Tension and Static Friction


by bigbaddrumlad
Tags: friction, static, tension
bigbaddrumlad
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#1
May12-10, 08:10 PM
P: 4
1. The problem statement, all variables and given/known data
There is a 40 newton beam being held by a rope. The rope is attached at the top of the beam and the ground making a 25 degree angle at the top of the beam. There is a force applied to the beam with 200 newtons in the positive x direction. Find the tension and static friction.

All I think I have figured out is:

Ftoty = T(cos 25) - 200

T = 200
cos 25

T = 220.676N


Ftotx = 200N - Fs - Tsin 25

Fs = 200N - (220.676 sin 25)

Fs = 106.738

Not sure if this is correct.
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Lil_Aziz1
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#2
May12-10, 08:45 PM
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Just to make sure, is this how it looks? http://i180.photobucket.com/albums/x...ntitled-21.png
bigbaddrumlad
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#3
May12-10, 09:27 PM
P: 4


No it's like this.

PhanthomJay
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#4
May12-10, 10:30 PM
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Tension and Static Friction


Where is the 200 N force applied? Halfway up the vertical member??
bigbaddrumlad
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#5
May12-10, 10:42 PM
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Quote Quote by PhanthomJay View Post
Where is the 200 N force applied? Halfway up the vertical member??
It does not state where it's applied. I'm guessing center of gravity?
PhanthomJay
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#6
May12-10, 10:49 PM
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The 200 N force is apparently an applied horizontal force....have you stated the problem correctly? Does the vertical member only weigh 40 N? Something is not right....
bigbaddrumlad
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#7
May12-10, 10:53 PM
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Quote Quote by PhanthomJay View Post
The 200 N force is apparently an applied horizontal force....have you stated the problem correctly? Does the vertical member only weigh 40 N? Something is not right....
Yes this is how it is stated. The applied force states it is in the positive x direction (horizontal). I just can't figure out how to set up the problem.
PhanthomJay
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#8
May13-10, 04:39 AM
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I guess you'll have to assume that the 200 N force is applied at the center of the vertical member, but the problem should have stated such. You should first find horizontal and vertical reactions at the ground on the vertical member and on the rope. Without taking a shortcut approach, try summing moments about the bottom of the rope = 0, to solve for the vertical reaction forces. The horizontal component of the reaction on the rope is trig related to the vertical component on the rope (Tx/Ty= tan 25).


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