Suppose you have a meter stick with two masses attached. The mass m1 is 0.60 kg and it is located at x1 = 30 cm from the left (zero) end of the meter stick. The pivot point is located at x = 45 cm. The mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.25 kg and it is located at x2 = 80 cm from the left end of the meter stick. Calculate the net torque (in N⋅m with the proper sign) due to these three weights. Use g = 9.8 m/s2.

Mass m1 exerts a force m1*g at a distance of 15 cm (0.15 m) left of pivot point. That creates a counterclockwise torque of 0.6*9.8*0.15 N-m. The mass of the meter stick exerts a gravitational force ms*g through a piint 5 cm to the right of the pivot, creating a clockwise torque. Mass m2 exerts a force m2*g that is applied 35 cm to the right of the pivot.

All you have to do is compute the three torques and add them. Treat the counterclockwise torque as negative.

If you need more assistance with this one, please show your work.

To calculate the net torque, we need the formula:

Net Torque = counterclockwise torque - clockwise torque

First, let's calculate the torques contributed by each mass:
1. Torque due to mass m1:
The force exerted by mass m1 is given by:
Force = m1 * g = 0.60 kg * 9.8 m/s^2 = 5.88 N
The distance from the pivot point to the force applied by mass m1 is:
Distance = x1 - x = 30 cm - 45 cm = -15 cm = -0.15 m (negative because it is to the left of the pivot point)
The torque created by mass m1 is:
Torque_m1 = Force * Distance = 5.88 N * (-0.15 m) = -0.882 N⋅m (negative because it is counterclockwise)

2. Torque due to the mass of the meter stick (mms):
The force exerted by the mass of the meter stick is given by:
Force = mms * g = 0.40 kg * 9.8 m/s^2 = 3.92 N
The distance from the pivot point to the force applied by the mass of the meter stick is:
Distance = xms - x = 50 cm - 45 cm = 5 cm = 0.05 m
The torque created by the mass of the meter stick is:
Torque_mms = Force * Distance = 3.92 N * 0.05 m = 0.196 N⋅m

3. Torque due to mass m2:
The force exerted by mass m2 is given by:
Force = m2 * g = 0.25 kg * 9.8 m/s^2 = 2.45 N
The distance from the pivot point to the force applied by mass m2 is:
Distance = x2 - x = 80 cm - 45 cm = 35 cm = 0.35 m (positive because it is to the right of the pivot point)
The torque created by mass m2 is:
Torque_m2 = Force * Distance = 2.45 N * 0.35 m = 0.8575 N⋅m

Now, we can calculate the net torque:
Net Torque = Torque_m1 - Torque_mms + Torque_m2
Net Torque = (-0.882 N⋅m) - (0.196 N⋅m) + (0.8575 N⋅m)
Net Torque = -1.0785 N⋅m

Therefore, the net torque due to these three weights is -1.0785 N⋅m (counterclockwise direction).

Please note that the final answer has the proper sign because counterclockwise torques are considered negative.