The Fp is correct, but the Wl is incorrect Map fat Sapling Learning macmillan le

Question

The Fp is correct, but the Wl is incorrect

Map fat Sapling Learning macmillan 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 95.30 kg per meter of length and the tension in the cable was T= 12690 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.522 m, x= 1.250 m and h = 1.980 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/s Number WL 2531 Number Fp 12049.88

Explanation / Answer

Sum the moments about the hinge:
0 = T(d - s)sin - W(d - x) - F(d/2)
where F is the weight of the beam.

= arctan(h/(d-s)) = arctan(1.98 / 5.058) = 21.378º
so
0 = 12690N*5.058m*sin21.378º - W*4.33m – 95.3kg/m*5.58m*9.81m/s²*2.79m
4.33W = 11257.0803 N
W = 2599.78 N
W = 2600 N

(b) sum the vertical forces:
Fv + Tsin - W – 95.3kg/m*5.58m*9.81m/s² = 0
Fv + 12690N*sin21.378 - 2600N – 5216.7N = 0
Fv = 3190.74 N

sum the horizontal forces:
Fh - Tcos = 0
Fh - 12690N*cos21.378º = 0
Fh = 11816.86 N

P = (11816.86² + 3190.74²) N

P = 12240 N

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