49 thoughts on “Homework 24.A”

    1. Yeah, there is a pin at C connecting the two bodies together. Because of that there exists 'two' reaction forces at C Cx and Cy. It should be noted that Cx and Cy of the weird half circle thing will be in the opposite direction of the Cx and Cy forces acting on the backwards J bar.

    1. I think you can assume that in this case. Since AC is a two force member, you can combine the reaction forces at A and C into two collinear forces in opposite directions. Then the y component reaction force at A would just be the force along the line of action from A to C, and there would be no x reaction force.

        1. Caitlyn,
          There is certainly a reaction force Ax, however if you choose to treat member AC as a two-force member (which is what the explanation is regarding in this thread), then you wouldn't have to consider the reaction forces at A when drawing the individual free-body diagrams for AC and BD. Note that you should definitely have an Ax reaction force in your overall FBD.

          I hope this helps!

  1. I am confused on how to set up the forces on AC (the two-force member). If you don't need x and y components for points A and C, how would you simplify that?

    1. We do need x and y components for points A and C. In C-shaped part, if you set up a moment equation at points A and C, you will get Ax and Cx equal to 0 because AC is a two forces member. The next thing we need to do is to set up a moment equation and sum of force for this system. Since you already know the value of Ax, you are able to get the values of Ay, Bx, and By.

    1. Hi Dan,
      I would break the system into 2 parts by detaching the C shaped structure from the L shaped structure. You can then identify the forces that act on each of the separate structures.

    1. A two force member is a component that only has two points of load. In the diagram, we can see that AC only has two points of load, since it only has two pin joint connections

    1. In most problems, I have found that using moments is very helpful because it allows for you to disregard forces in the FBD at the point you are choosing to sum your moments around. For example, when I solved 24.A, I used sum of moments at B for the member DCB. This allows us to find Cy in terms of P right away since sum of moments isolates Cy and P into the same equation. I hope this helps!

    1. I took this to be because joint C is a pin. When you push on point D, imagine how things would start to move or flex. As such, think about what impending motion the system would have if the pin joint at A were to be released. A would feel no reaction in the X because this impending motion is vertical. Think about C as being tangent to the constraint circle through which the assembly attached at B can move.

    1. I agree, this can also be proven by taking the moment about point A, which would result in Cx = 0. This would also imply that Ax = 0, which creates the two force member.

        1. Hi Evan, yes AC is a two member force because there are only two force acting on that member, at point C and at point A. However, the "L" shape member has three forces acting on it, P at D, C, and at point B. Because there are more than two forces acting it is not a two force member. I hope this helps.

  2. Does the shape of the bar AC alter the problem in anyway? I understand we have to utilize radius instead of distance, but is there any other factors we need to consider when solving this problem?

    1. Hi Victoria! The shape of the bar does not affect the problem in any way. However, the fact that AC is a 2-force member is the most important part of the member

    2. Because the members are massless in our problems it doesn't matter what the shape of the member is, only the forces on it. I want to draw one as a giant mass of spaghetti.

    1. Yes, AC is a two force member. Therefore you can treat it as a truss component with Fac acting up and down at either ends of the arc. Using this, moment, and the sum of the forces, you should be able to find the reaction force you are looking for. Hope this helps!

    1. Hi Shreyas. AC is a two-force member because it would not be in equilibrium otherwise (you can try to do the sum of forces and sum of moments to convince yourself). And these forces shall have the same magnitude and same action line, but opposite direction. In general, a rigid body with applied forces only in its extremes is considered a two-force member.

    2. While AC has reaction forces Ax, Ay, Cx, and Cy, the reaction forces can also be written as two vectors each with two components. These vectors can be written as

      F(at A) = Ax i + Ay j
      and
      F(at C) = Cx i + Cy j

      The important thing to see is that the reaction forces can be written as two individual forces and are the only forces on member AC thus making it a two-force member.

    1. When drawing the reactions, they can be any way you like as long as their counterpart on the other FBD is equal in magnitude and opposite in direction. For example, if you have two members joined by a pin A, the reactions on one FBD will be in the positive x and y-directions and the reactions on the other FBD will be in the negative x and y-directions.

  3. I am confused on how to find Ay and By if there is no outside force acting on the structure in the y direction. Do I need to take the moment of part of the structure to solve for them?

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