After an unfortunate accident at a local warehouse you have been contracted to determine the cause. A jib crane collapsed and injured a worker. The horizontal steel beam had a mass of 88.50 kg per meter of length and the tension in the cable was T = 11.78 kN. The crane was rated for a maximum load of 1.000 × 103 lbs. If d = 5.290 m, s = 0.594 m, x = 1.450 m and h = 1.890 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.
3 answers
hvrxs
Weight of load: 2228 N
Magnitude of force: 11,191 N
1.
Write out the Parameters (Tension, beam mass, d, s, x, h)
m= beam mass*d= 88.5 * 5.290
=465.5 kg
find theta using arctan(opp/ adj)
arctan (h/ (d-s)) = arctan (1.890/ (5.290- 0.594)) = 21.9 degrees
T(d-s)sin theta - mg (d/2) - WL (d-x) = 0
11780 (5.290- 0.594)sin 21.9 - [465.5 * 9.81 (5.29/2)] - WL (5.290- 1.450) = 0
Solve for WL.
2. Magnitude of force at point P:
get the horizontal and vertical components
Px = T*cos theta
= 11780cos21.9 = 10930 N
Py = T*sin theta - WL - mg
= 11780sin21.9 - 2228 N - (465.5 * 9.81)
=-2401 N
Use pythagorean theorem to get the magnitude
Magnitude of force: 11,191 N
1.
Write out the Parameters (Tension, beam mass, d, s, x, h)
m= beam mass*d= 88.5 * 5.290
=465.5 kg
find theta using arctan(opp/ adj)
arctan (h/ (d-s)) = arctan (1.890/ (5.290- 0.594)) = 21.9 degrees
T(d-s)sin theta - mg (d/2) - WL (d-x) = 0
11780 (5.290- 0.594)sin 21.9 - [465.5 * 9.81 (5.29/2)] - WL (5.290- 1.450) = 0
Solve for WL.
2. Magnitude of force at point P:
get the horizontal and vertical components
Px = T*cos theta
= 11780cos21.9 = 10930 N
Py = T*sin theta - WL - mg
= 11780sin21.9 - 2228 N - (465.5 * 9.81)
=-2401 N
Use pythagorean theorem to get the magnitude
Magnitude of force at point P:
F = √(Px^2 + Py^2)
F = √(10930^2 + (-2401)^2)
F = 11223 N or 11.22 kN (rounded to two decimal places)
F = √(Px^2 + Py^2)
F = √(10930^2 + (-2401)^2)
F = 11223 N or 11.22 kN (rounded to two decimal places)
1 answer