Calculating Tension on 500kg Beam Supported by Cable

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Homework Help Overview

The problem involves a 500 kg beam that is free to pivot at its lower end and is supported by a horizontal cable at its upper end, with a specified angle of 35 degrees between the beam and the wall. The task is to determine the tension in the cable while considering the torque acting on the beam.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of torque using the pivot point at the end of the beam and the forces acting on it. Questions arise about the torque from the cable tension and how to relate it to the torque from the weight of the beam.

Discussion Status

The discussion has explored the relationship between the torques from the weight of the beam and the tension in the cable. Some participants have provided guidance on how to set up the equations for torque, while others express uncertainty about specific calculations and the setup of the problem.

Contextual Notes

There is a noted lack of information regarding the length of the beam, which is crucial for calculating the torques. Participants are also navigating the implications of the angles involved in the torque calculations.

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Homework Statement



A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
the cable?

Homework Equations



Torque = r * F

The Attempt at a Solution



Using pivot point at the end of the beam, the torque is 0.

weight acting down on the beam is mg*l/2*sin(35). L is not givin not sure where to go from there.
 
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yankees26an said:

Homework Statement



A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
the cable?

Homework Equations



Torque = r * F

The Attempt at a Solution



Using pivot point at the end of the beam, the sum of the torques from all forces [/color] is 0.

torque from [/color]weight acting down on the beam is mg*l/2*sin(35), clockwise[/color]. L is not givin not sure where to go from there.
What's the torque from the cord tension about the pivot?
 
PhanthomJay said:
What's the torque from the cord tension about the pivot?

you mean from the top of the beam to the bottom? :rolleyes:
 
yankees26an said:
you mean from the top of the beam to the bottom? :rolleyes:
yes.
 
PhanthomJay said:
yes.

Not sure how to find it.
 
yankees26an said:
Not sure how to find it.
You correctly identified the torque about the pivot from the weight force as mgL/2sintheta, that is Torque = r X F = rFsintheta, where F in this case is mg, and r is L/2, and theta is 35 degrees(the angle between the force and position vector). Now do the same for the torque about the pivot for the unknown horizontal tension force (call it T) at the top of the beam, that is, Torque from T = r X F = rFsintheta, where r is L, T is itself, and theta , the angle between T and L , is ?.. Solve for T by summing these 2 torques and setting that sum equal to 0. The 'L" term should cancel out.
 
mg*L/2*sin(35) = T*L*sin(55)

mg/2*sin(35) = T*sin(55)

(mg/2*sin(35))/sin(55) = T

T = 1.72 kN

Thanks :)
 

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