Map Sapling Learning After an unfortunate accident at a local warehouse you have

Question

Map Sapling Learning 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 86.80 kg per meter of length and the tension in the cable was T= 12690 N. The rane was rated for a maximum load of 4545 kg. If d = 5.580 m, s = 0.522 m, x 1.700 m and h 1.890 m, what was the magnitude of W (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 Number Number Previous Give Up & View Salution ) Check Answer Next xit

Explanation / Answer

s = 0.522 m

mass per meter = 86.8 kg/m

mass = 86.8 * d = 86.8 * 5.58 = 484.344 kg

crane load = 454.5 kg

tan theta = h/(d-s)

tan theta = 1.89/(5.58-0.522)

tan theta = 0.373

theta = 20.48 deg

at equillibirum condition of torque at point is

T sin theta* (d-s) - WL *(d-x) - mg (d/2) = 0

WL*(5.58-0.522) = (12690 * sin 20.48) * (5.58 -0.522) -(484.344* 9.81*5.58/2)

WL = 1819.1 N

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Horizontal Component of force F = T cos theta

Fx = 12690 * cos 20.48

Fx = 11887.92 N

using newtons 2nd law

Fy = Wl + mg - T sin theta

Fy = (1819.1) + (484.44 * 9.81) -(12690 * sin 20.48)

Fy = 2131.47

thus force F^2 = Fx^2 + Fy^2

F^2 = 11887.92^2 + 2131.47^2

F = 12077.49 N

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