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 90.20 kg per meter of length and the tension in the cable was T = 11.26 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 5.870 m, s = 0.630 m, x = 1.600 m and h = 2.160 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.

Explanation / Answer

Sum the moments about P:
0 = T(d - s)sin - W(d - x) - F(d/2)
= arctan(h/(d-s)) = arctan(2.16 / 5.094) =
where F is the weight of the beam.
0 = 11260N*5.094m*sin23º - W*3.98m - 90.20kg/m*5.870m*9.8m/s²*5.870m/2
3.98W = 8046.79 N
W = 2021 N load

(b) sum the vertical forces:
Fv + Tsin - W - 90.20kg/m*5.870m*9.8m/s² = 0
Fv + 11260N*sin23 - 2021N - 5188.84N = 0
Fv = 2810.27 N vertical force at P

sum the horizontal forces:
Fh - Tcos = 0
Fh - 11260N*cos23º = 0
Fh = 10364.8 N horizontal force at P

mag P = (10364.8² + 2810.27²) N = 10739.02 N total reaction at P

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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