A 37.2 kg beam is attached to a wall with a hinge and its far end is supported b
ID: 2242138 • Letter: A
Question
A 37.2 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 ? = 24.3
A 37.2 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 ? = 24.3 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
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)
Now cable makes 90 deg - theta with the horizontal.
Net horizontal force = 0
Fh - T cos(90 - theta) = 0
Fh - T sin(theta) = 0
Fh = T sin(theta)
So,
Fh = mg/2 * cos(theta) * sin(theta)
Fh = 37.2 * 9.8/2 * cos(24.3 deg) * sin(24.3 deg)
Fh = 68.365 N
B.
Force exerted by beam on hinge is
F = m g [sqrt( (sin theta)