69 thoughts on “Homework 25.B”

  1. 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?

    1. 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.

    1. 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.

        1. 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.

      1. 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.

    1. 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.

  2. 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?

    1. 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.

    1. 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!

  3. 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.

    1. 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.

  4. 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.

  5. 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...

    1. 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

  6. 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.

    1. 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.

  7. 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.

    1. 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.

  8. 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?

    1. 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.

    1. 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.

  9. 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?

  10. 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?

    1. 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.

    1. 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.

  11. 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?

    1. 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.

    1. 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.

  12. 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?

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