Chapter 7, Problem 46P

To determine

The location of centre of mass of three-mass system

Answer to Problem 46P

0.4375 m.

Explanation of Solution

Given:

Mass at left hand $=m_1=1\text{kg}$

Distance of centre of mass of mass at left hand $=r_1=0\text{m}$

Mass at middle $=m_2=1.50\text{kg}$

Distance of centre of mass of mass at middle $=r_2=0.50\text{m}$

Mass at right hand $=m_3=1.1\text{kg}$

Distance of centre of mass of mass at right $=r_3=0.50+0.25=0.75\text{m}$

Distance of centre of mass of the three-mass system from left $=r_{cm}$

Formula Used:

Centre of mass of system of masses is given as

  $\begin{array}{l}{r}_{cm}=\frac{m_1{r}_1+m_2{r}_2+m_3{r}_3+\mathrm{..}}{m_1+m_2+m_3+\mathrm{..}}\\ \text{Where }{m}_1,m_2\text{,}{m}_3\text{ are the masses and }{r}_1,r_2\text{,}{r}_3\text{ are their respective centre of masses}\text{.}\end{array}$

Calculation:

Centre of mass of three-mass system is given as

  $r_{cm}=\frac{m_1{r}_1+m_2{r}_2+m_3{r}_3}{m_1+m_2+m_3}$

Substituting the values

  $\begin{array}{c}{r}_{cm}=\frac{(1)(0)+(1.5)(0.5)+(1.1)(0.75)}{1+1.5+1.1}\\ r_{cm}=0.4375\text{m}\end{array}$

Conclusion:

Therefore, the centre of mass of system is 0.4375 m.