Question

Two bodies of masses 6 kg and 24 kg are placed at a distance 12 m. A third body of mass 2 kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point. [Ans. 4 m from 6 kg]​

Answer 1

Answer:

4+6+9+8+5=32

Explanation:

Answer 2

The position is 4 m from 6 kg.

Explanation:

Given that,

Mass of first body = 6 kg

Mass of second body = 24 kg

Mass of third body = 2 kg

Distance = 12 m

Let body C be placed at a distance x from A,

The distance of AC is x .

The distance of BC is (12-x).

We need to calculate the position of the third mass when the force acting on it is zero

As gravitational force on C due to A = gravitational force on C due to B,

$F_{1}=F_{2}$

$\dfrac{Gm_{1}m}{x^2}=\dfrac{Gm_{2}m}{(12-x)^2}$

$\dfrac{m_{1}}{x^2}=\dfrac{m_{2}}{(12-x)^2}$

Put the value into the formula

$\dfrac{6}{x^2}=\dfrac{24}{(12-x)^2}$

$(12-x)^2=4x^2$

$x=\dfrac{12}{3}$

$x=4\ m$

Hence, The position is 4 m from 6 kg.

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Topic : gravitational force

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