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 88.50 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 = 5.000 m, s = 0.522 m, x = 1.550 m and h = 2.250 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

mass m = 88.50 kg/m of length = 88.50*5 = 442.5 kg

from the figure

tan(theta) = h/d-s = 2.25/5-0.522 = 0.5024

theta = tan-(0.5014) = 26.67 deg

sum of moments about point P is 0 (system is in equilibrium)

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

F = weight of beam = m*g

0 = (11520*(5-0.522)*sin(26.67)) - (W*(5-1.55)) - (442.5*9.8*(5/2))

3.45*W = 12313.43

W = 3549.11 N

(b)

sum the vertical forces is 0 (system is in equilibrium)

so, Fv + T*sin(theta) - W - F*d = 0

Fv + 11520*(sin26.67) - 3549.11 - (442.5*9.8*5) = 0

Fv = 30422.375 N

sum of horizontal forces is 0 (system is in equilibrium)

so, Fh - T*cos(theta) = 0

Fh = 11520*cos(26.67) = 10294.347

magnitude of force at point P

Fp = sqrt(Fv^2+Fh^2)

Fp = sqrt [(30422.375)^2 + (10294.347)^2]

Fp = 32116.88 N

answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.