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 90.20 kg per meter of length and the tension in the cable was T = 12690 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.000 m, s = 0.450 m, x = 1.400 m and h = 1.890 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

According to the given problem,

Sum the moments about P:
0 = T(d - s)sin - W(d - x) - F(d/2)
= tan-1(h/(d-s)) = tan-1(1.89 / 4.55) = 22.557°
where F is the weight of the beam.
0 = 12690N*4.55m*sin22.557º - W*3.6m - 90.2kg/m*5.0m*9.8m/s²*2.5m
3.6W = 11099.73 N
W = 3083.3 N load

(b) sum the vertical forces:
Fv + Tsin - W - 90.2kg/m*5.0m*9.8m/s² = 0
Fv + 12690N*sin22.557° - 3083.3N - 90.2kg/m*5.0m*9.8m/s² = 0
Fv = 2635.15 N vertical force at P

sum the horizontal forces:
Fh - Tcos = 0
Fh - 12690N*cos22.557º = 0
Fh = 11719.2 N horizontal force at P

mag P = (11719.2² + 2635.15²) N = 12011.8 N total reaction at P

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