Answer to Question #189502 in Classical Mechanics for Jethro

Question

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.

Answer 1

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

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

$Nb-\text{rope reaction force}$

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

$\text{for hinge A}$

$\Sigma M_a(\vec{F_i})=0;$

$N_b*AB*\cos30\degree-F_b*AB*sin30\degree-F_c*AC*\sin30\degree=0$

$N_b= \frac{F_b*AB*sin30\degree+F_c*AC*\sin30\degree}{AB*\cos30\degree}=$

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

$\text{Aswer: } 2662\ N$