Question

A meter stick is suspended by two vertical cables and is parrallel to the ground. Cable 1 has a tension of 332N and is attached at the 9.00cm mark on the meter stick. A 60.0 Kg block hanging vertically from the meter stick at the 54.0cm mark is in static equillibrium, ignore the mass of the meter stick.

A. What is the tension on cable 2?

B. At what point on the meter stick is cable two attached?

Answer 1

here we need to apply force balance for first part and torque balance for second part

net force in vertical direction = 0

T1 +T2 -60*9.8 = 0

T2 = 60*9.8 - 332 = 256 N

torque balance about center of meter stick which is 50 cm away from both end

total clockwise (CW) torque = total ACW torque

332 * 0.09 + 60*9.8*(0.54-0.50) = T2* r = 256 *r

from here   

so mark on scale = 0.50+0.208594 = 0.708594 or 70.8 cm

let me know in a comment if there is any problem or doubts