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 97.00 kg per meter of length and the tension in the cable was T = 11260 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.558 m, x = 1.850 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.

Explanation / Answer

s = 0.558 m , x = 1.85 m , h = 2.16 m

(a) Net torque about P is zero

Torque = rxF

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

theta = arctan(h/(d-s)) = arctan(2.16/(5.58 -0.558))

theta = 23.3 degrees

0 = (11260*(5.58 -0.558)*sin(23.3)) - W*(5.58-1.85) - 97*5.58*9.8*5.58/2

WL = 2029 N

(b) net force along vertical direction is zero

Fv +Tsin(theta) -W - F =0

FV +11260*sin(23.3) - 2029 - 97*5.58*9.8 =0

vertical force Fv = 2880 N at P

net force along horizontal direction is zero

Fh - Tcos(theta) = 0

Fh - 11260*cos(23.3) =0

Fh = 10341.7 N

P = (10341.7^2 +2880^2)^0.5

P = 10735 N

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