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You have been hired to determine the cause of an accident at a factory warehouse

ID: 1471127 • Letter: Y

Question

You have been hired to determine the cause of an accident at a factory warehouse. A crane collapsed and injured a worker. An image of the particular 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 = 1.100 Times 10^4 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.486 m, x = 1.350 m and h = 2.250 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.

Explanation / Answer

A)

from figure, theta = tan^-1(h/(d-s))

= tan^-1(2.25/(5.58 - 0.486))

= 23.83 degrees

As the beam is in equilibrium, net force and net torque acting on th bem must be zero.

Apply net torque on beam about point P = 0

Tmax*(d-s)*sin(23.83) - 97*d*d/2 - WL*(d-x) = 0

11000*(5.58-0.486)*sin(23.83) - 97*5.58*5.58/2 - WL*(5.58 - 1.35) = 0

21129 - 4.23*WL = 0

==> Wl = 21129/4.23

= 4995 N

B) Let Fx and Fy are force exerted by wall on th beam.

Apply, Fnetx = 0

Fx - T*cos(23.83) = 0

Fx = T*cos(23.83)

= 11000*cos(23.83)

= 10062 N

Apply, Fnety = 0

Fy + T*sin(23.83) - 97*5.58*9.8 - 4995 = 0

Fy = 97*5.58*9.8 + 4995 - T*sin(23.83)

= 97*5.58*9.8 + 4995 - 4995*sin(23.83)

= 8281 N


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

= sqrt(10062^2 + 8281^2)

= 13031 N

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