A meter stick is supported at the 50-cm mark. A 0.40 kg mass is hanging at the 20 cm mark and a 0.60 kg hanging at the 80-cm mark. Where should a third mass that weighs 0.30 kg be hung to keep the stick balanced?

To solve this problem, we need to understand that for the meter stick to be balanced, the total torque on the meter stick must be zero. Torque (τ) is calculated by the equation τ = r * F, where r is the distance from the pivot point and F is the force applied.

The force due to gravity on an object is calculated by F = m * g, where m is the mass of the object and g is the acceleration due to gravity. We'll use g = 9.8 m/s² for this problem.

First, let's calculate the torques due to the 0.40 kg and 0.60 kg masses:

Torque due to 0.40 kg mass = r * F = (50 cm - 20 cm) * (0.40 kg * 9.8 m/s²) = 30 cm * 3.92 N = 117.6 N*cm.

Torque due to 0.60 kg mass = r * F = (80 cm - 50 cm) * (0.60 kg * 9.8 m/s²) = 30 cm * 5.88 N = 176.4 N*cm.

The total torque due to these two masses is 117.6 Ncm - 176.4 Ncm = -58.8 Ncm. The negative sign indicates that the torque is in the opposite direction, so to balance the meter stick, we need to add a torque of 58.8 Ncm in the positive direction.

Now, let's find where to place the 0.30 kg mass. We'll call the distance from the 50 cm mark to where we place the 0.30 kg mass r.

Setting up the equation for the torque due to the 0.30 kg mass gives us:

r * (0.30 kg * 9.8 m/s²) = 58.8 N*cm.

Solving for r gives us:

r = 58.8 N*cm / (0.30 kg * 9.8 m/s²) = 20 cm.

So, the 0.30 kg mass should be placed 20 cm from the 50 cm mark. Since we need to add positive torque, we place it on the side of the 0.60 kg mass, which is at the 80 cm mark. Therefore, the 0.30 kg mass should be placed at the 70 cm mark.