Question

Figure shows a safe (M = 430 kg) hanging by a rope (negligible mass) from a boom (a = 1.9 m and b = 2.5 m) that consists of a uniform hinged beam (m = 85 kg) and horizontal cable (negligible mass). a) What is the tension in the cable? In other words, what is the magnitude of the tension force on the beam from the cable? b) Find the magnitude of the net force on the beam from the hinge.

Answer 1

from the figure

sintheta = a/L

costheta = b/L

net torqe about the hinge = 0

T*L*sintheta - Mg*L*costheta - mg*L/2*costheta = 0

T*a - Mg*b - mg*b/2 = 0

T = (M + m/2)*g*(b/a)

T = ( 430 + (85/2))*9.8*2.5/1.9

T = 6092.76 N <<<<=========ANSWER

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along horizantal

Fh - T = 0

Fh = T = 6092.76 N

alongvertical Fv - Mg - mg = 0

Fv = (M+m)*g = (430+85)*9.8 = 5047 N

net force on beam from hinge = F = sqrt(Fh^2+Fv^2)

F = 7911.63 N <<<========ANSWER