Question

Three particles of masses $1\;{\rm{kg}}$, $2\;{\rm{kg}}$ and $3\;{\rm{kg}}$ are subjected to forces $\left( {3i - 2j + 2k} \right)\;{\rm{N}}$, $\left( { - i + 2j - k} \right)\;{\rm{N}}$ and $\left( {i + j + k} \right)\;{\rm{N}}$ respectively. Find the magnitude of the acceleration of the CM of the system.

Answer 1

The above problem can be resolved using the concept and fundamentals involved in the analysis of the centre of mass. The problem is given with the system of three particles, such that each particle is experienced with some force. The magnitude of the acceleration of the centre of mass is utilised by taking the ratio of the magnitude of the net force on the system and total mass of the system.

Complete step by step answer:
Given:
The mass of particle 1 is, ${m_1} = 1\;{\rm{kg}}$.
The mass of particle 2 is, ${m_2} = 2\;{\rm{kg}}$.
The mass of particle 3 is, ${m_3} = 3\;{\rm{kg}}$.
The magnitude of force of particle 1 is, ${F_1} = \left( {3i - 2j + 2k} \right)\;{\rm{N}}$.
The magnitude of force on particle 2 is, ${F_2} = \left( { - i + 2j - k} \right)\;{\rm{N}}$.
The magnitude of force on particle 3 is, ${F_3} = \left( {i + j + k} \right)\;{\rm{N}}$.
Then, the expression for the magnitude of acceleration of the centre of mass (CM) of the system
${a_{CM}} = \dfrac{{{F_{net}}}}{M}$……………………………………….. (1)
Here, ${F_{net}}$ is the net force on the system and its value is,
${F_{net}} = {F_1} + {F_2} + {F_3}$ ………………………………... (2)
And M is the total mass of the system and its value is,
$M = {m_1} + {m_2} + {m_3}$……………………………… (3)
Solve by substituting the values of equation 2 and 3 in equation 1 as,

{a_{CM}} = \dfrac{{{F_{net}}}}{M}\\\ \Rightarrow {a_{CM}} = \dfrac{{{F_1} + {F_2} + {F_3}}}{{{m_1} + {m_2} + {m_3}}}\\\ \Rightarrow {a_{CM}} = \dfrac{{\left( {3i - 2j + 2k} \right)\;{\rm{N}} + \left( { - i + 2j - k} \right)\;{\rm{N}} + \left( {i + j + k} \right)\;{\rm{N}}}}{{1\;{\rm{kg}} + 2\;{\rm{kg}} + 3\;{\rm{kg}}}}\\\ \Rightarrow {a_{CM}} = \dfrac{{\left( {3i + j + 2k} \right)\;{\rm{N}}}}{{6\;{\rm{kg}}}} \end{array}$$ The above result shows the net force in vector form. Now taking out the magnitude of net force as, $$\begin{array}{l} \Rightarrow \left| {{F_{net}}} \right| = \sqrt {{3^2} + {1^2} + {2^2}} \\\ \Rightarrow \left| {{F_{net}}} \right| = \sqrt {14} \;{\rm{N}} \end{array}$$ Now substituting the value as, $$\begin{array}{l} \Rightarrow {a_{CM}} = \dfrac{{\left| {{F_{net}}} \right|}}{{6\;{\rm{kg}}}}\\\ \Rightarrow {a_{CM}} = \dfrac{{\sqrt {14} }}{6}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}$$ Therefore, the magnitude of the acceleration of the CM of the system is $$ = \dfrac{{\sqrt {14} }}{6}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$. **Note:** Try to understand the problem involved in the analysis of the centre of mass. The centre of mass is that point where all the weight is supposed to be a focus on. It is of vital significance to learn the fundamentals of centre of mass along with the acceleration of the centre of mass.