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 95.30 kg per meter of length and the tension in the cable was T = 12.56 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 6.160 m, s = 0.486 m, x = 1.050 m and h = 1.980 m, what was 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.

Answer 1

Ty = T*h/(d-s) = 12560*1.980/(6.160-0.486) = 4383 N
4383 = (6.160*95.30*9.81/2+ WL*5.11)/(6.160-0.486)
WL = (24,869-2,879.5)/5.110 = 4003.22N = 4.0032 kN ...Ans.

(b) Θ = arctan(h/(d-s)) = arctan(1.980 / (6.160-0.486)=19.23

sum the vertical forces:
Fv + TsinΘ - W - 95.30kg/m*6.160m*9.81m/s² = 0
Fv + 12560N*sin19.23 - 4003N - 5759N = 0
Fv = 2381N ← vertical force at P

sum the horizontal forces:
Fh - TcosΘ = 0
Fh - 12560N*cos19.23º = 0
Fh = 11 859 N ← horizontal force at P

mag P = √(11859² + 2381²) N = 12096 N ← total reaction at P