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.
FREE Expert Solution
(a) Calculate net torque about pivot (make sure to convert cm → m)
τnet = τ1 + τms + τ2
= m1 g r1 + mms g rms + m2 g r2
= g( m1 (0.60 - x1) + mms (0.60 - xms) + m2 (0.60 - x2) )
(Subtracting each x from pivot means CCW torques are + and CW torques are −)
Problem Details
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.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Center of Mass & Simple Balance concept. You can view video lessons to learn Center of Mass & Simple Balance. Or if you need more Center of Mass & Simple Balance practice, you can also practice Center of Mass & Simple Balance practice problems.
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Based on our data, we think this problem is relevant for Professor Acosta & Hamlin's class at UF.