May have made an arithmetic error. Please help! After an unfortunate accident at
ID: 1504809 • Letter: M
Question
May have made an arithmetic error. Please help!
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 97.00 kg per meter of length and the tension in the cable was T = 11780 N. The crane was rated for a maximum load of 454.5 kg. If d = 5.580 m, s = 0.522 m, x = 1.800 m and h = 1.980 m, what was the magnitude of W_L (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/s^2. W_L = F_P =Explanation / Answer
here,
mass of beam , mb = 97 kg
T = 11780 N
d = 5.58 m
s = 0.522 m
x = 1.8 m
h = 1.98 m
theta = arctan( h/(d-s))
theta = 21.38 degree
taking moment of force about the point P
( m * g * d) * d/2 + wl * ( d - x ) - T * sin(theta) * ( d - s) = 0
( 97 * 9.8 * 5.58) * 5.58/2 + wl * ( 5.58 - 1.8) - 11780 * sin(21.38) * 5.058 = 0
wl = 1831.23 N
equating the forces horizontally
Fp - T * cos(theta) = 0
Fp - 11780 * cos(28.31) = 0
Fp = 10371.05 N