Question

Three particles of masses 8 kg, 4 kg and 4 kg situated at (4,1),(−2,2) and (1,−3)m are acted upon by external forces 6ˆj,−6ˆi and 14ˆiN respectively . The acceleration of the center of Mass of the system is​

Answer 1

Answer:

The acceleration of the center of mass of the system is $\dfrac{6j+8i}{16}$.

Explanation:

Given that,

Force is

$F_{1}=6j\ N$

$F_{2}=-6i\ N$

$F_{3}=+14i\ N$

Masses of three particles

$m_{1}=8\ kg$

$m_{2}=4\ kg$

$m_{3}=4\ kg$

Situated points are

$x_{1},y_{1}=(4,1)$

$x_{2},y_{2}=(-2,2)$

$x_{3},y_{3}=(1,-3)$

We need to calculate the acceleration of the center of mass of the system

Using formula of center of mass

$\vec{a_{cm}}=\dfrac{m_{1}a_{1}+m_{2}a_{2}+m_{3}a_{3}}{m_{1}+m_{2}+m_{3}}$

Put the value into the formula

$\vec{a_{cm}}=\dfrac{8\times\dfrac{6j}{8}+4\times\dfrac{-6i}{4}+4\times\dfrac{14i}{4}}{8+4+4}$

$\vec{a_{cm}}=\dfrac{6j-6i+14i}{16}$

$\vec{a_{cm}}=\dfrac{6j+8i}{16}$

Hence,  The acceleration of the center of mass of the system is $\dfrac{6j+8i}{16}$.

Answer 2

Answer: 0.625

Explanation:

36+64÷256=0.625