HELP! can you answer any of these questions?

Question

a mass of 3 kg is placed at the 90 cm mark on a .5 kg meter stick. the beam is 100 cm long and the pivot is at the 30 cm mark.

1. how much torque is present due to the 3 kg mass?

2. how much torque is present due to the center of gravity of the meter stick?

3. how much torque must be added to balance the beam?

4. how much mass should be placed at the 20 cm mark to balance the beam?

all answers should be in Nm and kg. thank you!!!

Answer 1

T= F * length of the moment arm (L)

F = MA

N = Kg-m/s^2

A= 9.8 m/s^2

#1 I assume you want the torque around the pivot

L= 90 - 30 = 60 cm = 0.6m

F= 3 kg * 9.8 m/s^2

F= 29.4 N

T= 29.4N * 0.6m

T= 17.64 N-m

QED