A 2.0 m long uniform beam is supported by a cable as shown below. The cable...

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Question

Answer 1

Answer #1

since the beam is in equilibrium

net force acting on the beam is zero

and also Net torque acting on the beam is zero

T_net = 0 and F_net = 0

T_net = 0

(300*sin(35)*2) -(w_beam*1) = 0

w_beam = 344.2 N

F_net = 0

Fv + T*sin(35) = w_beam

Fv = w_beam - T*sin(35)

Fv = 344.2-(300*sin(35)) = 172.3 N is the vertical force at the hinge

Fh = T*cos(35) = 300*cos(35) =245.8 N is the horizontal force at the hinge

and mass pf the beam is m = w_beam/g = (344.2)/9.8 = 35.12 Kg

Answer 2

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