Correction: **"All VERTICAL members of length 3d"**

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Correction: **"All VERTICAL members of length 3d"**

We are encouraging you to communicate with us and with your colleagues in the class through the threaded discussions on the course blog. If you have questions on this homework, please ask here.

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Since the block is held by a pulley, is it exerting a downward force of 2W or 0.5W?

Which 'free body' are you talking about? (or exerting on which object?)

So the diagram tells you that the block has a weight of W, so that will remain constant but I think what you want to find is the downward tension acting on K. To find this I made a FBD of only the block and the pulley it was attached to. The pulley had a 2 tension forces in the upward direction. I found the sum of forces in the y direction which gave me the force of the tension, which was 0.5 W. This is the same force that acts on K.

You would set up the equation SUM Fx = 2T - W = 0

Then, solving for T gives you T = 0.5W

Can we use the pulley system to solve for the reaction forces at K and M to solve for the forces QK and MR?

Hi I think you could but what I would do is section off the truss so it cuts at 1, EH,2,3 and cable K to ground. . So you have four unknowns. But you can solve for force EH by using method of joints at the joint E. So you have three unknowns for the section. Then you can use the Fx, Fy, and a moment equation(i did abut H) to solve the forces 1,2,3. With regards to the pulley, you can solve for the tension from K to the ground using the given W. And W = 2P. This is what professor Krousgrill suggested in his lecture

I'm struggling to visualize where you're cutting it. How is the cable K included?

If we take the moment at points A or B, do we have to take into consideration the dimension (r/2) due to the offset of the weighted block because of the radius of the pulley?

Yes, that is what I did.

So are us supposed to use the same distance as before 4d for horizontal and 3d for vertical, because they both say horizontal again.

Sorry I missed the correction.

Can we assume the pulleys are massless/frictionless? I'm assuming since no coefficient of friction is mentioned we are meant to neglect those.

Yes I assumed that they are massless/frictionless.

I'm confused on how to go about solving the pulley system to get values of P so I can continue solving the problem.

You can separate the right most pulley and weight as its own independent system to find the relationships between T, P, and W. From there, the tensions act on the structure and you can consider them in your X, Y, and Moment equations.

What is T? There is no T in this problem. Are you referring to tension in the cable at K?

Never mind, I was able to figure it out.

How do we go about solving the pulley system in this problem without using tension?

I used the M pulley to solve for that downward force (the tension) in terms of P, and then by understanding the cable set up was able to find the downward force at K in terms of P as well.

What is r in terms of d? I am taking the moment about a point to solve for one of the three forces, so my answer becomes much cleaner if I am able to cancel all of the dimensional constants from the answer.

I do not believe you actually need to do this calculation. If you continue solving for the moment, you will see what I mean. Just keep the 'r' by itself and you shouldn't have to do any tricky manipulations in order to get the value only in terms of P and not with 'd' or 'r'.

Your r terms should cancel out when your are calculating your moment since one is +r from a point and the other is -r from a point.

I believe W is at a distance + r/2 from the point while T is - r from the point, but they equal and cancel in the moment equation because they are multiplied by W = 2P and T = P respectively .

Hope this clarifies the canceling out of r

I believe this is correct, if you treat tension as an internal force then the external forces with respect to the truss will be the reaction force caused by tension at the unlettered pin (-r from K) and W (+r/2 from M).

I found out that r is its own term, but r will cancel out during calculations. Just stick with the math and the terms Pr and -Pr will cancel in the moments equation.

I know it's mentioned above, but what is the best way to split the truss? Is it down the middle of QR, down the middle of NO, or something different? Thanks!

I cut diagonally through IE, EH, DH, and CH. This will expose EH, which is a member you don't want but by drawing a FDB at point E, you can see this member is 0. Thus, you are left exposing IE, DH, CH: 1,2,3 respectively.

I also cut diagonally like Radhika but I also cut through the cable that connected to the ground, opposite of the pulley. This makes it easier to solve for the forces because you don't need the reaction force and can simply put it in terms of W.

Also, how would you use the pulleys in a moment equation about A? This part is confusing me a bit.

You take the cable and pulley as the system then the force becomes internal.

Cut EI,EH,HC,HD and all the cables, consider the upper right side. Do free body diagram.

So I was able to solve for member 3 (Fch) but am confused how to get the other two members. I think that to get the other two members, I have to solve for Fdb. Is this necessary or is there another way I am not seeing? Thanks!

Cut EI,EH,HC,HD and all the cables, consider the upper right side. Do free body diagram.

What are the directions of the 3 tension forces at K and M after sectioning through the cables?

Tension, as you said, always point away from the cut. The arrow is pulling the cutting point.

I know that the tension on cable around pulley at K is a half of W, but how can I get the force act on K?

You take the cable and pulley as the system then the force becomes internal. So no need to consider it.

Can we cut diagonally and then consider the left half?

Yes, but the right half would be easier.

Is it possible to consider either side from the diagonal cut?

yes. Usually the side without external pin or fixed constraint is better.

For solving the pulley system, I solved the tension in each section of cable to be .5W. I know this acts partially on M because the cable connects to that point on the truss. But does the cable also exert a force on K or is the rest of the tension all supported by the unlettered pin?

I would recommend that you choose the cable, the pulley and the trusses as you system. then the only external force/constraint except gravity comes from the unlettered pin that fixes the cable.

Im a bit confused with this. Do i assume that the pulleys are massless and frictionless or not?

If the problem does not state friction constant (like this problem), assume it's frictionless.

Is force 2 supposed to be Fdh or Feh?

I believe it's supposed to be Fdh

Is loading on joint "H" supposed to be loading on joint "N"?

Yes you are right! It's on N as shown.

To find the unknowns, will it work to just use Moment at H, Sum of the forces in x, and sum of the forces of y?

It depend on how you draw your system. For the system I drew (excluding the points A, B, C, and D) then you will need a fourth equation because there would be four unknowns. But knowing that the only internal force in the x direction at E is HE then to keep E at equilibrium in the x direction all of 2P must be accounted for by the truss of HE. Hope that helps

What is the net cable force on the center of a pulley when the cable runs over 1/4 of the pulley?

I don't understand your question. But why do you need to consider that? You can make both the pulley and the cable as the system such that their interaction would become internal forces. Then you only cut the cables to expose the tension in the cable and no need to study 'net cable force on the center of a pulley '.

The friction of the pulley does not need to be accounted for in this problem. Draw an FBD at the pulley holding up the block to find the tensions.

How do you determine where the cut should be made on the diagram? Can it be where ever we think it would make it easiest to solve?

Anywhere you like (So yes, choose the cut that makes it easiest). Also you want to cut across the cables in which the forces are asked.

Is the tension in each cable over the pulley between M and W equal to T or T/2?

What do you mean by T? You can isolate the system consisting of the weight W and the lower pulley. Then the FBD tells you how to relate W to tension in the cable.

Does the point the pulley is grounded to play a role in solving the problem?

Isolate the system that you wish to study (cut all the cables and replace them by a tension force; cut all trusses and replace them by a assumed tension; erase all the constraint and replace them by proper reactions)

Then you will see whether that point plays a role.

Is Fei equal to Fed? If so, can I cut from the middle of line CD and line HC as my section?

You can do analysis of joint E and see.

Can I assume that half of the weight W is carried by the pin joint below K and half is carried by point M?

That's very hard to answer. But what i can tell you is that the weight W is hold by 2 cables around the small pulley. Force in the cable is the same everywhere.

Where do we split up the image here? This part has always confused me.

I asked a similar question earlier and here is the answer I received: "Anywhere you like (So yes, choose the cut that makes it easiest). Also you want to cut across the cables in which the forces are asked."

Honestly you can cut anywhere that is through the forces you are trying to determine.

As stated, you can choose to cut the system wherever you want. Personally, I chose to cut the system right across each of the components we needed to find. Once this cut was made, I chose to do my calculations using the side with the pulley, as I didn't have to find the reaction forces at A and B by choosing this side.

Principle 1, cut through the section where most of the truss members are asked;

Principle 2, if one cut gives you more than 3 forces, do a joint analysis to simplify a bit.

Principle 3, if one cut gives too many forces, do multiple cuts.

Principle 4, once cut, use the side with no constraint forces and do moment/force balance eqs. to solve for forces.

Whenever you cut something, replace it by a force;

Whenever you replace a constraint, replace it by proper reaction forces.

Apply:

Cut EI, EH, HC, HD; get 4 forces; use joint E; do force equilibrium eqs.

From what I understand, I

*From what I understand, as a general rule of thumb it is easiest to start on a truss problem from the side that is not directly fastened to the wall. In this case the bottom left is held in place through a pin joint and roller which leads to reaction forces. Due to this, it is better to focus on the right section of the diagram.

For the moment equation about point H after the cut, would both the tension and weight of the block need to be included? Or would it simply be the tension in terms of 1/2 W? Also when using the tension would the radius of K only effect the x-component of HK or would it effect the y-component too?

If you include W in your free body, include W and tension near K;

If you do not include W and cut the cables around the small pulley, do not include W but include 2 tensions. Also, remember the tension near K.

I am struggling to understand how the moment operates differently within trusses when you cut where you want to focus on within the FBD. Are there some forces we can ignore because they are not within this focus area?

Are there some forces we can ignore because they are not within this focus area?

Yes, try to eliminate those forces/trusses that you are not interested in. This can be done by

1) choose the side without constraint

2) choose proper cut

3) use weight rather than cut the cables around pulley or consider pulley-cable-nodes interactions

Is it recommended to "slice" the FBD diagonally slicing through 4 members?

That's what I did, I cut through 1, EH, 2, and 3.

See my comments to Kiernan Luke Schuerman above

Is the reaction Ax equal to 4P or 6P? or none?

I am also confused about the direction of Ax should it be to the right or left?

There must be Ax. and Ax must be -6P (to left) to balance the applied forces to the right. Think about the FBD for the entire thing.

When I cut through EI, EH, DH, and CH, would the right side with the pulley include the unlettered pin on the ground that holds the pulley system in place? Or would I be cutting the cable between K and the unlettered pin as well?

I cut through the cable between K and the pin.

Since it's a cable, wherever you cut (unlettered pin or in the middle between unlettered pin and K), FBD would not change.

Can someone clarify what direction the tension force (just to the left of K) would point?

From my understanding of the pulley system, tension would be directed from the pin to W, but the reaction force on the pin is downwards. Thanks.

Does anybody know what's the relationship between the radius of the pulley "r" and bar? I mean what's the relation between r and d?

Feraas,

It actually does not matter what the radius of the pulley is, as they will cancel out in the moment equation. When considering the distance of the two cables from H, the closer one is acting at (4d-r) and the further on is acting at (8d+r). Because both moments are acting in the same direction and with the same force, the two uses of "r" cancel out.

In terms of the forces and reaction forces acting on the whole FBD of the truss, is there only 1 force acting on the truss due to the weight and pulley system which would be at M?

I think for the whole FBD of the truss, besides from the weight, there will be four applied forces on the left side at the four joints, a reaction in the X and Y at A and a normal force at B.

So to set up a FBD at point E, we would show F_EI, F_EH, and F_DE, correct? If we just show F_EI and F_EH, then both are zero force members.

Hi, yes to find F(eh) and to set up a FBD at joint E, those three forces would need to be included. Then a sum of forces equilibrium equation can just be applied to solve for one of the loads.

Am I correct in thinking the reaction forces do not need to be calculated in order to solve?

Yes, the reaction forces shouldn't have to be calculated

Am I correct in assuming that we will not need to use radius r if we do not calculate the reaction forces at A or B? This way we simply take the forces (tensions) of each cable of the pulley and solve from that way?

Hi Dawes. No, you cannot assume that way. Radius r of pulleys shows up in the moment equation of the body while solving for the load carried by members 1,2 and 3.

Would tension force P at K and 2 tension forces (both P) at M due to the pulley all be in downward direction?

The tensions of the cables from the pulley M are in the upward direction.

Thank you. Would it be upwards at K also?

Can we consider the pulley as a different system that has an internal Fx and Fy sum of 0 and calculate the values of Ax, Ay, and By just based on the forces applied on the left portion of the truss?

Hi Arthur, If you remove the pulleys from your truss you would need to consider the reaction of the pulleys on the truss. I suggest considering the pulleys as part of the truss. On the other hand, the calculations of the reactions at A and B would only be needed if you choose to work with the structure's bottom section.

Thank you for the response Diego, I just have one more question. I saw on the discussion that EH is a zero-force member, but I do not understand why.

Nvm, I misread the discussion