Problem: In the figure: the mass m1 is 0.70 kg and it is located at x1 = 25 cm.The pivot point is represented by the solid triangle located at x = 60 cm. The center of mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.34 kg and it is located at x2 = 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/s2.(b) True/False -- The apparatus described in the previous question is in equilibrium.

🤓 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 = τ₁ + τ_ms + τ₂

= m₁ g r₁ + m_ms g r_ms+ m₂ g r₂

= g( m₁ (0.60 - x₁) + m_ms (0.60 - x_ms)+ m₂ (0.60 - x₂) )

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

In the figure:

the mass m₁ is 0.70 kg and it is located at x₁ = 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 m₂ is 0.34 kg and it is located at x₂ = 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/s².

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