Question

A meter stick on earth made from a uniform-density material rests on a fulcrum positioned precisely at its middle. If a box of mass m = 3 kg is suspended from the right end of the meter stick, the system would not be in equilibrium. Where and how could a force be applied to keep the system in equilibrium? (For ease of calculation, consider g = 10 N/kg.)

Answer 1

Answer:

A force of 75 N placed at 0.7 m on the meter stick.

Explanation:

The weight of the box is equal to:

$W=mg=3*10=30N$

The net torque is equal to zero and is equal to:

$F(x-0.5)-(30*0.5)=0\\F(x-0.5)=15$

For a force with a value of 75 N that is placed at 0.70 m on the meter stick, it would produce a torque of 15 N m

If you replace that values in the equation:

$75(0.7-0.5)=15\\15=15$