82 thoughts on “HOMEWORK 25 - Fall 24”

    7. Yes, as the forces are located in the x and y axes, the moment would be in the z direction as it is perpendicular to both axes. Since only the z component contributes to the magnitude of the moment, your final vector answer would only contain the magnitude as the z component with the same sign resulting from the moment about point A equation.

  2. In the image, M is clockwise meaning that it's -z using RHR, so when we write the answer, (if the answer is clockwise), should we write it as positive or negative?

    1. since M is clockwise, my moment was negative, since it would be opposing the normal counter-clockwise rotation of M so instead you would -M in moment of b equation.

    2. My moment was negative because it opposes the counterclockwise direction of positive moments. Therefore, you should subtract M in the moment equation for point B.

    3. The sign of your answer would be whichever direction the z component of the moment points towards. If you found a clockwise moment, the z component would be negative according to the right hand rule, and you would get a vector that looks something like . Hope this helps!

    7. When you initially write your moment equation for joint B, it doesn't matter. Assume a direction (-z or +z) and then calculate M. If the value is negative, your assumption is wrong, and it is the opposite direction. If the value is positive, your assumption is right.

    8. When you write the answer as a vector it should be negative. Can always check that with the right hand rule. I found this to be tricky too, but since we assume M is negative you actually get a positive answer, that just means we're right in our assumption.

    1. The right angle at C just proves that the member BC is vertical, as we know that DC is horizontal. This allows us to get the vertical length of the bar BC with the given geometries, which helps us solve for the moment.

    1. I dont think that is necessary for this problem. In order to visualize the distances useful in this problem it helps to break up the overall system into smaller systems(do this by drawing FBD of the different bars). From there, think about what forces you would need in order to set up a moment equation to find the moment M and you might be able to narrow down what force balances or a moment balances are necessary to compute.

    2. When calculating the moment about A, it is best to only look at the FBD containing member AED. When you calculate the moment in this FBD, you do not need to include the reactions at B since they are not part of this FBD diagram. In this moment equation, you should only have one unknown which will then help you solve for other components in another FBD.

    3. For this problem, the length between in A and B is not needed. This is because it is best to take the moment about A when the free body diagram is just member AD, and not the entire machine. This eliminates the forces around point B.

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    4. You shouldn't need to calculate that length. If you break each segment of the free body diagram into separate joints, you should be able to solve for everything asked for and your moment about a shouldn't require any information from joint B.

    5. While there is not information given that tells you what the distance/length between A and B is, you do not specifically need this length to find M. You can use the FBD of specific segments to find the value of M.

    6. I do not think there is a need to calculate the length between A and B. However, we do need to calculate the perpendicular length between A and E, or the projection of AE on AB for determining the moment due to force 2P about point A. For this, we use the length d/2 and the given triangle with sides 3, 4, 5.

  5. Since there are many ways of solving this problem, which is the easiest way to go about it? And is there a strategy we could use to figure out which way to go about similar problems to save time?

    1. The one I found was the easiest was solving for the minimum moment required to keep the frame in equilibrium was to first start with the first free body diagram A E D, then moved to the next once I found force D then to the second free body and then once I got to the third there was only one unknown which was the moment at B. When approaching these problems, I typically go straight to the free body diagram that has the term i'm solving for and if I cant solve then I go to another one and find that extra unknown.

    2. The easiest way I found to go about this problem was to separate each member and create local equilibriums for all of them. However, you should notice that DC is a two-force member. After separating the members you can find the moment about A giving you Fdc. After you have Fdc all you need to do is just take the moment about B and you can solve for M now that you know Fdc. This is only one way to solve the problem but I found it to be easiest.

    3. Identifying the two-force member was the first step I took when solving the problem and it helped me greatly as I knew that the two forces had to have equal magnitudes. By identifying this in the beginning, it also helps solve for the values in the links attached to it because they all have to be in equilibrium. Remember, two-force members do not have individual x- and y- forces like multi-force members and reaction forces have, but instead have colinear forces. This will be helpful for setting your system in equilibrium. Hope this helps!

  6. Im having a hard time understanding how to draw the direction for the x and y components of each force from diagram to diagram. Anyone have any helpful pointers there?

    2. The way I approach it is drawing them in the positive direction for the first instance you see them and then in the negative direction for the second instance since they will be equal and opposite.

    4. From my point of view, I think that it is always good to start with the initial full body diagram. On there, I draw my reaction forces in the positive x and positive y directions, along with all the other forces that are already given in the free body diagram. Then, for this one, I broke it up into three different free body diagrams, one for each member of the frame. You can also notice that the horizontal member is a two force member, so I drew the same force in opposing directions through the axis of the member. Then, on member AED I drew the reaction forces, and the two member force opposing the direction of member DC. I did the same thing on member BC, making sure to also include the reaction forces and the moment.

    2. You should not have to solve for Dx or Dy to solve this problem. In fact, if you look at the far-right member and apply Newton's second law in the y-direction, you may find that you don't need to solve for any component other than By...

    2. No, it isn't a zero force member, but instead, you can treat it as a two force member because there are only two forces acting on it, in the x direction at D and C, which cancel out and you can separate the system into the components and slowly solve for each variable to find the overall value of M.

    1. I believe what is being asked for is that when you get the final value for M, they want it in standard vector notation, meaning i hat for x, j hat for y, and k hat for z. Hope this helps!

    2. Since the forces are in the x and y axis, the moment as a result of the forces should act in the z axis. So your x and y coordinates would be zero, while the magnitude of your z coordinate would be the magnitude of the moment and the sign would be positive or negative depending on clockwise or anticlockwise motion.

    1. Solving for the Moment at point A, will solve for the X component of D, which than you can use to solve for the X component of C, which than can be used to solve for the Moment about point B, Hope this helps.

  14. When I solved for M, I got a positive number, but I have reason to believe that it should be a negative number, I double checked and all my signs are correct, so I'm not sure where I went wrong.

    1. I think in this case, you may want to check your internal forces at joint D and C, because they are internal forces when you draw FBD respect to different members it will change direction. By doing this I think you will get the correct answer.

    2. You likely did not make any error. You probably assumed that M was negative to start in your calculations because that is how it was drawn. Then, when you get a positive number, it confirms that you were correct and M is indeed negative. It can be confusing, but essentially a positive number means that your initial assumption was correct.

    3. It may be the case that you found the magnitude for the moment, but the problem asks for the moment as a vector, so I would refer to the direction of the moment from the diagram.

    4. In your original equation, if you took M as clockwise (with a negative sign), then getting a positive answer just shows that your assumption was correct, and the sign for the final answer is indeed negative.

    5. Something I would recommend next time for Frames and Machines problems is to draw the complete problem in its different FBDs with their respective forces and opposite forces and this way you can keep a more organized problem and not mix up your signs. Also remember to have your force in the direction that you drew it to prevent messing up.

  15. I had the same issue. It makes more sense for the moment to be negative (meaning that it is opposite of the way it is draw in the diagram). I went back and reversed the direction of some of the reaction arrows I drew initially. When I went back through to correct the signs, I found the final moment to be negative. You can try this and see if it helps.

  16. When drawing the individual free body diagrams of each of the links is there some convention as to the order to go to make sure your forces all face the correct direction? For example, to get the correct direction for the moment the forces must be going into DC in the FBD so that they face the correct directions, is this because these are internal forces through that member?

    1. For this, you can actually pick any direction you want for an FBD, as long as you make the forces at that same couple in another FBD equal and OPPOSITE to the direction you chose at the first FBD

  19. Conceptually, does anyone know why the moment (M) on member CB doesnt translate to the other members when you make isolated FBD's? I would think an external moment being applied would have an effect on the whole system, no matter where its being applied.

    1. The isolated FBDs create isolated systems, so within the system for segment AD you will not consider any forces from the segment CB because you have broken them into two different systems. So for segment AD, you don't consider M or the forces at B, and for CB, you don't consider the forces at A.

    1. Unfortunately I don't think you can consider BC as a two force member since there is an external moment acting on it. Because of this DC is the only two force member.

    2. No, DC is not a two-force member because it has an additional connection at point E, where external forces P and 2P are applied. This results in more than two forces acting on DC, so it does not satisfy the conditions of a two-force member.

    1. Yes, it is correct to assume that there are no vertical forces acting on DC. Since no external loads are applied to the member and there are only internal forces at D and C, DC is a two-force member. Since DC is a two-force member, there are only two horizontal forces (one at D and another at C) and no vertical forces. The forces are horizontal because forces on a two-force member act along the member.

  23. If I am working the equilibrium for AED and considering the forces at D. Can I assume Dy equals 0 because Ay opposes the 2P in the y direction completely and if so why?

    1. Yes, you can assume Dy=0 because DC is a two-force member and can only exert a force along its horizontal line . So, Ay fully balances the 2P vertical force at E.

  25. Within this homework assignment, we treat member DC as a two-force member and make it have two unknowns. For any two-force member when turning the forces into two unknowns with equal magnitudes but opposite direction, do we always write it as being along the axis of the member?

    1. Yes BC is a two force member. Member BC is a two-force member because it is only connected at points B and C with no external loads or moments along its length, meaning the forces at B and C must be equal, opposite, and collinear.

  27. You can find the forces asked for without using the global FBD, and simply using the separate FBDs of each component. I found it easier to do the homework like this.

  28. How do we express the final vector for M? Is it solely in the z direction because it's a couple and doesn't have components in the x and y directions?

    1. Yes, my moment is expressed in vector form with the only component being in the z direction. You can also think about the definition of moment being the cross product of r x F

  29. On problem B, they explain in the video that the force F_AC will act on the main bar with point B with a downwards force, but I'm not sure why this is. To my understanding F_AC should already be zeroed out elsewhere, so why would it apply twice?

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