Map Sapling Learning macmillan learning After an unfortunate accident at a local

Question

Map Sapling Learning macmillan learning 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 80.00 kg per meter of length and the tension in the cable was x 1.550 m and h 2.160 m, what was the magnitude of (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 Number Number

Explanation / Answer

Given

tension T = 11780 N

length d = 5.870 m

mass of the beam m = (80.00 kg/m)(5.870 m) = 469.6 kg

distance s =0.594 m

x = 1.550 m

h =2.160 m

acceleration due to gravity g = 9.810 m/s^2

from the figure

tan (thita) = (h/(d - s))

this implies

thita = tan^-1 (h/(d - s))

= tan^-1 (2.160/(5.870 - 0.594))

thita = 22.26 degree

using equilibrium condation of torque at point p

T*sin(thita)*(d - s) - WL(d - x) - (m*g)*(d/2) = 0

WL = 11780*sin(22.26)*(5.870 - 0.594) - (469.6*9.81)*(5.870/2)/(5.870 - 1.550)

WL = 2324.05 N

the horizontal component of force at p is

Px = T*cos(thita)

= (11780)*cos(22.26)

= 10902.20 N

using newtons second law

Py = WL + (M*g) - T*sin(thita)

= 2324.05 + (469.6*9.81) - 11780*sin(22.26)

=2468.44 N

Thus the force is P

P = sqroot(Px^2 + Py^2)

= sqroot (10902.20^2 + 2468.44^2)

= 11178 N

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