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## 69 thoughts on “Homework 25.B”

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In order to find the reactions from A and B, couldn't you just look at the overall FBD and use moments, Fx, and Fy to solve? Is the reason you can't do this because you don't know the distance (r) for the W weight?

The reason is not that you don't know r, actually when you solve it you will find r cancelled out. The reason is actually that there are four unknowns, which is Ax, Bx, Ay, By, and there are only three equations which is moment, Fx, Fy. You cannot solve for four unknowns using just three equations and this is where you should do FBDs for each individual part of the structure.

I am assuming that there should be a Cx and a Cy force couple at point C?

And if you were to break up the whole arrangement into bar AD and CB, which one would these force couples act on?

Hi Alexander, there are x and y reactions at C, but I don't think there is a moment because C is a pin joint. There is a lecture example similar to this one that you can reference.

Is the tension from the cable on BC equal to W/2?

If you draw out the FBD of the small pulley and the block, you have ΣF = T + T - W = 0. If you rearrange that you do come up with T = W/2, yes.

Will there be reaction forces at the pin joint at D?

I don't see any myself. There's just the tensions from the pulley.

I have the same question, even though it is an ideal pulley, wouldn't rope pull it down left due to the tension in the cables?

when you draw the FBD for the pulley d, the horizontal tension and the vertical tension would be equal from moment equation, and from the fx and fy equation the reaction force Dx and Dy would equal each the vertical and horizontal tension.

The only reaction forces on the pulley would be from the string. Which will W/2 in both directions.

That is correct, when looking at the diagram I determined that the only reactions would be the weight of the pulley equally distributed from the string, giving us W/2 in both ways.

I drew all of my FBDs but I feel like I have too many unknowns. Any advice on the best place to start?

Joint C seemed like a good bridging point between the different FBD's for me. Try using this as the substitution starter.

I drew two FBD's (one for beam AD one for beam BC). I then used the FBD for AD and took the moment about A to get Fbc. You can then plug Fbc's components into your net force in the y and x-direction to get your reaction at A.

Is there a reaction in the y-direction at E? If the tension is pulling only in the x?

I believe we can ignore E and just focus on the horizontal tension coming off of D for our equations.

^ That is what I did and it worked out well for me

The one problem I'm having is the radius. I feel like the process to solve this problem is similar to the last example in lecture 24. However, the radius is now unknown and doing the moment of the global equalibrium in respect to B, I get a W(d-r) within the equation. Any suggestions on what I should do?

Hi Alexander,

I would take a look at some of the other forces that are on the global FBD; namely, there should be a force from the cable near the top of the diagram. Accounting for this should allow you to cancel out any r's in the equation and leave you with only d's and W's.

Should we split up C into multiple forces of CB, CA, and CD, or do we just need CB?

Hi Maya,

Since the two links are both multi-force members, there will just be two reactions at C: Cx, and Cy. One link will feel these reactions in their positive directions, and the other will feel them equally, but in opposite directions. Hope this helps!

Can you still solve for moments in the overall FBD even though there is a pin than can move internally at C?

I believe that if you take the global FBD you should not account for the forces on C

I'm struggling to find one of the reaction forces without a moment equation. I split up the members into AD and BC and have moment equations about C but I can't seem to figure out how to get one of the variables without these.

Hi Aaron,

Try and find reaction forces Ay and By from the global FBD. This will make solving for those moment equations about C possible. Remember to account for the tension force in the global FBD, and the r's should cancel out.

I just wanted to make sure, when drawing the FBD for member BC we only take into account one tension that is the one that is (d) away from C right?

yes

Is it possible to solve for the reaction forces without taking moment equations because when I drew the three FBDs I saw that I can solve for the unknowns with only the Fx and Fy equations.

Double check your process. Moment equation is usually needed.

Are there reaction forces at E? I'm thinking that there's only a reaction force in the x direction because of the rope that's pulling tension, but I may be wrong...

I believe you are correct. There should only be a reaction in the X direction at point E.

I keep getting my reaction forces to cancel out and getting 0, does anybody have any tips?

Is the member CB two-force member ??

forces are applied at C, B, and the middle. So No.

Member CB is not a 2 force member, as there are reaction forces at C and B as well as the tension force at the midpoint. This problem does not have and 2 force members

For the whole system's FBD, how would I cancel r when summing the moments at A? I only see one moment with r, -(3d+r)W, but I'm not sure what the other moment with r in the opposite direction would be.

I created a separate FBD diagram for the pulley system so I could get the value for the tensions in terms of W. Then when you do the whole system's FBD your r's should cancel out that you get from the pulley at point D.

When solving for the reactions at A and B, do we need to also find the reactions at C in order to solve? I was thinking I need to do a moment about C as opposed to solving for the reactions there, but I am not completely sure.

Hi Tessa,

For this problem, you won't need to find the reactions at C. After finding some reaction forces from the global FBD, you can go into local FBDs, take the moment about C, and find the remaining reaction forces.

Can we treat the pulley as a point when calculating the distance between A and E?

no

Is there a moment about D due to the pulley? and moment about the smaller pulley?

If you isolate the pulley, yes, there will be moment about D due to the tensions.

I understand the structure on how to solve this problem by breaking up the frame into separate FBDs, but I don't understand how the radius' cancel out within these equations. For some reason, I am continuously left with an r when I solve. Can someone help me explain where my problem might be stemming?

The radius cancels out in the moment equation of A. I did not use the radius after that equation. Ya still need help?

How do you find the tension on ED?

Ideal pulley, tension is the same everywhere.

Do FBD for the weight W, get tension in terms of W.

How did one solve for Bx? By using moment about C? Considering the whole body as the system?

Yes that is exactly how you do it; taking whole body as the system and taking moments around C.

I am still a little confused, but this may be answered above. What is the best way to start this problem?

FBD1: whole system. Remove pin at A, add 2 forces; remove pin at B, add 2 forces; cut cable between BD, add tension;

FBD2: cut two vertical cables, add 2 tensions. Get tension in terms of W.

For FED1, do moment about A and B respectively to get Ay, By.

For FBD3: isolate BC, do moment about C get Bx.

For FBD1: Do force balance in x get Ax.

Done.

I know there are Cx and Cy but is there also a moment about C?

Hi! Since C is a pin joint, there will not be a reaction moment at C. However, you could choose to calculate the moment about whatever point you would like.

For the moments included in the global fbd moment equations for A and B, I am kind of confused as to what tensions are included in those equations besides the reactions. Is it 2 tensions on the big pulley with one using the full distance from where ever the moment point is and one using just the radius of the pulley?

I think I figured out the tensions but would you need to include Ex in those moment equations as well?

I wouldn't think so, as the only force acting at E is the tension force, which we can consider at point D for the block.

How many tensions would there be? Would there be one each at Ex, 2r from D, and the middle of BC?

If we wanna do the moment equation for the entire frame, what all forces to we need to consider in the equation? Ax, Ay, Bx, By and Ey. Do we consider the weight W or either of the tensions in the moment equation?

It depends on what's in your system. If the block is in your system, of course there'll be weight to consider. If the block is not in your system, use tension instead of weight for certain points.

If you're taking the moment for the entire frame you wouldn't have an Ey, and yes you should consider the weight of the block and the tension in the pulley at D

The pulley ratio would be 1:2 correct?

Of the radii, yes. However, this should ideally not be relevant in solving the problem. If you set up moment equilibrium equations for the whole system, you'll see what I mean.

I thought bar BC was a zero force truss for a little while. What's the difference with this case? Is it because there are non-parallel forces acting on AD?

Machines like the one in this problem are different from trusses. In this case, while BC is in static equilibrium like all other elements of the machine, it is not a zero force member. The simplest way to justify this is because there are outside reaction forces acting at point B from the mounting hardware. It's also important to remember that AC and CD are not separate members like in a truss, where one could bend the joint at C.

Is the tension at E 1/2 W? Since it is connected to an ideal pulley with only one rope?

Not just the tension at E, but the tension throughout! An ideal pulley is frictionless and simply redirects tension- so whatever tension goes from the end of the cable along BC will be the same as the tension as is felt at point E. That tension is derived from the weight W, which is divided between two cable sections collinear with gravity.

I feel like we don't need the FBD for AD since we could get y reactions for B for global FBD and Bx through BC FBD. And then we can return to the global FBD and get A_x. So it's completely ignoring reactions at C. Am I missing something?