A crate with a mass of 173.5 kg is suspended from the end of a uniform boom with

Question

A crate with a mass of 173.5 kg is suspended from the end of a uniform boom with a mass of 86.9 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable. 12 E 10 9 5 4 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Horizontal position, x, (m) 2877 N You'll need to get the various positions from the graph. Many are exactly on one of the tic marks.) Submit Answer You have entered that answer Incorrect. Tries 6/12 Previous before

Explanation / Answer

The values given are as:

m = mass of the crate = 173.5 kg
m = mass of the boom = 86.9 kg
g = acceleration due to gravity = 9.81 m/s²
L = length of the boom
= angle of the boom to horizontal
= angle of the cable to horizontal

From the graph, we calculate the angle as
tan() = (6-2)/(15-2) = 4/13 = 0.31
so
= arctan(0.31) = 17.10°
also
angle can be calculated as
tan() = (10-6)/(15-2) = 4/13 = 0.31
so
= arctan(0.31) = 17.10°

since the system is in equilibrium, then the clockwise torque (caused by the weight of the boom and crate) must be equal in magnitude to the counterclockwise torque (caused by the tension in the cable)
So:
m×g×cos()×L/2 + m×g×cos()×L = T×sin( + )×L
(m/2 + m)×g×cos() = T×sin( + )
T = (m/2 + m)×g×cos() / sin( + )

T = [(173.5)/2 + (86.9)]×(9.81)×cos(17.10°) / sin(17.10° + 17.10°)

T = 1626.54/0.5621

T = 2893.68 N
hence the tension in the cable is 2877.36 N

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