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.The horizontal steel beam had a mass of 81.70 kg per meter of length and the tension in the cable was T = 1.100 × 104 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.000 m, s = 0.630 m, x = 1.450 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.

Answer 1

"m = 81.70kg/m, T = 1.100 times 10^4N, Max load = 45.5 kg d = 5.000m, s = 0.630m, x = 1.450m, h = 1.980m g = 9.810 m/s^2 Sum of moments about P: O = T(d-s) sin theta-W(d-x)-f(d/2) F rightarrow weight of weam 11000(4.370) sin 24.3 degree-W(5-1.450)-81.70 times 5 times 9.8 times 5/2 = 0 tan theta h/d-s = 1.980/5.000-0.630 theta = 24.3 degree 19781.49-3.55W-10008.25 = 0 3.5W = 9773.24 W = 2792.35N rightarrow load Sum of horizontal force F