Question

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 4.79 g coins stacked over the 28.8 cm mark, the stick is found to balance at the 38.4 cm mark. What is the mass of the meter stick?

Answer 1

To solve the exercise, the key concept to be addressed is the Mass Center.

The center of mass of an object is measured as,

$X_{cm} = \frac{\sum m_ix_i}{\sum m_i}$

$X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

Our values are given by,

$x_1 = 50cm$

$m_1 = ?$

$x_2 =28.8cm$

$m_2 = 4.79g$

$x_3 = 28.8cm$

$m_3 = 4.79g$

$X_{cm} = 38.4cm$

Replacing the values in our previous equation we have,

$X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

$38.4 = \frac{m_1(50)+2(28.8*4.79)}{m_1+2*4.79}$

$38.4(m_1+2*4.79)= m_1(50)+2(28.8*4.79)$

$38.4m_1 +367.872 = 50m_1+275.904$

$11.6m_1 = 91.968$

$m_1 = 7.928g$

Therefore the mass of the meter stick is 7.928g