an unfortunate accident at a local warehouse you have been contracted to determi
ID: 1506104 • Letter: A
Question
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 95.30 kg per meter of length and the tension in the cable was T = 12.69 kN. The crane was rated for a maximum load of 1.000 times 10^3 lbs. If d = 6.160 m, s = 0.594 m, x = 1.800 m and h = 1.890 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^2.Explanation / Answer
Sum the moments about P:
0 = T(d - s)sin - W(d - x) - F(d/2)
= arctan(h/(d-s)) = arctan(1.890 / 5.566) = 18.75
where F is the weight of the beam.
0 = 12690N*5.566m*sin18.75º - W*4.36m - 95.3kg/m*6.160m*9.8m/s²*6.160m/2
W*4.36 = 12734N
W = 2990 N load
(b) sum the vertical forces:
Fv + Tsin - W - 95.3kg/m*6.160m*9.8m/s² = 0
Fv + 12690N*sin18.75 - 2990N - 5754N = 0
Fv =4655 N vertical force at P
sum the horizontal forces:
Fh - Tcos = 0
Fh - 12690N*cos18.75º = 0
Fh = 12017 N horizontal force at P
mag P = (12017² +4655²) N = 12888 N total reaction at P
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