Question

Three small spheres, having masses m1 = 1 kg, m2 = 3 kg, and m3 = 4 kg, are held fixed on the x axis in deep space where the effects of earth's gravity can be neglected. They are positioned at x = 0, x = 3 m, and x = 6 m, respectively. What is the magnitude of net gravitational force on m2?

Answer 1

Answer: $F_{net} = 0.7411 N$ towards the mass $m_{3}$

Explanation: We know that the gravitational force is a long range force which is always attractive in nature.

Given that:

• mass $m_{1} = 1 kg$

• mass $m_{2} = 3kg$

• mass $m_{3} = 4kg$

The masses are positioned on X-axis at the following points:

• Position of mass $m_{1}$  $x_{1} = 0$

• Position of mass $m_{2}$  $x_{2} = 3$

• Position of mass $m_{3}$  $x_{3} = 6$

Mathematically:

Gravitational force on mass $m_{2}$ due to mass $m_{1}$ is given by

$F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}$...................(1)

• where: $(r_{21})^2$= the radial distance between masses $m_{2}$ & $m_{1}$=3

Similarly, gravitational force on mass $m_{2}$ due to mass $m_{3}$ is given by

$F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}$............................(2)

• where: $(r_{23})^2$= the radial distance between masses $m_{2}$ & $m_{3}$=3

Now, put the respective values in the above equations.

$F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}$

$F_{21} = 2.2233\times 10^{-11} N$

Again,

$F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}$

$F_{23} = 2.9644\times 10^{-11} N$

∵Mass $m_{2}$ is in the middle of the masses $m_{3}$ & $m_{1}$ therefore the forces  $F_{23}$ & $F_{21}$ will attract them in radially opposite direction.

∴$F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N$ towards the mass $m_{3}$

Answer 2

Answer:

The magnitude of the net force on $m_{2}$ is $7.411\times 10^{- 11}\ N$

Solution:

As per the question:

$m_{1} = 1\ kg$

$m_{2} = 3\ kg$

$m_{3} = 4\ kg$

Respective positions of the above mentioned masses are:

x = 0 m

x = 3 m

x = 6 m

Now,

We know that the gravitational force between two masses separated by some distance is given by:

$F_{G} = \frac{GMm}{x^{2}}$

Therefore. the net gravitational force on $m_{2}$ is given by:

$F_{G, net} = F_{G,2-3} - F_{G, 1-2}$

$F_{G, net} = \frac{Gm_{2}m_{3}}{x^{2}} - \frac{Gm_{1}m_{2}}{x^{2}}$

$F_{G, net} = \frac{G}{x^{2}}(m_{2}m_{3} - m_{1}m_{2})$

where

G = Gravitational constant

$F_{G, net} = \frac{6.67\times 10^{- 11}}{3^{2}}(4\times 3 - 1\times 2)$

$F_{G, net} = 7.411\times 10^{- 11}\ N$