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

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Homework Help Overview

The problem involves a 40 Newton beam supported by a rope at a 25 degree angle, with a horizontal force of 200 Newtons applied to the beam. The objective is to calculate the tension in the rope and the static friction acting on the beam.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the application point of the 200 N force and the weight of the beam. There are attempts to express the forces in terms of equations, but uncertainty remains regarding the correctness of the calculations and assumptions.

Discussion Status

The discussion is ongoing, with participants questioning the details of the problem statement and the assumptions made about the force application. Some guidance has been offered regarding the setup of the problem and the need to consider vertical and horizontal reactions.

Contextual Notes

There is ambiguity regarding the application point of the 200 N force, as it is not specified in the problem statement. Participants are also considering the implications of this missing information on their calculations.

bigbaddrumlad
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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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