Given Moment about a Pin and a Roller

[hdp12]

Homework Statement

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.

Homework Equations

ΣFx=0
ΣFy=0
ΣM=0

The Attempt at a Solution

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[haruspex]

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

 

[hdp12]

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

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 R_AY, R_BY, and R_BX

 

[SteamKing]

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 M_A = M_B = 0.

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

 

[hdp12]

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

 

[SteamKing]

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

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.

(Hint: There's nothing especially tricky about this problem, if you write the correct equilibrium equations.)

 

[hdp12]

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

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

 

[SteamKing]

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

I already told you what M_A and M_B 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.)

 

[hdp12]

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

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 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.

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

 

[SteamKing]

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)R_BY, but if that's that case, aren't we supposed to get the thing in terms of both support reactions? so.. 0=M₂-(a⋅R_AY+b⋅R_BY)

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.

 

[hdp12]

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.

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

 

[SteamKing]

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

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.

 

[hdp12]

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.

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?

 

[SteamKing]

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?

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.

 

[hdp12]

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.

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

 

[SteamKing]

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

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

 

[hdp12]

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

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.

 

[SteamKing]

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.

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.

 

[hdp12]

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.

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

 

[SteamKing]

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

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

 

[hdp12]

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

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.

 

[SteamKing]

I'm not familiar with the abbreviation OP,

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

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.

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

 

[hdp12]

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?

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?

 

[SteamKing]

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..

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

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.

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

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?

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.

 

[hdp12]

@A, ΣM=0: M2-(a+b)RBY -> RBY=M2 / (a+b)

yes. R_BY=M₂/(a+b)

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

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

 

[SteamKing]

yes. R_BY=M₂/(a+b)

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

The equation above for ΣFy = 0 is not correct.

 

[hdp12]

The equation above for ΣFy = 0 is not correct.

ΣFy = 0: 0 = R_AY - R_BY is?

why is the position of the moment on the rigid body insignificant?

 

[SteamKing]

ΣFy = 0: 0 = R_AY - R_BY is?

why is the position of the moment on the rigid body insignificant?

Because the applied moment M₂ in this case is a special kind of moment known as a couple :

http://en.wikipedia.org/wiki/Couple_(mechanics)

A couple is produced by two forces of equal magnitude but acting in opposite direction, and these forces are separated by a certain distance.

Since the forces are acting in opposite directions and are equal in magnitude, their net force equals zero, thus they cannot influence the values of the reactions on the beam which keep it in equilibrium.

The couple could be applied anywhere on the beam without affecting the value of the reactions.

 

[hdp12]

Because the applied moment M₂ in this case is a special kind of moment known as a couple :

Since the forces are acting in opposite directions and are equal in magnitude, their net force equals zero, thus they cannot influence the values of the reactions on the beam which keep it in equilibrium.

The couple could be applied anywhere on the beam without affecting the value of the reactions.

isn't the moment acting about the rigid body, like along the x-axis? The couple would be if it was acting into or out of the page.. unless I'm analyzing the diagram incorrectly...

 

[SteamKing]

isn't the moment acting about the rigid body, like along the x-axis? The couple would be if it was acting into or out of the page.. unless I'm analyzing the diagram incorrectly...

You're interpreting the diagram incorrectly.

If the moment were acting about the x-axis, how could vertical reactions develop at the supports?