91Ó°ÊÓ

The given data can be listed below as,

• In the figure E13.6 (a), the distance between the mass A and C is${\mathrm{r}}_{\mathrm{AC}1}=10\mathrm{cm}$.
• In the figure E13.6 (a),the distance between the mass C and B is${\mathrm{r}}_{\mathrm{CB}1}=40\mathrm{cm}$.
• In the figure E13.6 (a),the distance between the mass A and B is${\mathrm{r}}_{\mathrm{AB}1}=50\mathrm{cm}$.
• In the figure E13.6 (b), the distance between the mass C and A is${\mathrm{r}}_{\mathrm{AC}2}=10\mathrm{cm}$.
• In the figure E13.6 (b),the distance between the mass A and B is${\mathrm{r}}_{\mathrm{AB}2}=40\mathrm{cm}$.
• In the figure E13.6 (b),the distance between the mass C and B is${\mathrm{r}}_{\mathrm{CB}2}=50\mathrm{cm}$.
• All the masses are in both the figures of,$\mathrm{m}={\mathrm{m}}_1={\mathrm{m}}_2=2\mathrm{kg}$.

The law describes that any matter in the world attracts a different particle with the help of a force which varies directly with the masses’ product and it is also inversely proportional with their distances’ square.

The ratio of the product of masses and the square root of their distances multiplying with the gravitational force constant provide the magnitude and the direction of the net gravitational force.

From Newton’s law of gravity, the force required to find the net gravitational force on a mass can be expressed as,

$F=G\frac{m_1{m}_2}{r^2}$

Here, $m_1$is the mass of Aand is the mass of B and Cas both the masses are acting from the right of A. Moreover, r is distance between the masses A, B and C. The value of G which is the gravitational constant is $6.67×{10}^{-11}N.m^2$. The direction of the mass A is positive as both the forces are acting from the right of the mass A.

The net force equation will be given as,

$$\begin{gathered} ∑F=F_c-F_B \\ ∑F=Gm\left(\frac{m_1}{r_{AC}}-\frac{m_2}{r_{AB}}\right) \\ =\frac{Gm{m}_1}{r_{AC}}+\frac{Gm{m}_2}{r_{AB}} \end{gathered}$$

Here $F_c$is the force working due to mass C and $F_B$ is the force working due to mass B. m is the mass of C and $m_{1}and{m}_2$are the mass of A and B respectively.

Substitute the value in the equation.

$$\begin{gathered} ∑\mathrm{F}=\left(6.67×{10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)×\left(2.0\mathrm{kg}\right)×\left(\frac{2.0\mathrm{kg}}{{\left(0.10\mathrm{m}\right)}^2}-\frac{2.0\mathrm{kg}}{{\left(0.50\mathrm{m}\right)}^2}\right) \\ ∑\mathrm{F}=6.67×{10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2×375{\mathrm{kg}}^2/\mathrm{m} \\ ∑\mathrm{F}=2.5×{10}^{-8}{\mathrm{Nm}}^2/{\mathrm{kg}}^2/{\mathrm{kg}}^2/{\mathrm{m}}^2 \\ ∑\mathrm{F}=2.5×{10}^{-8}\mathrm{N} \end{gathered}$$

Thus, the magnitude of the net gravitational force on mass A due to masses B and C is $2.8×{10}^{-8}\mathrm{N}$and it is in the positive direction.

In the question diagram (b), $m_1$is the mass of Aand is the mass of B and Cas both the masses are acting from the right and also the left of A. The direction to the mass A is negative as the higher force is acting from the left of the mass.

Thenet force equation will be given as,

$$\begin{gathered} ∑F=F_B-F_C \\ ∑F=Gm\left(\frac{m_1}{r_{AB}}-\frac{m_2}{r_{AC}}\right) \\ ∑F=\frac{Gm{m}_1}{r_{AB}}-\frac{Gm{m}_2}{r_{AC}} \end{gathered}$$

where$F_c$is the force working due to mass C and$F_B$is the force working due to mass B. m is the mass of C and$m_{1}and{m}_2$are the mass of A and B respectively.

The value of G is$6.67×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2$,thus substituting the value in the equation, we get-

$$\begin{gathered} ∑F=\left(6.67×{10}^{-11}N.m^2/k{g}^2\right)×\left(2.0kg\right)×\left(\frac{2.0kg}{{\left(0.40m\right)}^2}-\frac{2.0kg}{{\left(0.10m\right)}^2}\right) \\ ∑F=6.67×{10}^{-11}N.m^2/k{g}^2×375k{g}^2/m^2 \\ =2.5×{10}^{-8}N.m^2/k{g}^2/k{g}^2/m^2 \\ =2.5×{10}^{-8}N \end{gathered}$$

Thus, the magnitude of the net gravitational force on mass A due to masses B and C is ${\text{2.5×10}}^{-8}\mathrm{N}$and it is in negative direction.