Question

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

Answer 1

Answer:C

Explanation:

Given

mass $m_1=400\ gm$ is at $x=20\ cm$ mark

mass $m_2=320\ gm$ is at $x=75\ cm$ mark

Scale is Pivoted at $x=50\ cm mark$

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

Taking ACW as positive thus

$T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)$

$T_{net}=0.12g-0.08g=0.04g$

Therefore a net torque of 0.04 g is required in CW sense which a mass $400\ gm$ can provide at a distance of $x_o$ from pivot

$0.04g=0.4\times g\times x_o$

$x_o=0.1\ m$

therefore in meter stick it is at a distance of $x=60\ cm$