Respuesta :
Answer: [tex]F_{net} = 0.7411 N[/tex] towards the mass [tex]m_{3}[/tex]
Explanation: We know that the gravitational force is a long range force which is always attractive in nature.
Given that:
- mass [tex]m_{1} = 1 kg[/tex]
- mass [tex]m_{2} = 3kg[/tex]
- mass [tex]m_{3} = 4kg[/tex]
The masses are positioned on X-axis at the following points:
- Position of mass [tex]m_{1} [/tex] [tex] x_{1} = 0[/tex]
- Position of mass [tex]m_{2} [/tex] [tex] x_{2} = 3[/tex]
- Position of mass [tex]m_{3} [/tex] [tex] x_{3} = 6[/tex]
Mathematically:
Gravitational force on mass [tex]m_{2}[/tex] due to mass [tex]m_{1}[/tex] is given by
[tex]F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}[/tex]...................(1)
- where: [tex](r_{21})^2[/tex]= the radial distance between masses [tex]m_{2}[/tex] & [tex]m_{1}[/tex]=3
Similarly, gravitational force on mass [tex]m_{2}[/tex] due to mass [tex]m_{3}[/tex] is given by
[tex]F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}[/tex]............................(2)
- where: [tex](r_{23})^2[/tex]= the radial distance between masses [tex]m_{2}[/tex] & [tex]m_{3}[/tex]=3
Now, put the respective values in the above equations.
[tex]F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}[/tex]
[tex]F_{21} = 2.2233\times 10^{-11} N [/tex]
Again,
[tex]F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}[/tex]
[tex]F_{23} = 2.9644\times 10^{-11} N[/tex]
∵Mass [tex]m_{2}[/tex] is in the middle of the masses [tex]m_{3}[/tex] & [tex]m_{1}[/tex] therefore the forces [tex]F_{23}[/tex] & [tex]F_{21}[/tex] will attract them in radially opposite direction.
∴[tex]F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N[/tex] towards the mass [tex]m_{3}[/tex]
Answer:
The magnitude of the net force on [tex]m_{2}[/tex] is [tex]7.411\times 10^{- 11}\ N[/tex]
Solution:
As per the question:
[tex]m_{1} = 1\ kg[/tex]
[tex]m_{2} = 3\ kg[/tex]
[tex]m_{3} = 4\ kg[/tex]
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:
[tex]F_{G} = \frac{GMm}{x^{2}}[/tex]
Therefore. the net gravitational force on [tex]m_{2}[/tex] is given by:
[tex]F_{G, net} = F_{G,2-3} - F_{G, 1-2}[/tex]
[tex]F_{G, net} = \frac{Gm_{2}m_{3}}{x^{2}} - \frac{Gm_{1}m_{2}}{x^{2}}[/tex]
[tex]F_{G, net} = \frac{G}{x^{2}}(m_{2}m_{3} - m_{1}m_{2})[/tex]
where
G = Gravitational constant
[tex]F_{G, net} = \frac{6.67\times 10^{- 11}}{3^{2}}(4\times 3 - 1\times 2)[/tex]
[tex]F_{G, net} = 7.411\times 10^{- 11}\ N[/tex]
