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 83.40 kg per meter of length and the tension in the cable was T = 12170 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.522 m, x = 1.800 m and h = 1.980 m, what was the magnitude of WL (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

Taking the sum the moments about P:
0 = Tsin * (d - s) -  WL* (d - x) - F * (d/2)
Where,
= tan^-1(h/(d-s))
= tan^-1(1.980 / (5.580 - 0.522))
= 21.38o

WL is the Load of Weight.
F is the weight of the beam.
T is the tension.

Substituing Values -

0 = 12170 * sin(21.4º) * (5.580 - 0.522) - WL* (5.580 - 1.8) - 83.4*5.58 *9.8 * (5.58/2)
WL = 2575.7 N


(b)
Sum the vertical forces: = 0
Fv + T*sin - WL - 83.4*5.58 *9.8 = 0
Fv + 12170 * sin(21.4º) - 2575.7 - 83.4*5.58 *9.8 = 0
Fv = 2696 N


Sum the horizontal forces:
Fh - Tcos = 0
Fh - 12170 * cos(21.4º) = 0
Fh = 11331 N

|F| = (11331^2 + 2696^2) N = 11 647 N

Magnitude of force at the attachment point P, = 11 647 N

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