Question

What do you mean by couple? Write any two properties of a couple. Find the position vector of the center of mass of$1Kg$,$2Kg$,$3Kg$at the points$\left( {3\hat i + 2\hat j} \right)$,$\left( {5\hat j + \hat k} \right)$and $\left( {2\hat i + \hat k} \right)$respectively.

Answer 1

In order to solve this question, we are going to first define the couple of the forces after which we state the properties. After that the three given masses and the points are taken and the center of mass is found using the formula for the coordinates of center of mass.

Formula used: The center of mass.
${C_{CM}} = \dfrac{{{m_1}{c_1} + {m_2}{c_2} + {m_3}{c_3}}}{{{m_1} + {m_2} + {m_3}}}$
Where,${c_1}$,${c_2}$and ${c_3}$are the coordinates of the masses.

Complete step-by-step solution:
The couple is defined as the system of the forces that have a resultant moment and not a resultant force. It mainly affects the system by creating a rotation without translation, or more generally without any acceleration of the center of mass.
Properties of a couple:
As the two properties forming a couple act equal and opposite, they do not produce any resultant translation motion but rotational motion.
When a couple force acts the resultant force is equal to zero.
It produces the rotational motion because the algebraic sum of any two forces about any point in their plane is not zero.
We are given three masses$1Kg$,$2Kg$,$3Kg$at the points $\left( {3\hat i + 2\hat j} \right)$,$\left( {5\hat j + \hat k} \right)$and $\left( {2\hat i + \hat k} \right)$respectively. We need to find the center of mass.
${x_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$
Putting the values, we get
${x_{CM}} = \dfrac{{1 \times 3 + 2 \times 0 + 3 \times 2}}{{1 + 2 + 3}} = \dfrac{9}{6} = \dfrac{3}{2}$
Now,
${y_{CM}} = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}}$
Putting values, we get
${y_{CM}} = \dfrac{{1 \times 2 + 2 \times 5 + 3 \times 0}}{{1 + 2 + 3}} = \dfrac{{13}}{6}$
And the third coordinate
${z_{CM}} = \dfrac{{{m_1}{z_1} + {m_2}{z_2} + {m_3}{z_3}}}{{{m_1} + {m_2} + {m_3}}}$
Putting the values, we get
${z_{CM}} = \dfrac{{1 \times 0 + 2 \times 1 + 3 \times 1}}{{1 + 2 + 3}} = \dfrac{5}{6}$
Therefore, center of mass is$\left( {\dfrac{3}{2},\dfrac{{13}}{6},\dfrac{5}{6}} \right)$

Note: It is important to note that the moment of a couple in its plane about any point is constant, both in magnitude and direction. The center of mass depends on the coordinates of the three or more masses whose center of mass is to be calculated and also the values of the number of the masses as given.