Determine Moment by Resolving the Force.

  • Thread starter celtics777
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  • #1
celtics777
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Homework Statement



A 20-lb force is applied to the control rod Ab as shown. Knowing that the length of the rod is 9 in. and that α=25°, determine the moment of the force about point B by resolving the force into horizontal and vertical components.

The first image I provided was given.

Homework Equations



M=Fd
Law of Sines and other basic trig

The Attempt at a Solution



The second Image I provided is as far as I can get. Assuming what I did so far is correct I'm not sure where to go from here. I know that M=Fd but can someone tell me what I am supposed to use for F and d in my cases. Do I also need to calculate the length between point A and the 90 degree angle?

Thank you.
 

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Answers and Replies

  • #2
Doc Al
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M=Fd
In order to find the moment using that relationship, F and d must be perpendicular.

Find the moment of each component of F separately and add them up, being mindful of sign.
 
  • #3
celtics777
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I attached an update to my work. Am I doing this right? Where should I go from here?
 

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  • #4
Doc Al
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I attached an update to my work. Am I doing this right? Where should I go from here?
The distance 'd' needed to find the moment due to the vertical force Fv would be the horizontal distance between the pivot and the point of application of the force. What would the sign of that moment be? (Is it clockwise or counterclockwise?)
 
  • #5
celtics777
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i'm not sure how to tell if it would be cw or ccw.
 
  • #6
Doc Al
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i'm not sure how to tell if it would be cw or ccw.
Just imagine which way the rod would tend to turn if you were to push it vertically downward or horizontally to the right.
 
  • #7
cepheid
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i'm not sure how to tell if it would be cw or ccw.

To determine this, try looking at the picture. Which way would the rod rotate it if you pulled on it in that way?
 
  • #8
celtics777
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The original picture right? The rod would rotate clockwise.
 
  • #9
Doc Al
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The original picture right? The rod would rotate clockwise.
Good. But since you are separately calculating the moment from each component of the force, use that same reasoning for each component. For example: If you pushed vertically downward (the Fv component), which way would the rod rotate?
 
  • #10
celtics777
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If I push down it would rotate cw. I'm having a harder time picturing what would happen if I push "to the right" would it also be cw?
 
  • #11
Doc Al
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If I push down it would rotate cw.
No, that's not right. Grab a pencil (or anything comparable) and let it represent the rod at its angle. Hold it by the pivot point and push down at the tip (where the force is applied). That should help you visualize how the rod will rotate.
I'm having a harder time picturing what would happen if I push "to the right" would it also be cw?
In that case it would rotate cw. But grab that pencil and really convince yourself.
 
  • #12
celtics777
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Oooh I see. That makes sense now. What do I do now that I know pushing down is ccw and to the right is cw?

Does this mean cw is pos and ccw is neg?
 
  • #13
Doc Al
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What do I do now that I know pushing down is ccw and to the right is cw?

Does this mean cw is pos and ccw is neg?
The usual convention is for ccw to be positive and cw to be negative.
 
  • #14
celtics777
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Oh, that seems fairly odd. Where ccw is counterclockwise and cw is clockwise?
 
  • #15
Doc Al
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Oh, that seems fairly odd. Where ccw is counterclockwise and cw is clockwise?
Right. At least that's the convention used in most physics books. You can think of it as coming from the right hand rule. When you curve the fingers of the right hand in the direction of the rotation, if the thumb points up the sign is positive. (Try it.)
 
  • #16
celtics777
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Ok. I remember the right hand rule now. With my problem does this mean (-68.952 lb-in)+(68.894 lb-in) = -.058 lb-in cw?
 
  • #17
Doc Al
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Ok. I remember the right hand rule now. With my problem does this mean (-68.952 lb-in)+(68.894 lb-in) = -.058 lb-in cw?
Review my comments in post #4. I believe you had the distances paired with the wrong force components.
 
  • #18
pongo38
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The reason for the acw convention is that rotations and moments are positive when looking FROM the origin in the direction of the positive direction of the axis. When you consider your problem, you are looking TOWARDS the origin, as an external observer.
 

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