Question #189502

A crane boom 10 m long, weighing 870 N and having its center of gravity 3m from the bottom, is hinged at the lower end and makes an angle of 30° with the vertical. It is held in position by a horizontal cable fastened at the upper end. The boom supports a 4350-N load at its upper end. Find the tension in the cable.


1
Expert's answer
2021-05-09T15:33:29-0400

let A lower point of the crane boom\text {let A lower point of the crane boom}

B the upper point of the crane boom;Fb=4350\text{B the upper point of the crane boom};F_b= 4350

Nbrope reaction forceNb-\text{rope reaction force}

C point of the center of gravity of the crane boom;Fc=870\text{C point of the center of gravity of the crane boom};F_c=870

for hinge A\text{for hinge A}

ΣMa(Fi)=0;\Sigma M_a(\vec{F_i})=0;

NbABcos30°FbABsin30°FcACsin30°=0N_b*AB*\cos30\degree-F_b*AB*sin30\degree-F_c*AC*\sin30\degree=0


Nb=FbABsin30°+FcACsin30°ABcos30°=N_b= \frac{F_b*AB*sin30\degree+F_c*AC*\sin30\degree}{AB*\cos30\degree}=


=4350100.5+87030,5100.8672662=\frac{4350*10*0.5 +870*3*0,5}{10*0.867}\approx 2662


Aswer: 2662 N\text{Aswer: } 2662\ N





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