Chapter 10, Problem 18PQ

Two metersticks are connected at their ends as shown in Figure P10.18. The center of mass of each individual meterstick is at its midpoint, and the mass of each meterstick is m.

a. Where is the center of mass of the two-stick system as depicted in the figure, with the origin located at the intersection of the sticks?

b. Can the two-stick system be balanced on the end of your finger so that it remains lying flat in front of you in the orientation shown? Why or why not?

FIGURE P10.18

(a) The center of mass of the stick on the x axis would be at (0.5 m, 0), and the center of mass of the stick on the stick on the y axis be at (0, 0.5 m), assuming the sticks are uniform. We can then use Equation 10.3 to find the x and y coordinates of the center of mass.

$\begin{array}{l}{x}_{\text{CM}}=\frac{1}{M}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{m}_j{x}_j=\frac{1}{2m}[m(0.50\text{m})]=0.25\text{m}}\\ y_{\text{CM}}=\frac{1}{M}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{m}_j{y}_j=\frac{1}{2m}[m(0.50\text{m})]=0.25\text{m}}\end{array}$

The location of the center of mass is $\boxed{(0.25\text{m,}0.25\text{m})}$

(b) No. The location of the center of mass is not located on the object, so your finger would not be in contact with the object. In a different orientation, balancing by applying a force at the center of mass might be possible, but not in the orientation shown.