Question

Two masses m and 9m are placed at a distance of 10m. How far must a third mass be placed from mass 'm' along the line joining them such that it is in equilibrium?


1.2.5 m
2.5 m
3.7.5 m
4.Not possible to attain equilibrium

Answer 1

The third mass should be placed at distance 2.5 m from mass m such that it is in equilibrium

Therefore, option (1) is correct.

Explanation:

Given:

Two masses m and 9m are placed at a distance of 10 m

To find out:

Where on the straight line joining the masses a third mass m be placed to be in equilibrium

Solution:

It is evident that the third mass should be placed in between the two masses so that the system is in equilibrium

Let the third mass m be placed at a distance x from mass m

distance of the third mass m from the mass 9m = 10 - x

The gravitational force acting on the third mass m due to mass 9m and m should be equal

Therefore,

$\frac{G\times 9m\times m}{(10-x)^2}=\frac{G\times m\times m}{x^2}$

$\implies 9x^2=(10-x)^2$

$\implies (3x)^2-(10-x)^2=0$

$\implies (3x-10+x)(3x+10-x)=0$

$\implies (4x-10)(2x+10)=0$

$\implies x=10/4=2.5$ or $x=-10/2=-5$

But the distance cannot be negative

Thus,

$x=2.5$ m

Thus, the third mass should be placed at distance 2.5 m from mass m

Hope this answer is helpful.

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