# Homework Help: Given Moment about a Pin and a Roller

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1. Mar 2, 2015

### hdp12

1. The problem statement, all variables and given/known data
For my statics homework, we are directed to:
Draw the free body diagram and use the force and moment equilibrium to determine the support reactions for the following systems

I'm having a bunch of trouble with problem 4.10, which is displayed in the following picture

I'm really stuck so if someone could kind of guide me in the correct direction, I'd really appreciate it.

2. Relevant equations
ΣFx=0
ΣFy=0
ΣM=0

3. The attempt at a solution

2. Mar 2, 2015

### haruspex

The diagram alone is not clear to me. Any words to go with it?

3. Mar 2, 2015

### hdp12

no there were not, simply the moment is acting on that point at the rigid body between the roller, A, and the pin, B. It's then necessary to find the support reactions at A and B, which are RAY, RBY, and RBX

4. Mar 2, 2015

### SteamKing

Staff Emeritus
The beam appears to be pinned and simply supported at the ends. Pin connections typically cannot support a moment (i.e., the ends of the beam are free to rotate), therefore MA = MB = 0.

You have written several different moment equations for this beam. There is only one correct equation.

5. Mar 2, 2015

### hdp12

okay, so if the moments at the ends are 0, how does that help me get to the support reactions at each point?

6. Mar 2, 2015

### SteamKing

Staff Emeritus
Like I said at the end of my post, you can have only one correct moment equation, but you have written at least two. Knowing what I said, is one of your moment equations invalid?

Write the correct moment equation, and you can find the reactions at the supports.

7. Mar 2, 2015

### hdp12

okay... I think the correct momentum equation is going to have to do with the sum of moments... so
ΣM=0 : 0 = |MA+MB| - |M2|
is this the correct one?

8. Mar 2, 2015

### SteamKing

Staff Emeritus
I already told you what MA and MB were.

How is this moment equation going to give you the reactions?

Try again. (You had written the correct equation down already. I thought you would recognize it once I told you about pinned connections.)

9. Mar 2, 2015

### hdp12

okay, my bad. I'm not very good with moment problems yet, but I am trying. I see that if what I wrote last post was true, then M2 would be equal to zero, which is not the case.

the equation you must be referring to is 0=M2-(a+b)RBY, but if that's that case, aren't we supposed to get the thing in terms of both support reactions? so.. 0=M2-(a⋅RAY+b⋅RBY)

10. Mar 2, 2015

### SteamKing

Staff Emeritus
If you select one of the supports as the reference point for calculating moments, the resulting moment equation will contain only the other support reaction. Once you obtain this support reaction, then the sum of the forces equation is used to find the other reaction.

11. Mar 2, 2015

### hdp12

okay, I think I'm starting to see it now...
@A, ΣM=0: M2-(a+b)RBY -> RBY=M2 / (a+b)
and also
@B, ΣM=0: M2-(a+b)RAY -> RAY=M2 / (a+b)
yeah?

12. Mar 2, 2015

### SteamKing

Staff Emeritus
You can only write one moment equation. You can use this equation to solve for one reaction. As I explained, you use the sum of the forces to find the remaining reaction.

13. Mar 2, 2015

### hdp12

I really appreciate your time and patience, thank you so much.

now.. I have a solution for RBX and RBY, but something tells me that RAY is opposite to RBY... whether or not I am correct in that feeling how do I determine their directions mathematically?

14. Mar 2, 2015

### SteamKing

Staff Emeritus
This is a consequence of the sum of the forces being equal to zero. Since the only forces acting on this beam are the reactions, they must sum to zero.

15. Mar 2, 2015

### hdp12

right.. but is the support reaction at A the same direction as the support reaction at B? How to write out this equation properly

16. Mar 2, 2015

### SteamKing

Staff Emeritus
Ask yourself, if the two reactions are in the same direction, can they ever sum to zero?

17. Mar 2, 2015

### hdp12

no, so they must be in opposite directions...
must you determine the force from the moment? I'm trying to figure out how the distances come into play, because my next problem has two moments at different distances along the rigid body.

18. Mar 2, 2015

### SteamKing

Staff Emeritus
A moment is a force multiplied by a distance. If you divide a moment by a distance, you get a force. You can check the units.

As for your second problem, just write the equations of static equilibrium and solve for the unknown reactions. That's all there is to it.

19. Mar 2, 2015

### hdp12

right.. so is this or is this not correct,
ΣFy = 0: 0 = RBY + RAY -(Force determined from Moment)
Force determined from moment = M2/b
RAY=M2/b - RBY

20. Mar 2, 2015

### SteamKing

Staff Emeritus
It's not clear what these equations represent. Are they supposed to show the solution to the reactions for the OP?

21. Mar 2, 2015

### hdp12

I'm not familiar with the abbreviation OP, but the equations are supposed to determine the force from the moment and it's distance from point B, then conclude that the support reaction at A is the support reaction at B minus the force caused by the moment.

22. Mar 2, 2015

### SteamKing

Staff Emeritus
OP means "Original Post" or "Original Poster"; i.e., the first post in the thread or the person who created the thread.

You just solved a problem where you calculated the reactions for such a situation. Why are you chasing your tail with this stuff?

23. Mar 2, 2015

### hdp12

yes, it's supposed to be a solution to the original post. I didn't think I was chasing my tail, and sorry if that's actually the case.. I just don't understand why the position of the moment on the body doesn't matter. There is significance in it's position, or the problem wouldn't have the first distance a and the second distance b, it would just make the whole length a single variable.

In order to solve for the support reaction at A, do you or do you not need to determine the force from the moment by dividing length b?

24. Mar 2, 2015

### SteamKing

Staff Emeritus
What happened to this solution to RBY?
@A, ΣM=0: M2-(a+b)RBY -> RBY=M2 / (a+b)

I thought you had solved the problem posted in the OP. Perhaps you should show your calculations for RAY and RBY here.

I think you confused my general observation about a moment divided by a distance = a force with being the solution to the reactions of the beam. That is not the case.

25. Mar 2, 2015

### hdp12

yes. RBY=M2/(a+b)

for the force equilibrium equations...
ΣFy=0: 0 =RAY + RBY - (M2/b)
RAY = (M2/b) - RBY

do you understand what I think I'm supposed to do?