How Do You Calculate Tension and Static Friction in a Beam Supported by a Rope?

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SUMMARY

The discussion focuses on calculating tension and static friction in a beam supported by a rope, specifically a 40 Newton beam with a 200 Newton horizontal force applied. The tension (T) is calculated as 220.676 N using the equation T = 200 / cos(25). Static friction (Fs) is determined to be 106.738 N using the equation Fs = 200 - (220.676 * sin(25)). Participants emphasize the importance of correctly identifying the point of force application and suggest summing moments about the bottom of the rope to find vertical reaction forces.

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Homework Statement


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 = 220.676NFtotx = 200N - Fs - Tsin 25

Fs = 200N - (220.676 sin 25)

Fs = 106.738

Not sure if this is correct.
 
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No it's like this.
 
Where is the 200 N force applied? Halfway up the vertical member??
 
PhanthomJay said:
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?
 
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...
 
PhanthomJay said:
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.
 
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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