Question 7 of 12 Map sapling learning After an unfortunate accident at a local w

Question

Question 7 of 12 Map sapling learning 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 11520 N. The crane was rated for a maximum load of 454.5 kg. If d 5.290 m, s 0.522 m, x 1.150 m and h 2.250 m, what was the magnitude of W (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 Number W. Number di W.

Explanation / Answer

here,

m = 97 kg

maximum load , lm = 454.5 kg

tension , T = 11520 N

theta = arctan(h/(d-s))

theta = arctan( 2.25 /( 5.29 - 0.522))

theta = 25.26 degree

taking moment of force about point P

t * sin(theta) * (d-s) - Wl * (d - x) - m*g*d/2 = 0

11520 * sin(25.26)* ( 4.768) - Wl *( 4.14) - 97*9.8*5.29/2 = 0

solving for Wl

Wl = 1076.52 N

equating the forces horizontally

Fp = t * cos(theta)

Fp = 11520 * cos(25.26)

Fp = 10418.5 N

Wl = 1076.52 N and Fp = 10418.5 N

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