PHYSICS TORQUE

[pizza1994]

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Daves friend Rockson (m= 90 kg) sits 2/3 of the way along a steel beam (1000 kg) that sticks out horizontally from a vertical wall. The beam is supported by a cable that connects the end of the beam to the wall making a 30 degrees angle with the beam. If the cable can hold up to 15,000 N of tension then is Rockson safe?

Answer= 11200 N

Please can someone help me on this? you are suppose to use torques and the point where the wall and beam form 90 degrees is the pivot point but in terms of calulcation how is everyone doing this? thanks

 

[pepocho]

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[pizza1994]

that is the force of tension

 

[DrknoSDN]

Heh, this problem is annoying. I gave up when doing a similar TPR problem and had to read the solution.

So to start you have a beam bolted to the wall and attached to the cable at radius (r).
There is no net horizontal movement.

To determine how much tension is in the cable you need to balance the clockwise and counterclockwise torques.

Use the point where the beam is bolted to the wall as your point of rotation.
Set Mg (beam mass) and mg (Rockson mass) as clockwise torques.

The counterclockwise torque is the vertical component of the tension of the cable expressed as (ForceTorque(y) = Ft sin(theta))
Balance the torques:
Ft sin(theta) = Mg + mg

Adjust for position of torques along the beam. (Rockson is 2/3 from the wall, the beam center of gravity is 0.5 radius from the wall, and the cable is attached a 1 radius from the wall.)
1r * Ft sin(theta) = (1/2)r Mg + (2/3)r mg.

Drop the excess r variable because it is in every part of the equation:
Ft sin(theta) = (1/2)Mg + (2/3)mg.

Solve for Force of Torque on cable by dividing everything by sin (theta)
Ft = [(1/2)Mg + (2/3)mg] /sin(theta)

Plug --> Chug
Ft = (1/2)(1000kg)(10m/s) + (2/3)(90kg)(10m/s) / (sin(30))
Ft = (5000N + 600N) / (0.5) = 11,200N


This example is very similar to Princeton Review example 4-16 on page 111. Hope it helps.