A uniform rod 8m long weighing 8kg is supported horizontally by two vertical parallel strings at P and Q and at distances of 2m and 6m from one end. Weights of 1kg,1.5kg and 2kg are attached at distances of 1m,5m and 7m respectively from the same end. Find the tension in each vertical strings.

Answer 1

total force down = 1 g + 8 g + 1.5 g + 2 g = 12.5 g
total force up = p+q
so p + q = 12.5 g

moments about left end
clockwise = 1*1 g + 8*4 g + 1.5*5 g + 2*7 g = 54.5 g
counterclockwise = 2 p + 6 q
so 2 p + 6 q = 54.5 g
=======================
p + q = 12.5 g
p + 3 q = 27.25 g
-----------------------subtract
-2 q = -14.75 g
q = 7.375 g
p = 5.125 g
use whatever you use for g to get Newtons instead of kilograms 9.8 m/s^2 or 9.81 m/s^2 or 10 m/s^2