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 80.00 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 = 0.486 m, x = 1.950 m and h = 2.160 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.

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 = 1 19 10 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.486 m, x = 1.950 m and h = 2.160 m, what was the magnitude of m (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 Number Number

Explanation / Answer

tan@ = h / ( d - s)   = 2.160 / ( 5.580 - 0.486) =0.424

@ = 23 deg

balancing torque about point P ,

Net torque = 0

(d/2 x (80d g) ) + ( (d -x) x WL ) - ( (d-s) x Tsin@ ) = 0

(5.580/2 x (80 x 5.580 x 9.81)) + ( (5.580 - 1.950) x WL) - ( (5.580 - 0.486) x 11910 sin23) = 0

12217.92 + 3.63WL - 60669.54 = 0

WL = 13347.55 N .......Ans

balancing forces on beam in Horizontal direction,

Px - Tcos23 = 0

Px = 11910cos23 =10963.21 N

in vertical,

Py + 11910sin23 - WL - (80 x 5.58 x 9.81) = 0

Py   = - 4653.61 + 13347.55 + 4379.18

Py = 13073.12 N

magnitude of P = sqrt(Px^2 + Py^2) = 17061.61 N ......Ans

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