Chapter 9, Problem 8Q

A uniform meter stick supported at the 25-cm mark is in equilibrium when a 1 -kg rock is suspended at the 0-cm end (as shown in Fig. 9-37).Is the mass of the meter stick greater than, equal to, or less than the mass of the rock? Explain your reasoning,

To determine

Is the mass of the meter stick greater than, equal to or less than the mass of the rock?

Answer to Problem 8Q

Solution:

The mass of the meter stick ( $m_{stick}$ )is less than the mass of the rock ( $m_{rock}$ ).

Explanation of Solution

Given info:

The mass of the rock ( $m_{rock}$ )is 1 kg. The lever arm of the rock ( $r_{rock}$ )is 0.25 m. The lever arm of center of mass of the meter stick ( $r_{stick}$ )is 0.50 m.

Formula used:

A system is in equilibrium when it satisfies:

${\displaystyle \sum \overrightarrow{F}}=0$                 (1)

And

${\displaystyle \sum \overrightarrow{\tau }=0}$                 (2)

Calculation:

Applying the equation (2)we have:

$-{\tau }_{rock}+{\tau }_{stick}=0$                 (3)

Or

$-m_{rock}.g.r_{rock}+m_{stick}.g.r_{stick}=0$                 (4)

Isolating ( $m_{rock}$ )from equation (4)we obtain:

$m_{rock}=\frac{r_{stick}}{r_{rock}}{m}_{stick}$                 (5)

Inserting data given info into equation (5)we have:

$m_{rock}=\frac{0.50}{0.25}{m}_{stick}$                 (6)

Solving we obtain:

$m_{rock}=2m_{stick}$                 (7)