Question

two particles each of mass 1 kg are placed at P and Q such that PO is equals to a cube is equal to N with gravitational force experienced by another one 1kg misplaced it or where R is equal to L is what

Answer 1

Answer:

The answer will be g(2 x(L^2+N^2)/(L+N)^2(L-N)^2)

Explanation:

According to the problem both the particles of with their respective mass are place d in P and Q position.

Now there is an another particle which is placed in R

we need to find the gravitational force of this particle

Now This particle is placed L-N distance from P and L+N distance from Q

Therefore for P the gravitational force on R will be

f(P) = gm^2/(L-N)^2

So as for Q it will be

f(Q) = gm^2/(L+N)^2

Therefore the net force will be, f = f(P)+ f(Q) = gm^2/(L-N)^2+ gm^2/(L+N)^2

                                                        = g(2 x(L^2+N^2)/(L+N)^2(L-N)^2) [ m = 1 kg for both]