# Equilibrium Tension Torque

1. Nov 29, 2009

### joemama69

1. The problem statement, all variables and given/known data

Note Picture...

A uniform beam, 5.00 m long with a mass of 2.00 x 103 kg, is held against a wall in the position shown by a hinge and a horizontal steel wire attached to its end.

a) Find the tension in the wire.
b) Find the magnitude of the total force supplied by the hinge.

3. The attempt at a solution

the only forces for the torque is the mass of the beam perpendicular to the beam as well as the tension of the cable

beam torque = mgdcosQ = 2000(9.8)(2.5)(cos20)
tension torque = TxsinQ this is where im a littel confused because the tension is in the x direction but the torque should be perpindicular to the beam so its confusing me.

2000(9.8)(2.5)(cos20) = TxsinQ.....Tx = 135426.28

is this correct
T

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2. Nov 29, 2009

### rl.bhat

Torque due to the tension is T*sinθ*L where L is the length of the beam.

3. Nov 29, 2009

### joemama69

oh right right...

2000(9.8)(2.5)(cos20) = Tx(5)sin20.....Tx = 26926.87 N

is this correct