Find magnitude of a force on a ring.

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Homework Statement


http://session.masteringengineering.com/problemAsset/1127430/5/Probs.2-83_84.jpg
Refer to image for problem.
I have to find the magnitude of the force F3. I'm stuck on where to begin.

Homework Equations


Don't know.


The Attempt at a Solution


I've done. cos(y)=sqrt(1-cos^2(60)-cos^2(45))
I got the y angle to be 60. I don't know what that means so if someone could explain that would be great. I just used this because I thought I needed to know that. I then tried to break down each force vector into their components.
F = x, y, z
F1=80(4/5), 0, 80(3/5)
F2=0, 0, -110
F3=?
FR=120cos(60), 120cos(60), 120cos(45)

Thats all I got. I don't know what to do now.
 

Answers and Replies

  • #2
TSny
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I've done. cos(y)=sqrt(1-cos^2(60)-cos^2(45))
I got the y angle to be 60. I don't know what that means so if someone could explain that would be great. I just used this because I thought I needed to know that.
I don't understand what you did above. It looks like you might be trying to find the angle between FR and either the x or y coordinate axis. But FR does not make a 60 degree angle with respect to either the x or y axis.

I then tried to break down each force vector into their components.
F = x, y, z
F1=80(4/5), 0, 80(3/5)
F2=0, 0, -110
F3=?
FR=120cos(60), 120cos(60), 120cos(45)
Your expressions for the x and y components of FR are incorrect. It will help to first find the projection of FR onto the x-y plane (shown in red in my attachment). Then use this projection to find the x and y components of FR.
The ring remains stationary. What does that tell you about the forces?
 

Attachments

  • #3
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To me it looks like the red vector is 60 degrees from the positive x axis, because it shows that vector is 30 from the positive y. What am I missing?
 
  • #4
TSny
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Yes, that's correct. The red vector is 60 degrees from the x axis. But FR is not 60 degrees from the x axis.
 
  • #5
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Ok. That makes sense. So how do you find what that angle is?
 
  • #6
TSny
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I wouldn't bother with trying to find the angle that FR makes to the x and y axes. The x-component of FR is just the x-component of the red vector, and you know the angle that the red vector makes to the x-axis.
 
  • #7
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Is the x component 120(3/5) and the y is 120(4/5)?
 
  • #8
TSny
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Is the x component 120(3/5) and the y is 120(4/5)?
No. What reasoning did you use to get that?

What angle does the red vector make to the x-axis? How would you use this angle to find the x-component of the red vector?
 
  • #9
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I thought it might be a 3,4,5 triangle. The angle is 60. Would the x component be 120cos(60)? I thought you said that was wrong though.
 
  • #10
TSny
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Is the magnitude of the red vector 120N?
 
  • #11
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Ah. Its not. I think its 84.8? So would the x be 84.8cos(60) and y be 84.8sin(60)?
 
  • #12
TSny
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Yes, that looks correct. (Maybe use 84.9 rather than 84.8)
 
  • #13
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Ok. So this is what I now have as broken into their components.
F = x, y, z
F1= 80(4/5), 0, 80(3/5)
F2= 0, 0, -110
F3=?
FR= 84.9cos(60), 84.9sin(60), 120cos(45)

Attempt at solution:
80(4/5) + Fx3 = 84.9cos(60)
Fx3 = 21.55
Fy3 = 84.9sin(60)
Fy3 = 73.52
80(3/5) - 110 + Fz3 = 120cos(45)
Fz3 = 146.85
F3 = sqrt(21.55^2+73.52^2+146.85^2)
F3 = 165.6

That look right?
 
  • #14
TSny
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Ok. So this is what I now have as broken into their components.
F = x, y, z
F1= 80(4/5), 0, 80(3/5)
F2= 0, 0, -110
F3=?
FR= 84.9cos(60), 84.9sin(60), 120cos(45)
The above looks good :smile:
Attempt at solution:
80(4/5) + Fx3 = 84.9cos(60)
Fx3 = 21.55
That look right?
No. Why should the x-component of FR equal the sum of the x-components of the other forces? That's not correct.
[EDIT: Actually you are correct! I did not realize that FR is the resultant of the other 3 forces.]

Note that the ring is at rest and remains at rest. So, what does Newton's first (or second) law of motion tell you about the forces?
 
Last edited:
  • #15
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I thought the components of the other forces would add up and equal the resultant vector component.

The forces should cancel out. Or I thought equal each other, which is what I did. So are you saying the x components from all the vectors should add up to 0? and y and z components?
 
  • #16
TSny
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Ah, I see. FR represents the resultant of the other 3 forces. I was thinking it was just one out of 4 forces acting on the ring. There was no statement of the problem. But I guess the subscript R should have clued me in! Sorry about that.

Good! Your work now looks correct to me.
 
  • #17
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Ok. Thank you. Its my fault. Should have put more of a description for the problem!
Thank you for the help. I really appreciate it.
 
  • #18
TSny
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Glad to be of help.

Note that there would actually have to be another force acting on the ring, because the net force must be zero if the ring is to remain at rest. (This could be a force coming from the board that the ring is mounted in.) But as long as the problem stated that FR is the resultant of the three given forces, then your work looks correct to me.
 

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