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 86.80 kg per meter of length and the tension in the cable was T = 12170 N. The crane was rated for a maximum load of 454.5 kg. If d = 6.160 m, s = 0.594 m, x = 1.100 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

W_b = weight of the beam = mass per unit length x g x d = 86.80 x 6.16 x 9.8 = 5240 N

tan = h /(d - s)

tan = 2.16/(5 - 0.522)

= 25.75 deg

Using equilibrium of torque about P

T Sin (d - s) = W_b (d/2) + W_L (d - x)

(11.52 x 10³) Sin25.75 (5 - 0.522) = (5240) (5/2) + W_L (5 - 1.450)

W_L = 4720.68 N

using equilibrium of force in vertical direction

P_y + T Sin = W_b + W_L

P_y + (11.52 x 10³) Sin25.75 = 5240 + 4720.68

P_y = 3293 N

using equilibrium of force in horizontal direction

P_x = T Cos

P_x = (11.52 x 10³) Cos25.75

P_x = 9394.2 N

Net force is given as

P = sqrt(P_y² + P_x²)

P = sqrt((9394.2)² + (3293)²)

P = 9954.64 N