09.4 A sign of mass M- 20 kg hangs from the end of a uniform, horizontal beam of
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Question
09.4 A sign of mass M- 20 kg hangs from the end of a uniform, horizontal beam of mass m 5.0 kg and length L 2.0 m. The beam is attached to the wall on the other end, and is supported from below by a strut (see diagram). The strut is attached to the beam a distance of d 1.2 m from the wall and makes an angle of 0- 40° with the beam. (a) What is the force of the strut acting on the beam? Assume that this force acts along the direction of the strut. (b) What is the magnitude and direction of the force of the wall acting on the beam?Explanation / Answer
the sign is in equilibrium
along vertical
net torque = 0
T*sintheta*d - Mg*L - mg*L/2 = 0
T*sin40*1.2 - 20*9.8*2 - 5*9.8*2/2 = 0
T = 572 N
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alon horizontal Fnet = 0
Fx - T*costheta = 0
F = T*costheta = 572*cos40 = 438.2 N
along vertical Fnet = 0
Fy + T*sintheta - Mg - mg = 0
Fy + 572*sin40 - 20*9.8 - 5*9.8 = 0
Fy = -122.7 N
magnitude F = sqrt(F^2+Fy^2)
F = sqrt(438.2^2+122.7^2) = 455.1 N
direction = tan^-1(Fy/Fx) = 15.6 below the horizontal
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