A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the stick and a 320.0 gram object is hung at the 75.0 cm mark. Drawing is approximate. The meter stick is unbalanced. Determine the cm-mark on the meterstick that a 400 gram object needs to be hung to achieve equilibrium. A) 10.0 B) 40.0 C) 60.0 D) 90.0 E) none of the above is within 10% of my answer

C

Explanation:

Given

mass is at mark

mass is at mark

Scale is Pivoted at

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

Therefore a net torque of 0.04 g is required in CW sense which a mass can provide at a distance of from pivot

therefore in meter stick it is at a distance of

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