🤓 Based on our data, we think this question is relevant for Professor Acosta & Hamlin's class at UF.

(a) Calculate net torque about pivot (make sure to convert cm → m)

**τ _{net} = τ_{1} + τ_{ms} + τ_{2}**

**= m _{1} g r_{1} + m_{ms} g r_{ms }+ m_{2} g r_{2}**

**= g( m _{1} (0.60 - x_{1}) + m_{ms} (0.60 - x_{ms})_{ }+ m_{2} (0.60 - x_{2}) )**

(Subtracting each x from pivot means CCW torques are + and CW torques are −)

In the figure:

- the mass
*m*_{1}is 0.70 kg and it is located at*x*_{1}= 25 cm. - The pivot point is represented by the solid triangle located at
*x*= 60 cm. - The center of mass of the meter stick (
*m*_{ms}= 0.40 kg) is located at its geometric center,*x*_{ms}= 50 cm. - The mass
*m*_{2}is 0.34 kg and it is located at*x*_{2}= 85 cm.

(a) Calculate the net torque about the pivot (in N**·**m with the proper sign) due to these three weights. Use *g* = 9.79 m/s^{2}.

(b) True/False -- The apparatus described in the previous question is in equilibrium.