Given Moment about a Pin and a Roller

In summary, for the statics homework problem 4.10, the student is asked to draw a free body diagram and use force and moment equilibrium to determine the support reactions for a pinned and simply supported beam. The student initially struggles with writing the correct moment equation, but after some guidance, realizes that the correct equation is 0 = |MA+MB| - |M2|, where MA and MB are both equal to zero due to the pinned connections at the ends of the beam. The student then correctly determines that RBY = M2 / (a+b) and RAY = M2 / (a+b) and that the direction of RAY must be opposite to RBY in order for the sum of the forces to
  • #1
hdp12
68
2

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
HW4:4-10.png


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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IMG_2756.JPG
 
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  • #2
The diagram alone is not clear to me. Any words to go with it?
 
  • #3
haruspex said:
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 RAY, RBY, and RBX
 
  • #4
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
okay, so if the moments at the ends are 0, how does that help me get to the support reactions at each point?
 
  • #6
hdp12 said:
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.)
 
  • #7
SteamKing said:
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 = |MA+MB| - |M2|
is this the correct one?
 
  • #8
hdp12 said:
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?
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
SteamKing said:
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.

SteamKing said:
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)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
hdp12 said:
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)

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
SteamKing said:
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)RBY -> RBY=M2 / (a+b)
and also
@B, ΣM=0: M2-(a+b)RAY -> RAY=M2 / (a+b)
yeah?
 
  • #12
hdp12 said:
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?

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
SteamKing said:
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?
 
  • #14
hdp12 said:
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.
 
  • #15
SteamKing said:
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
 
  • #16
hdp12 said:
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?
 
  • #17
SteamKing said:
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.
 
  • #18
hdp12 said:
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.
 
  • #19
SteamKing said:
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 = RBY + RAY -(Force determined from Moment)
Force determined from moment = M2/b
RAY=M2/b - RBY
 
  • #20
hdp12 said:
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

It's not clear what these equations represent. Are they supposed to show the solution to the reactions for the OP?
 
  • #21
SteamKing said:
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.
 
  • #22
hdp12 said:
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?
 
  • #23
SteamKing said:
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?
 
  • #24
hdp12 said:
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 RBY?
@A, ΣM=0: M2-(a+b)RBY -> RBY=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 RAY and RBY 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.
 
  • #25
SteamKing said:
@A, ΣM=0: M2-(a+b)RBY -> RBY=M2 / (a+b)

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?
 
  • #26
hdp12 said:
yes. RBY=M2/(a+b)

for the force equilibrium equations...
ΣFy=0: 0 =RAY + RBY - (M2/b)
The equation above for ΣFy = 0 is not correct.
 
  • #27
SteamKing said:
The equation above for ΣFy = 0 is not correct.

ΣFy = 0: 0 = RAY - RBY is?

why is the position of the moment on the rigid body insignificant?
 
  • #28
hdp12 said:
ΣFy = 0: 0 = RAY - RBY 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.
 
  • #29
SteamKing said:
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...
 
  • #30
hdp12 said:
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?
 
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  • #31
oh, alright. Well thank you so much, I really appreciate it!
 
  • #32
hdp12 said:
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
I'm glad SteamKing seems to understand it because I still don't. If A is a roller, how is it going to supply the downward force necessary to oppose the applied moment? I see no mention of mass or gravity.
I note that the bases of the supports are drawn differently, but not according to any diagrammatic convention I know.
 
  • #33
haruspex said:
I note that the bases of the supports are drawn differently, but not according to any diagrammatic convention I know.

It's possible the support at A was drawn slightly differently to permit the x-axis to be drawn on the diagram w/o interference.

To me, it looked like both supports were pinned connections with no indication of any rollers present.
 
  • #34
SteamKing said:
It's possible the support at A was drawn slightly differently to permit the x-axis to be drawn on the diagram w/o interference.

To me, it looked like both supports were pinned connections with no indication of any rollers present.
The text says A is a roller.
 
  • #35
haruspex said:
The text says A is a roller.
What text? The OP simply posted a pitcher out of his textbook. He put the word "roller" in the thread title, but ... :rolleyes:
 
<h2>1. What is a given moment about a pin and a roller?</h2><p>A given moment about a pin and a roller refers to the force that is applied to a pin and roller connection, causing a rotation around the pin. This moment is typically measured in Newton-meters (Nm) and is an important concept in mechanics and engineering.</p><h2>2. How is a given moment about a pin and a roller calculated?</h2><p>A given moment about a pin and a roller is calculated by multiplying the force applied to the pin by the perpendicular distance from the pin to the line of action of the force. This distance is known as the moment arm and is typically measured in meters (m).</p><h2>3. What is the difference between a pin and a roller in terms of given moments?</h2><p>A pin is a fixed connection that allows rotation around the pin, while a roller is a connection that allows both rotation and translation. This means that a given moment about a pin only causes rotation, while a given moment about a roller can cause both rotation and translation.</p><h2>4. How does the direction of the force affect the given moment about a pin and a roller?</h2><p>The direction of the force affects the given moment about a pin and a roller by determining the direction of the rotation. If the force is applied in a clockwise direction, the rotation will be clockwise, and if the force is applied in a counterclockwise direction, the rotation will be counterclockwise.</p><h2>5. What are some real-world applications of given moments about a pin and a roller?</h2><p>Given moments about a pin and a roller are commonly used in mechanical and structural engineering, as well as in physics and other fields of science. They are important in understanding the stability and strength of structures, such as bridges and buildings, as well as in designing machines and other mechanical systems.</p>

Related to Given Moment about a Pin and a Roller

1. What is a given moment about a pin and a roller?

A given moment about a pin and a roller refers to the force that is applied to a pin and roller connection, causing a rotation around the pin. This moment is typically measured in Newton-meters (Nm) and is an important concept in mechanics and engineering.

2. How is a given moment about a pin and a roller calculated?

A given moment about a pin and a roller is calculated by multiplying the force applied to the pin by the perpendicular distance from the pin to the line of action of the force. This distance is known as the moment arm and is typically measured in meters (m).

3. What is the difference between a pin and a roller in terms of given moments?

A pin is a fixed connection that allows rotation around the pin, while a roller is a connection that allows both rotation and translation. This means that a given moment about a pin only causes rotation, while a given moment about a roller can cause both rotation and translation.

4. How does the direction of the force affect the given moment about a pin and a roller?

The direction of the force affects the given moment about a pin and a roller by determining the direction of the rotation. If the force is applied in a clockwise direction, the rotation will be clockwise, and if the force is applied in a counterclockwise direction, the rotation will be counterclockwise.

5. What are some real-world applications of given moments about a pin and a roller?

Given moments about a pin and a roller are commonly used in mechanical and structural engineering, as well as in physics and other fields of science. They are important in understanding the stability and strength of structures, such as bridges and buildings, as well as in designing machines and other mechanical systems.

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