Abar of uniform cross section is 86.9 in longand weighs 89.1 lb. a weight of 79.0 lb is suspended from one end. the bar and weight combination is to be suspended from a cable attached at the balance point. how far (in) from the weight should the cable be attached, and what is the tension (lb) in the cable?

y = 20.41 in

T= 168.1 lb

Explanation:

From diagram

Total force balance in vertical direction

T= 89.1 + 79 lb

 T= 168.1 lb

Now taking moment about point m

Mm= 0 Because system is in equilibrium position

 79 x 43.45 = T x y

Now by putting the value of T

79 x 43.45 = 168.1 x y

y = 20.41 in

So the cable attached at distance of 20.41 in from the mid point of bar.

Tension in the cable = 168.1 lb