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 = 11910 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.630 m, x = 1.400 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 in N Solution Given that tention in string, T = 11910 N theta = arctan(1.98/(5.58 - 0.63)) theta = 21.8° Also mass of beam = 93.6* 5.58 m = 522.288 Kg Now , balancing moment about point P 11910 * (5.58 - 0.63) * sin(21.8) - 522.288 * 9.8 * 5.58/2 - WL * (5.58 - 1.4) = 0 WL = 1821.389 N ----------------------- Now , at Point P Equilibrium of forces along x axis Px = 11910 * cos(21.8) Px = 11058.266 N Equilibrium of forces along y axis Py + 11910 * sin(21.8) - 1821 - 454.5 * 9.8 = 0 Py = 1852.11 N Net force at P P = sqrt(10980^2 + 1852.11^2) P = 11135.11 N the magnitude of force at P is 11135.11 N .

1. 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 = 11910 N. The crane was rated for a maximum load of 454.5 kg. If d
= 5.580 m, s = 0.630 m, x = 1.400 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 in N
Solution
Given that tention in string, T = 11910 N
theta = arctan(1.98/(5.58 - 0.63))
theta = 21.8°
Also
mass of beam = 93.6* 5.58
m = 522.288 Kg
Now , balancing moment about point P
11910 * (5.58 - 0.63) * sin(21.8) - 522.288 * 9.8 * 5.58/2 - WL * (5.58 - 1.4) = 0
WL = 1821.389 N
-----------------------
Now , at Point P
Equilibrium of forces along x axis

2. Px = 11910 * cos(21.8)
Px = 11058.266 N
Equilibrium of forces along y axis
Py + 11910 * sin(21.8) - 1821 - 454.5 * 9.8 = 0
Py = 1852.11 N
Net force at P
P = sqrt(10980^2 + 1852.11^2)
P = 11135.11 N
the magnitude of force at P is 11135.11 N