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. The horizontal steel beam had a mass of 88.50 kg per meter of length and the tension in the cable was and injured a worker 1 1780 N. The crane was rated for a maximum load of 4545 kg.lf d-5290 m, s x = 2.000 What was the magnitude of force at the attachment point P? The acceleration due to gravity is g = 9.810 m/s2 m and h = 1.890 m, what was the magnitude of m (the load on the crane) before the collapse? Number Number

Explanation / Answer

a)

Sum of moments about P

0=T*(d-s)*sin() -W*(d-x) -F*(d/2) {where F is the weight of the beam}

tan()=h/(d-s) = 1.89 /(5.29 - 0.558)

=21.77 degrees

0=11780*(5.29-0.558)*sin(21.77) - W*(5.29-2) - 88.5*5.29*9.8*(5.29/2)

W=2595.360 N

b)

let the vertical component of force on P be Fv and horizontal component be Fh

Sum of vertical forces,

Fv+ Tsin() - W- 88.5*5.29*9.8 =0

Fv= 2814.39

Sum of horizontal forces,

Fh-Tcos()=0

Fh=10939.85 N

magnitude of force at P=sqrt(2814.392 + 10939.852)

mag P=11296.066 N

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