Question

A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a vertical cable at the other end. Calculate the magnitude of the tension in the wire if the angle between the beam and the horizontal is θ = 33°.

Answer 1

T = 1351 N

Step-by-step explanation:

Weight = mg = 150 x 9.81 = 1471.5 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, Let’s determine the clockwise and counter clockwise torques by letting the pivot point be at the end of the beam that is attached to the wall.

Thus, the weight of the beam will produce clockwise torque and the vertical component of the tension will produce counter clockwise torque.

At the weight of the beam is at its center;

Clockwise torque = WL/2

Counter clockwise torque = TL sinθ

Now, clockwise torque will be equal to anti clockwise torque and thus;

TL sinθ = WL/2

Thus;

T sin 33 = 1471.5/2

T sin 33 = 735.75

T = 735.75 ÷ sin 33 = 735.75/0.5446

T = 1350.99

This is approximately 1351 N.