A 21.8 kg beam is attached to a wall with a hinge and its far end is supported b

Question

A 21.8 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 ? = 27.0

A 21.8 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 90 degree. If the beam is inclined at an angle of ? = 27.0 degree with respect to horizontal. A. 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.) The Net torque and the Net Force on the hinge must be zero since it is in equilibrium. What is the magnitude of the force that the beam exerts on the hinge?

Explanation / Answer

A)

T*L = m*g*L/2*cos(27)

so T = m*g/2*cos(27) = 21.8*9.80/2*cos(27) = 120N

Now the horizontal force at the hinge equals the horizontal component of T

so H = T*cos(90 - 27) = 120*cos(90 - 27) = 54.48N

B) Now sum vertical forces V - m*g + T*sin(90 - 27) = 0 (Here is where the sign was wrong--Note the vertical component is the weight minus the upward tension component)

So V = m*g - T*sin(90 - 27) = 21.8*9.8 - 120*sin(90 - 27) = 106.72N

so F = sqrt(54.48^2 + 106.72^2) = 120N

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