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

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To calculate the tension in the cable supporting a beam and crate, the sum of forces and torques must equal zero, using the hinge as the axis of rotation. The torque from the crate and beam is calculated, and the correct angle between the wire and beam is determined to be 80 degrees, derived from the wire's 50-degree angle with the horizontal. The torque from the tension in the wire is then expressed as T*L*sin(80), ensuring the force is perpendicular to the lever arm. The discussion emphasizes the importance of correctly identifying angles and applying trigonometric relationships to solve for tension. The final tension value is confirmed to be 2251 N.
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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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