Question 336: Gravitation

Question

A system consists of three particles each of mass '$$m_1$$' placed at the corners of an equilateral triangle of side '$$\frac{\mathrm{L}}{3}$$', A particle of mass '$$\mathrm{m}_2$$' is placed at the mid point of any one side of the triangle. Due to the system of particles, the force acting on $$\mathrm{m}_2$$ is

Accepted Answer

From the fig., we can see that the forces due to masses at $$\mathrm{Q}$$ and $$\mathrm{R}$$ cancel each other as they are equal and opposite. The force at $$\mathrm{P}$$ is only due to $$\mathrm{m}_1$$.
In $$	riangle \mathrm{PQS}$$,
$$\mathrm{h}=\frac{\mathrm{L}}{3} \cos 30^{\circ}=\frac{\mathrm{L} \sqrt{3}}{6}$$
$$	herefore \quad$$ Force on $$\mathrm{m}_2$$ due to $$\mathrm{m}_1$$ at $$\mathrm{P}$$ is
$$\begin{aligned}
F & =\frac{G m_1 m_2}{\left(\frac{L \sqrt{3}}{6}ight)^2} \
& =\frac{G m_1 m_2 \cdot 12}{L^2} \
& =\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}
\end{aligned}$$

Answer

$$\frac{3 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}$$

Answer

$$\frac{6 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}$$

Answer

$$\frac{9 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}$$

Answer

$$\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}$$

Gravitation

A system consists of three particles each of mass '$m_1' placed at the corners of an equilateral triangle of side '\frac{\mathrm{L}}{3}', A particle of mass '\mathrm{m}_2' is placed at the mid point of any one side of the triangle. Due to the system of particles, the force acting on \mathrm{m}_2$ is

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