236 Computation of the resultant of parallel forces acting on the lever

$R = \Sigma F$

$R = 30 + 60 - 20 + 40$

$R = 110 \, \text{ lb}$       downward
 

 

$M_A = \Sigma xF$

$M_A = 2(30) + 5(60) - 7(20) + 11(40)$

$M_A = 660 \, \text{ ft}\cdot\text{lb}$     clockwise
 

$Rd = M_A$

$110d = 660$

$d = 6 \, \text{ ft}$       to the right of A
 

Thus, R = 110 lb downward at 6 ft to the right of A.           answer