The uniform horizontal beam supporting the sign is 1.40 m long, has a mass of 15
ID: 1420179 • Letter: T
Question
The uniform horizontal beam supporting the sign is 1.40 m long, has a mass of 15.0 kg, and is hinged to the wall. The sign itself is uniform with a mass of 35.0 kg and over-all length of 1.20 m. The two wires supporting the sign are each 30.0 cm long, are 85.0 cm apart, and are equally spaced from the middle of the sign. The cable supporting the beam is 2.00 m long. What minimum tension must your cable be able to support without having your sign come crashing down? What minimum vertical force must the hinge be able to support without pulling out of the wall?Explanation / Answer
A)
let T1 and T2 are tensions in the strings connected to sign.
T1 = T2 = 35*9.8/2
= 171.5 N
angle made by cable with horizontal, theta = cos^-1(1.4/2)
= 45.6 degrees
now Apply net torque about hinge = 0
Tmin*1.4*sin(45.6) - T1*1.4 - T2*(1.4 - 0.85) = 0
Tmin = (T1*1.4 + T2*0.55)/(1.4*sin(45.6))
= (171.5*1.4 + 171.5*0.55)/(1.4*sin(45.6))
= 334.3 N
B) Apply, Fnety = 0
F_hingey + Tmin*sin(45.6) - T1 - T2 = 0
F_hingey = T1 + T2 - Tmin*sin(45.6)
= 171.5 + 171.5 - 334.3*sin(45.6)
= 104.1 N
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