How Do You Calculate the Tension in a Cable Supporting a Beam and Crate?

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Homework Help Overview

The problem involves calculating the tension in a cable that supports a beam with a crate hanging from it. The scenario includes a uniform beam, a vertical wall, and forces acting on the system, including the weight of the beam and the crate.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of torque from various forces acting on the beam and question the correct angle to use for the tension in the wire. There are attempts to clarify the relationship between the angles involved and how they affect the torque calculations.

Discussion Status

The discussion is ongoing with participants exploring different interpretations of the angles involved in the problem. Some have suggested methods for calculating torque, while others are questioning the assumptions made regarding angles and their relationships.

Contextual Notes

Participants are working with a diagram that provides angles, but there is some confusion about how to derive the angle between the wire and the beam. The problem context includes specific angles (50 degrees and a derived angle of 30 degrees) that are being debated.

elitespart
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A 1200 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1960 N crate hangs from the far end of the beam.

Picture: http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c09/ch09p_20.gif

Calculate the magnitude of the tension in the wire.


So I know that sum of forces and torques equals 0. Using the hinge as the axis of rotation:

Torque from crate = -1960Lcos(30)
Torque from beam = -1200(L/2)cos(30)
What would the torque from the tension in the wire be? It's directed 50 degrees above the horizontal so I thought it would be T*Lcos(50) but that is wrong. Can someone point me in the right direction? Thanks.
 
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It's directed 50 degrees above the horizontal so I thought it would be T*Lcos(50) but that is wrong.
To find the torque from the tension find the horizontal component of the T and find its vertical distance from the hinge. The product will give you the required torque.
 
Sorry I'm not quite sure what you mean. In the solution it divided the sum of the torques from the crate and the beam by cos(10). Where did they get that from?
 
Torque due to the tension is Tcos50*Lsin30
 
rl.bhat said:
Torque due to the tension is Tcos50*Lsin30

no that's not right. the answer is 2251N. Using your torque for the wire gives an answer of 6898.
 
elitespart said:
What would the torque from the tension in the wire be? It's directed 50 degrees above the horizontal so I thought it would be T*Lcos(50) but that is wrong. Can someone point me in the right direction?
What's the angle between the wire and the beam? Use that angle to determine the torque.
 
Doc Al said:
What's the angle between the wire and the beam? Use that angle to determine the torque.

Well I'm guessing it's cos(10) but I don't know where that is coming from.
 
Why guess? You have the diagram. Figure out the angle.
 
Doc Al said:
Why guess? You have the diagram. Figure out the angle.

That's what they had in the solution. From the pic I'm getting an angle of 30 between the wire and beam.
 
  • #10
elitespart said:
That's what they had in the solution.
We'll get to the solution soon enough. First find the angle.

From the pic I'm getting an angle of 30 between the wire and beam.
How do you get that? The diagram states that the wire is 50 degrees from the horizontal, so its angle with the beam must be even greater.
 
  • #11
Doc Al said:
We'll get to the solution soon enough. First find the angle.How do you get that? The diagram states that the wire is 50 degrees from the horizontal, so its angle with the beam must be even greater.

It would be 80 yes?
 
  • #12
elitespart said:
It would be 80 yes?
Yes! Now how do you use that to calculate torque?
 
  • #13
oh okay. so it's T*Lsin80 since force has to be perpendicular to the lever arm.
 
  • #14
elitespart said:
so it's T*Lsin80 since force has to be perpendicular to the lever arm.
Good! And to relate it to your solution, sin80 = cos ?
 
  • #15
Doc Al said:
Good! And to relate it to your solution, sin80 = cos ?

=cos10. Thank you very much. Much appreciated.
 

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