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W in N. ___________ F in N. ___________ After an unfortunate accident at a local

ID: 1280854 • Letter: W

Question


W in N. ___________ F in N. ___________ 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 93.60 kg per meter of length and the tension in the cable was T= 11910 N. The crane was rated for a maximum load of 454.5 kg. if d = 5.290 m, s= 0.522 m, x= 1.200 m and h = 2,250 m, what was the magnitude of W 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 in N. ___________ F in N. ___________

Explanation / Answer

Sum the moments about P:
0 = T(d - s)sin? - W(d - x) - F(d/2)
? = arctan(h/(d-s)) = arctan(2.250 / (5.29 - 0.522)) = 25.26 degree
where F is the weight of the beam.
0 = 11910N*(5.29 - 0.522)m*sin25.26 - W*(5.29 - 1.20)m - 93.6kg/m*5.29m*9.8m/s*5.29m/2
W = 2786.76 N (load)

(b) sum the vertical forces:
Fv + Tsin? - W - 93.6kg/m*5.29m*9.8m/s = 0
Fv + 11910N*sin25.26 - 2786.76N - 93.6*5.29*9.8N = 0
Fv = 2556.86 N (vertical force at P)

sum the horizontal forces:
Fh - Tcos? = 0
Fh - 11910N*cos25.26 = 0
Fh = 10771.2 N ? horizontal force at P

mag P = ?(10771.2^2 + 2556.86^2) N = 11070.5 N ( total reaction at P)