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 88.50 kg per meter of length and the tension in the cable was T = 11130 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.000 m, s = 0.522 m, x = 1.800 m and h = 2.070 m, what was the magnitude of W_L (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^2. W_L = F_P =

Explanation / Answer

Sum the moments about P:
0 = T(d - s)sin - W(d - x) - F(d/2)
= arctan(h/(d-s)) = arctan(2.070 / 4.478) = 24.81 degree
where F is the weight of the beam.
0 = 11130N*4.478m*sin24.81º - W*3.2m - 88.50kg/m*5.00m*9.8m/s²*5.00m/2

W = 3147.56 N load

(b) sum the vertical forces:
Fv + Tsin - W - 88.50kg/m*5.0m*9.8m/s² = 0
Fv + 11130N*sin24.81 - 3147.56N - 4336.5N = 0
Fv = 2813.79 N vertical force at P

sum the horizontal forces:
Fh - Tcos = 0
Fh - 11130N*cos24.81º = 0
Fh = 10102.75 N horizontal force at P

mag P = (10102.75² + 2813.79²) N = 10487.275 N total reaction at P

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