Tension in a cable supporting a beam

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Homework Help Overview

The problem involves a construction worker sitting on a steel beam supported by a cable, with the goal of determining the tension in the cable to assess safety. The context includes the weight of the worker and the beam, as well as the cable's rated capacity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating torque about the attachment point of the beam, considering the weights involved and the tension in the cable. Questions arise regarding the interpretation of the problem setup, particularly the worker's position on the beam.

Discussion Status

Multiple interpretations of the problem setup are being explored, particularly concerning the worker's position. Some participants have provided calculations and reasoning regarding the tension, while others question the assumptions made in the calculations.

Contextual Notes

There is uncertainty about the worker's exact position on the beam, which affects the calculations. The cable's rated capacity is noted but clarified as not being the working tension. Participants are also navigating language comprehension issues that impact their understanding of the problem.

aligass2004
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Homework Statement



An 88kg construction worker sits down 2.0 m from the end of a 1430 kg steel beam to eat his lunch. The cable supporting the beam is rated at 15,000 N. Should the worker be worried? To answer this, determine the tension in the cable.

Homework Equations





The Attempt at a Solution



First, I have no idea how to start this problem. Second, I don't know what they mean by "the beam is rated."
 
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do they give a picture?
 
Take the torque about the point where the beam is attached to the wall... take clockwise positive just like the other problem...

so you have these forces causing torque: the weight of the beam... the weight of the man... and the tension...

divide the tension up into the horizontal and vertical component... think about the torque due to each component separately...

the net torque about the point is 0...

you should be able to solve for the tension...
 
I tried solving this like the other one. The torque of the beam alone is 42042. Then the T with the man is T = (88)(9.8)(6) + 42042. For the tension, does the tension in the x-direction play a part?
 
aligass2004 said:
I tried solving this like the other one. The torque of the beam alone is 42042. Then the T with the man is T = (88)(9.8)(6) + 42042. For the tension, does the tension in the x-direction play a part?

no tension in the x-direction doesn't exert a torque about that point... one way to see this is that torque = r X F... when two vectors are parallel, their cross product is 0. r is in the x-direction...Tx is in the x-direction... so torque due to Tx is 0.

so you just need to consider Ty.
 
Ty = 7500. Was the first part right?
 
aligass2004 said:
Ty = 7500. Was the first part right?

No, that's not what I'm getting...

Also here:

T = (88)(9.8)(6) + 42042

it should be 88*9.8*4 since the man is 2m away from the end...

how did you get the 7500?
 
Oh. I did Ty = Tsin(theta) = 15000sin(30).
 
  • #10
aligass2004 said:
Oh. I did Ty = Tsin(theta) = 15000sin(30).

oh! yeah, don't use the 15000N... that's just a safety rating... means that beyond that the cable might snap...

Use the sum of the torques about the pin = 0

Tsin(theta) would appear in the above equation... letting you solve for T.
 
  • #11
So T = 45491.6/sin theta?
 
  • #12
aligass2004 said:
So T = 45491.6/sin theta?

You forgot to multiply Tsin(theta) times the distance to give torque due to Tsin(theta)...
 
  • #13
T = 45491.6/(6m x sin30)?
 
  • #14
aligass2004 said:
T = 45491.6/(6m x sin30)?

exactly.
 
  • #15
Got it. Finally. T = 15163.867. Thanks! :)
 
  • #16
aligass2004 said:
Got it. Finally. T = 15163.867. Thanks! :)

no prob.
 
  • #17
My english is ok but I have hard time comprehending some questions.

How did you know that it meant he was sitting 2m from the right end, and not the left end?

The picture doesn't help me determine this either, looks like he was sitting in the middle there. thanks.

If it was from the left end i got 14.606kN, therefore the man should not be worried.
 
  • #18
pooface said:
My english is ok but I have hard time comprehending some questions.

How did you know that it meant he was sitting 2m from the right end, and not the left end?

The picture doesn't help me determine this either, looks like he was sitting in the middle there. thanks.

If it was from the left end i got 14.606kN, therefore the man should not be worried.

Yeah, I assumed it was from the right end... I think I looked at the diagram and thought he was leaning a little to the right so the center of mass would be slightly to the right...

But you're right it could be either way from the wording of the question.
 

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