Question

Three equal masses of 1 kg each are placed at the
vertices of an equilateral triangle of side 1 m, then
the gravitational force on one of the masses due to
other masses is (approx.)
(1) 6.67 x 10-11 N (2) 10-2 N
(3) 11.5 x 10-11 N (4) 11.5 x 10-3 N​

Answer 1

Answer:11.5 ×10⁻11

Explanation:

Given :

Mass of each particle at the vertex of the triangle= 1 kg

Length of side of equilateral triangle= 1 m

We know that from Newton's Law of gravitation the  gravitational force between two particles of masses  and having distance r between their centres is given by

F=Gm1m2/R

The force exerted by first on second and second on first is equal but opposite in direction.

F12=F21

Let consider the force exerted by particle and particle 2 on third particle  then according to the question we have

F13=G*1*1/1=G

F23=G*1*1/1=G

Since it is from the geometry it can be seen that that angle between  and  is

Let  be the resultant force magnitude given by

Fnet=G2+G2+2G2*cos60

=11.5*10-11

Since the magnitude of each mass is same and they are placed a the same distance from each other so magnitude of net force on each particle by other two is same

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Answer 2

Answer:

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