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Jump Sapling Learning After an unfortunate accident at a local warehouse you hav

ID: 2035743 • Letter: J

Question

Jump Sapling 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 steel beam had a mass of 88.50 kg per meter of length and the tension in the cable was T- 11390 N. The crane was rated for a maximum load of 454.5 kg. If d 5.290 m, s 0.450 m, x 1.900 m and h 1.890 m, what was the magnitude of Wi (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? Number

Explanation / Answer

total mass of the beam=88.5*d

center of mass lies at a distance of d/2 from P.

tension in the cable =T=11390 N

let force at P in vertical direction be Py , in upward direction

let force at P in horizontal direction be Px, towards right.

theta=arctan(h/(d-s))=21.33 degrees

balancing moment about P:

T*sin(theta)*(d-s)=W1*(d-x)+88.5*d*9.8*(d/2)

using the given values,

11390*sin(21.33)*(5.29-0.45)=W1*(5.29-1.9)+88.5*5.29*9.8*(5.29/2)

==>W1=2335.441 N

balancing forces in horizontal direction:

Px=T*cos(theta)=10609.763 N

balancing forces in vertical direction:

Py+T*sin(theta)=W1+weight of the beam

==>Py=5549.9+88.5*5.29*9.8-11390*sin(21.33)=2780.4 N

net force magnitude=sqrt(Px^2+Py^2)

=10968 N