A 25,0 kg beam is attached to a wall with a hinge while it\'s far end is support

Question

A 25,0 kg beam is attached to a wall with a hinge while it's far end is supported by a cable such that the beam is horizontal . If the angle between the beam and the cable is=70.0 degrees. What is the vertical component of the force exerted by the hinge on the beam ?

A 25,0 kg beam is attached to a wall with a hinge while it's far end is supported by a cable such that the beam is horizontal . If the angle between the beam and the cable is=70.0 degrees. What is the vertical component of the force exerted by the hinge on the beam ?

component of the force exerted by the hinge on the beam ?

Explanation / Answer

Let T is the tension in the string.

as the beam is in equilibriumnet force and net torque acting on the beam should be equal to zero.

Let L is the length of the beam.

Net torque about = 0

25*9.8*(L/2) - T*L*sin(70) = 0

==> T = 25*9.8/(2*sin(70))

= 130.4 N

now Apply, Fnet = 0

Fy - 25*9.8 + T*sin(70) = 0

Fy = 25*9.8 - T*sin(70)

= 25*9.8 - 130.4*sin(70)

= 122.5 N (upward) <<<<<<---------Answer

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