91Ó°ÊÓ

i) The mass of particle A is $m_A=0.006\mathrm{kg}$

ii) The mass of particle B is$m_B=0.0012\mathrm{kg}$

iii) The mass of particle C,$m_C=0.008\mathrm{kg}$$m_C=0.008\mathrm{kg}$

iv) The distance between particle A and B,$r_{AB}=0.5\mathrm{m}$

v) The net force acting on A,role="math" localid="1657262151484" $F_{net}=2.77×{10}^{-14}\mathrm{N}\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{of}-163.8\mathrm{from}+\mathrm{x}\mathrm{ax}\mathrm{is}$

We can find the force exerted on particle A due to the particle C from the net force acting on A and the force exerted on particle A due to the particle B from the x component on the net force acting on A, we can find the x component of position vector which is the x-coordinate of particle C. Then we can easily find its y-coordinate.

Formula:

Gravitational force between particles, $F=\frac{GMm}{r^2}$

Using equation (i), the force acting on particle A due to particle B is

$$\begin{gathered} F_{AB}=\frac{G{m}_A{m}_B}{r_{AB}^2} \\ =\frac{6.67×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\left(6×{10}^{-3}\mathrm{kg}\right)\left(12×{10}^{-3}\mathrm{kg}\right)}{{\left(0.5\mathrm{m}\right)}^2} \\ =1.92×{10}^{-14}\mathrm{N} \end{gathered}$$

The net force acting on particle A due to particle B and C can be written as

$$\begin{gathered} F_{net}=\sqrt{F_{AB}^2+F_{AC}^2} \\ {F}_{AC}^2=F_{net}^2-F_{AB}^2 \\ ={\left(2.77×{10}^{-14}\mathrm{N}\right)}^2-{\left(1.92×{10}^{-14}\mathrm{N}\right)}^2 \\ =3.99×{10}^{-28}\mathrm{N} \end{gathered}$$

Similarly using equation (i), the force acting on particle A due to particle C is

$$\begin{gathered} F_{AC}=\frac{G{m}_A{m}_C}{r_{AC}^2} \\ 2×{10}^{-14}\mathrm{N}=\frac{6.67×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\left(6×{10}^{-3}\mathrm{kg}\right)\left(8×{10}^{-3}\mathrm{kg}\right)}{{\mathrm{r}}_{\mathrm{AC}}^2} \\ {\mathrm{r}}_{\mathrm{AC}}^2=0.16 \\ {\mathrm{r}}_{\mathrm{AC}}^{}=0.4\mathrm{m} \end{gathered}$$

The x-component of net force acting on particle A can be written as

$$\begin{gathered} F_{net,x}=\frac{G{m}_A{m}_B{x}_B}{r_{AB}^3}+\frac{G{m}_A{m}_C{x}_C}{r_{AC}^3} \\ =2.77×{10}^{-14}\cos \left(-163.8°\right)\left(\mathrm{ii}\right) \end{gathered}$$

Where,

$$\begin{gathered} X_B=r_{AB}\cos 150° \\ =-043\mathrm{m} \end{gathered}$$

Hence, substituting all given and derived values in equation (ii), we get

$$\begin{gathered} \frac{6.67×{10}^{-11}\left(6×{10}^{-3}\right)\left(12×{10}^{-3}\right)\left(-0.43\right)}{0.5^3}+\frac{6.67×{10}^{-11}\left(6×{10}^{-3}\right)\left(8×{10}^{-3}\right)x_c}{0.4^3}=-2.66×{10}^{-14} \\ -1.652×{10}^{-14}+\frac{6.67×{10}^{-11}\left(6×{10}^{-3}\right)\left(8×{10}^{-3}\right)x_c}{0.4^3} \end{gathered}$$

$$\begin{gathered} \mathrm{Again}, \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1.652×{10}^{-14}-2.66×{10}^{-14}×0.4^3}{6.67×{10}^{-11}\left(6×{10}^{-3}\right)\left(8×{10}^{-3}\right)} \\ =-0.2\mathrm{m} \end{gathered}$$

Therefore, the x-coordinate of particle c is -0.2 m

$x_c$And $y_c$are components of $r_{AC}$.

So, $r_{AC}$can be written as:

$$\begin{gathered} r_{AC}=\sqrt{x_C^2+y_C^2} \\ {y}_C^2=r_{AC}^2+x_C^2 \\ =0.4^2-{\left(-0.2\right)}^2 \\ {y}_C^{}=Â\pm 0.35\mathrm{m} \end{gathered}$$

Since$r_{AC}$is in third quadrant,

$y_C=-0.35\mathrm{m}$

Therefore, the y-coordinate of particle c is $-0.35\mathrm{m}$