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 = 11.26 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 5.870 m, s = 0.594 m, x = 1.900 m and h = 2.160 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

a) Sum the moments about P:

0 = T(d - s)sin theta - W(d - x) - F(d/2)

theta = arctan(h/(d-s))

= arctan(2.16 /(5.870 - 0.594)) = 22.26 degree

where F is the weight of the beam.

0 = 11260 N*(5.870 - 0.594)m *sin 22.26 degree - W*(5.870 - 1.9)m - 93.60kg/m*5.87m*9.8m/s^2*5.87m/2

W = 22504.26 - 15803.31)/3.97

= 1688 N. load

(b) sum the vertical forces:

Fv + Tsin theta - W - 93.60 kg/m*5.87 m*9.8m/s^2 = 0

Fv + 11260N*sin22.26 degree - 1688 N - 5384 N = 0

Fv = 2806 vertical force at P

sum the horizontal forces:

Fh - Tcostheta = 0

Fh - 11260N*cos 22.26 degree= 0

Fh = 10421 N horizontal force at P

mag P = sqrt(10421^2 + 2806^2) N = 10792 N total reaction at P

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