A 35.6 kg beam is attached to a wall with a hinge and its far end is supported b

Question

A 35.6 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of ? = 17.5

A 35.6 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of ? = 17.5 degree with respect to horizontal. What is the horizontal component of the force exerted by the hinge on the beam? (Use the 'to the right' as + for the horizontal direction.) What is the magnitude of the force that the beam exerts on the hinge?

Explanation / Answer

Let L = length of the beam
T = tension in the cable
Fh = horizontal component of force by the hinge
Fv = vertical component of force by the hinge

Take counterclockwise torque as positive. Let us find torques around the hinge.
Torque by T = T L
Torque by the force of gravity = -m g L/2 * cos(theta)
Torque by Fh = 0
Torque by Fv = 0

Net torque = 0
Therefore, T L - m g L/2 * cos(theta) = 0
Dividing both sides by L,
T - m g/2 * cos(theta) = 0
T = mg/2 * cos(theta) --------------------(1)

The cable makes 90 deg - theta with the horizontal.
Net horizontal force = 0
Therefore Fh - T cos(90 deg - theta) = 0
Fh - T sin(theta) = 0
Fh = T sin(theta) --------------------------(2)

Substituting the value of T from equation (1) into equation (2):
Fh = mg/2 * cos(theta) * sin(theta)
Fh = 35.6 * 9.8/2 * cos(17.5 deg) * sin(17.5 deg)
Fh = 50.027 N

b)

V - m*g + T*sin(90 - 17.5) = 0

So V = m*g - T*sin(90 - 17.5) = 35.6*9.8 - 166.36*sin(90 - 17.5) = 190.21N

so F = sqrt(50.027^2 + 190.21^2) = 196.6822N

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