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 81.70 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 = 5.870 m, s = 0.486 m, x = 1.100 m and h = 2.160 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 = N F_P = N

Explanation / Answer

from figure

tantheta = h/((d-s)

tantheta = 2.16/(5.87-0.486)

theta = 21.86 degrees

net torque about the point p = 0

T*sintheta*(d-s) - Mbeam*g*d/2 - WL*(d-x) = 0

(11130*sin21.86*(5.87-0.486))-(81.7*9.810*5.87/2) - (WL*(5.87-1.1)) = 0

WL = 4184.43 N <<<---------answer

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along horizontal

-T*costheta + Fx = 0

Fx = T*costheta = 11130*cos21.86 = 10329.71 N

along vertical

Fy - WL - Mbeam*g = 0

Fy - 4184.43 - (81.7*9.81) = 0

Fy = 4986 N

Fp = sqrt(Fx^2+Fy^2)

Fp = sqrt(10329.71^2+4986^2) = 11470.1 N <<<<-------answer

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