# Find magnitude of a force on a ring.

## 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.

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.

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

• forces.jpg
3.2 KB · Views: 330
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?

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.

Ok. That makes sense. So how do you find what that angle is?

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.

Is the x component 120(3/5) and the y is 120(4/5)?

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?

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.

TSny
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Is the magnitude of the red vector 120N?

Ah. Its not. I think its 84.8? So would the x be 84.8cos(60) and y be 84.8sin(60)?

TSny
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Yes, that looks correct. (Maybe use 84.9 rather than 84.8)

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?

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 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:
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?

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.

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.

TSny
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