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 90.20 kg per meter of length and the tension in the cable was T = 11.52 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 5.870 m, s = 0.558 m, x = 1.200 m and h = 1.980 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.

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 90.20 kg per meter of length and the tension in the cable was T11.52 kN. The crane was rated for a maximum load of 1.000 x 103 lbs. If d 5.870 m, s 0.558 m, x = 1.200 m and h = 1.980 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/s Number w1786 Number Fp 235019 Incorrect

Explanation / Answer

Solve this problem by taking moments of the forces about point P.

Under the balaced condition, sum the moments about P = 0

Therefore -

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

where F is the weight of the beam.

And -

? = arctan(h/(d-s)) = arctan(1.98 / (5.87 - 0.558)) = arctan(1.98 / 5.312) = 20.4 deg

Put these value in equation (i).

We have -

0 = 11520 N* 5.312 m* sin20.4º - W* (5.87 - 1.20) m - 90.2 kg/m*5.87 m*9.8m/s²*5.87 m/2

=> 0 = 21330 - W * 4.67 - 15229

=> W = 6101 / 4.67 = 1306 N

Hence, the load on the crane before the collapse, W(L) = 1306 N

Again, sum the vertical forces:
Fv + Tsin? - W - 90.20kg/m*5.87m*9.8m/s² = 0

=> Fv + 11520 N*sin20.4 - 1306 N - 5189 N = 0

=> Fv + 4016 - 6495 = 0

=> Fv = 2479 N ? vertical force at P

And, sum the horizontal forces:
Fh - Tcos? = 0
Fh - 11520 N*cos20.4º = 0

=> Fh = 10797 N ? horizontal force at P

Therefore, magnitude of force at the attachment P is -

Fp = ?(10797² + 2479²) N = 11078 N

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