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 = 12.95 kN. The crane was rated for a maximum load of 1.000

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 = 12.95 kN. The crane was rated for a maximum load of 1.000 times 103 lbs. If d = 5.580 m, s = 0.630 m, x = 1.400 m and h = 2.250 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

theta = arctan(h/(d-s)) = arctan(2.25/(5.58-0.63)) = 0.426627493 rad

taking moment about P

we have

W*(d-x) = T*cos(theta)*(d-s)

so,

W = T*sin(theta)*(d-s)/(d-x) = 12.95*10^3*sin(0.426627493)*(5.58-0.63)/(5.58-1.4) = 6345.885 N = 6.34 kN

balancing forces

Fy =  T*sin(theta)-W = 12.95*10^3*sin(0.426627493)-6345.885=-987.136 N

Fx = T*cos(theta) = 12.95*10^3*cos(0.426627493) = 11789.24 N

so,

F = sqrt(11789.24^2+(-987.136)^2) = 11830.495 N = 11.83 kN

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