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 93.60 kg per meter of length and the tension in the cable was T = 12.56 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 5.580 m, s = 0.450 m, x = 1.850 m and h = 1.890 m, what was 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/s2.

Explanation / Answer

Sum the moments about P:
0 = T(d - s)sin - W(d - x) - F(d/2)
= arctan(h/(d-s)) = arctan(1.89/5.094) = 20.36 degree
where F is the weight of the beam.
0 = 12560N*5.094m*sin20.36º - W*3.73m – 93.6kg/m*5.58m*9.8m/s²*5.58m/2
3.73W = 7979.6 N
W = 2139.3 N load

(b) sum the vertical forces:
Fv + Tsin - W – 93.6kg/m*5.58m*9.8m/s² = 0
Fv + 12560N*sin20.36 - 2139.3 N - 5118.42 N = 0
Fv = 2887.87 N vertical force at P

sum the horizontal forces:
Fh - Tcos = 0
Fh - 12560N*cos20.36º = 0
Fh = 11775.32 N horizontal force at P

mag P = (11775.32² + 2887.87²) N = 12124.27 N total reaction at P

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