at the diagram, that means that the moment is counter clockwise. Thus, your vector will be in the positive K direction.

In 25.A do we need to account for the reaction force By when taking the sum of moments around point A or is By's affect on the moment around point A already included in moment M?

Moment M is a force couple pure moment, meaning it can be applied at any location on the beam and still retain its value. Therefore, I do not believe this couple includes the moment created by By

No one say anything about them missing HW 26

Because there is a reading day, they are dropping one topic and one homework. HW 26 is the one they are dropping

I know we have talked about creating shear force and moment diagrams at an internal point in the system, but what do these graphics look like for an overall system? On H27.A we are asked to find the overall shear force and bending moment diagram

In the lecture we found shear force and bending moment diagrams at internal points inside the system in terms of a distance "x" from the origin. We found equations instead of raw values and these equations can be applied for each section of the member. By graphing out these equations using the x axis and the x variable in each equation, we can create a diagram of shear force and bending moments along the member, with each section of the member having its own set of equations that should connect to make an overall diagram that is consistent.

Since H26 was dropped, are we okay to skip the lecture 26 altogether? Or does watching lecture 26 aid in the work we will need to do in H27, along with lecture 27?

I think it is recommended to watch all the lectures. The homeworks do not always cover just that single lecture, but rather multiple lectures. After watching both lectures, you may be able to solve the homework more effectively.

I would recommend to watch the lectures since all lectures do help in understanding the topics better.

Lecture 26 still provides important information and helped my understanding of Homework 27 so I recommend watching it. Also, if you are doing the honors extension for this class it provides information needed to complete your project.

In lecture, professor mentioned that V'(x) should equal p(x). However, according to both his lecture example and the answer I got for H28.A, V'(x) = -p(x). What is the reason for the flipped sign?

I believe the reason for the flipped sign is that the force is generally pointing in the -y direction.

In H29.A, upon first glance it looks like the distributed load equates to the downward force P and cancels it out. How should I approach this? When I work out the segments, should I account for portions of the distributed load or just use the equivalent load in the middle?

The way I went about this problem is I drew the overall FBD and found the moment about point B. I then broke it up into two segments from 0 <x<2 and 2<x<4. For the first segment, you should account for the portion of the distributed load, and for the second segment account for the equivalent load. By doing this you can determine the moment and shear equations which makes it easier to graph the shear and moment.

Why is the V(x) graph not linear with distributed loads?

You can find V(x) by solving V'(x)=-p(x) where p(x) is the magnitude of the distributed load. The negative sign comes from the fact that the load is usually pointing in the -y-direction.

This will make the V(x) graph linear if the distributed load is constant, it will just not be parallel to the x-axis.

On stress and strain problems, are there any scenarios where it would help to section off a certain segment to analyze the axial load, or does it not matter which side you analyze after making a section.

On H30.B, there isn't a diameter value given for the cable even though we need it's cross-sectional area to solve for stress. How should we go about finding it?

Just leave your answer in terms of d.

If you watch the online-Krousgrill lecture, He states the value of d is .5 inches. Therefore, I solved in terms of d first than at the end just inserted that value of d in. I did this because the find doesn't state to leave your answer in terms of d.

If it is not too late I also just left my answer in terms of d. I saw that the online lecture that the value for d was given, however I did not use this value as he also had different numbers for other values that we do have. I assume that it was just a mistake for them to either not give us the value of d or to not ask for the answer in terms of d.

Hi John, I was confused about that as well. Since the value of d was not shared with everyone I believe it is safe to assume leaving your answer in terms of d will not result in a deduction of points. Hope that helps ðŸ™‚

Yes I left my answer in terms of d so I believe that is fine.

Leaving it in terms of d is fine thats what i did

I think its alright to leave answers in terms of d since it wasn't specified

On homework 30.B I cut the pulley in half to draw my FBD because that is what we did earlier in the semester. Is that correct?

Hey Moyosoreolu,

I believe that is the correct method for this problem. Professor Krousgrill advised us to do a similar thing in problem 30A during his lecture.

Sorry I meant to say he advised us to do that for problem 30B

Yes it is

Will every cable have a circular cross section and every member a square one unless otherwise specified?

I would assume so. There's no way to tell the area otherwise

For 30B do we leave our answer in terms of d?

Hi Daniel. I left my answer to 30B in terms of d because we were not given a numerical value for diameter. Hope this comment helps!

For 31A do we have to look up a specific value for the aluminum alloy?

Yes, you will find the value you need for the aluminum alloy on page 395 of the lecture book.

Hi Ryan! Yes, I attended Tutorial Room Office Hours and the TA said we must look up the tensile yield strength for aluminum 6061-T6. Here's the source I referenced, although apparently this tensile yield strength value can also be found in our lecturebook. http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA6061T6 Hope my comment helps!

That's a great material data website that could be handy in the future. Thanks for sharing.

Is the Young's modulus value needed to do 31b?

I didn't use it for the specific way I approached the problem, but I know some people said they did. Basically, it is up to you to choose the way you want to do it, and that will dictate what values you need to solve.

For 32A, do we need to draw a FBD of the plate? I don't remember drawing FBDs in class for problems like this.

Yes Ryan it can't never hurt to draw a quick and easy FBD.

Thank Sid! That's what I ended up doing.

Your welcome Ryan Tischler. Godspeed!

Hello,

In 32b fo my shear calculation I multiplied the yield strength in shear by 2 since there are two shear planes. I think this is right but if anyone can confirm I would appreciate it.

This multiplication is technically correct. Since there are two areas of pure shear, the area should be multiplied by two, but, you can move this multiplication to the other side and instead multiply your yield strength by two. The most important thing is that this factor of two is only multiplied into either the yield strength or the area, and not both.

For 32.A does anyone know why the area we use isn't just b^2?

For 32.B do we even need to solve for the reactions at A and B or can we just use method of joints/sections?

I believe either will work to reach the correct answer. I computed the reactions at A and B for my work, however that is only because it looked like the easiest way to solve the problem at the time.

For problem 34.B why is rho 3D/2?

Hi Andrew! rho is 3d/2 only for segment 1. This is because rho is equal to the radius of the shaft and since the diameter is equal to 3d, the radius would be 3d/2

I know it's past due but does anyone know why I kept getting the wrong answer for 33B for the shear stress when I was using all of the length measurements in feet?

I believe you needed to convert everything to inches since ksi is measured in square inches. I could be wrong though its been awhile since I looked at it.

For 34.A I know in the picture it looks like rb is greater than rc but is it ok to assume that this will hold true when solving?

In general for the maximum shear stress of a system, does the sign matter? Is it right to believe that the answer should be merely the largest absolute value of calculated stresses?

Hi Jadah, to the best of my knowledge the sign is only important in determining whether the component is in tensile stress or compression. Secondly, it is correct to use the absolutel value of the shear stress to determine the maximum shear stress of the system.

Can anyone help me with figuring out the correct torque load for the components in 34A? I am quite confused about how to solve it.

Hi Sera, I am not sure if you are refering to 34B instead, it doesn't seem as though we need to calcualte torque load in 34A. I would suggest going to the main homework discussion page and reading thorugh the comments of the specific homework assaignment you have questions about. There are very good answer and tips there. If you do mean 34B then I would suggest cutting the system into sections, with each cut exposing an internal torque load of one of the components. For example if you make a cut between C and D you will expose the torque load of component three. Then you can do a summation fo moments to solve for the torque load of component 3 (the only other moment in that case would be the applied moment 2T at D). Then you would repeat this process with cuts in the other sections to solve for the remaining torque loads. I hope this helps.

Thanks for your help!