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 85.10 kg per meter of length and the tension in the cable was T = 12.04 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 5.870 m, s = 0.522 m, x = 1.450 m and h = 2.070 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(2.16 / 5.348) = 22 degrees             

where F is the weight of the beam.

0 = 12040*5.348*sin22º - W*4.42 - 85.1*5.87*9.8*5.87/2

W = 2206.5 N

(b) sum the vertical forces:

Fv + Tsin - W - 85.1*5.87*9.8 = 0

Fv + 12040*sin22 - 2206.5 - 85.1*5.87*9.8= 0

Fv = 2591.7 vertical force at P

sum the horizontal forces:

Fh - Tcos = 0

Fh - 12040cos22º = 0

Fh = 11163 N horizontal force at P

mag P = ((11163)^2+(2591.7)^2)^0.5 = 11460 N total reaction at P

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