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- May 10, 2013
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A 0.5-kg uniform meter stick is suspended by a single string at the 30-cm mark. A 0.2-kg mass hangs at the 80 cm mark. What mass hung at the 10-cm mark will produce equilibrium?
A) 0.3 kg
B) 0.5 kg
C) 0.7 kg
D) 1.0 kg
My reasoning was for the meter stick to be at equilibrium the left side must equal the ride side thus: (x)(20cm) = (.2kg)(50cm), solving for x you would get 0.5kg. However, answer choice B is wrong, D is the correct answer. Their explanation states that there is two torques acting on the right side, the .2kg hanging mass and the weight of the meter stick.
How do you figure the weight of the meter stick is acting on the right and not left? Second, they set the torque of the meter stick to (0.5 kg)g(20 cm), "where 20cm represents the distance from the suspension to the center of mass of the meter stick." How did they derive that 20cm was the distance from the center of mass of the meter stick, does that make the meter stick 100cm in length?
A) 0.3 kg
B) 0.5 kg
C) 0.7 kg
D) 1.0 kg
My reasoning was for the meter stick to be at equilibrium the left side must equal the ride side thus: (x)(20cm) = (.2kg)(50cm), solving for x you would get 0.5kg. However, answer choice B is wrong, D is the correct answer. Their explanation states that there is two torques acting on the right side, the .2kg hanging mass and the weight of the meter stick.
How do you figure the weight of the meter stick is acting on the right and not left? Second, they set the torque of the meter stick to (0.5 kg)g(20 cm), "where 20cm represents the distance from the suspension to the center of mass of the meter stick." How did they derive that 20cm was the distance from the center of mass of the meter stick, does that make the meter stick 100cm in length?