After an unfortunate accident at a local warehouse you have been contracted to d

Question

After an unfortunate accident at a local warehouse you have been contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure below. The horizontal steel beam had a mass of 88.50 kg per meter of length and the tension in the cable was T = 12820 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.290 m, s = 0.558 m, x = 1.350 m and h = 2.250 m, what was the magnitude of W_L (the load on the crane) before the collapse? What was the magnitude of force at the attachment point P? The acceleration due to gravity is g = 9.810 m/s^2.

Explanation / Answer

Sum the moments about P:
0 = T(d - s)sin - W(d - x) - F(d/2)
= arctan(h/(d-s)) = arctan(2.25 / 4.732) = 25.43 deg
where F is the weight of the beam.
0 = 12820N*4.732m*sin25.43º - W*3.94m - 88.5kg/m*5.29m*9.8m/s²*5.29m/2
3.94W = 13914.41 N
W = 3531.577 N load..............Ans.

(b) sum the vertical forces:
Fv + Tsin - W - 88.5kg/m*5.29m*9.8m/s² = 0
Fv + 12820N*sin25.43 - 3531.577N - 4588.017N = 0
Fv = 2614.58 N vertical force at P

sum the horizontal forces:
Fh - Tcos = 0
Fh - 12820N*cos25.43º = 0
Fh = 11 577.87 N horizontal force at P

mag P = (11577.87² + 2614.58²) N = 11 869.42 N total reaction at P

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