After an unfortunate accident at a local warehouse you have been contracted to d
ID: 1908033 • Letter: A
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 80.00 kg per meter of length and the tension in the cable was T = 11.26 kN. The crane was rated for a maximum load of 1.000 103 lbs. If d = 5.870 m, s = 0.450 m, x = 1.400 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.
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 80.00 kg per meter of length and the tension in the cable was T = 11.26 kN. The crane was rated for a maximum load of 1.000 103 lbs. If d = 5.870 m, s = 0.450 m, x = 1.400 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
Tan = h/(d-s) = 2.07/(5.87-0.45) = 0.3819
= atan(0.3819) = 20.9 deg
Weight of beam Wb = (80*d)*g = 80*5.87*9.81 = 4606.776 N
This weight will act from mid-point of beam, i.e from d/2.
Balancing moments about P, Wl*(d-x) + 4606.776*(d/2) = (TSin)*(d-s)
Wl*(5.87 - 1.4) + 4606.776*(5.87/2) = (11.26*10^3 *Sin20.9)*(5.87 - 0.45)
Wl = 1846.38 N
Balancing forces in the vertical direction: Fv = Wl + Wb - TSin
Fv = 1846.38 + 4606.776 - (11.26*10^3 *Sin20.9) = 2436.286 N
Balancing force in horizontal direction: Fh = TCos = 11.26*10^3 *Cos20.9 = 10519.14 N
Net force = (10519.14^2 + 2436.286^2) = 10797.59 N
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