Question

Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the center of the triangle.

Answer 1

Thanks for asking the question!

ANSWER::

(a) 'm' is placed at mid-point of a side

Force in OA = 4Gm²/a²

Force in OB = 4Gm²/a²

As equal and opposite cancel each other

Force in OC = Gm² / [(√3/2)a]² = 4Gm² / 3a²

Net gravitational force on m = 4Gm²/a²

(b)at the center of the triangle ( Placed at O also called as centroid )

Force in OA = Gm² / [a/√3]² = 3Gm² / a²

Force in OB = 3Gm² / a²

Resultant force = [2(3Gm²/a²)² - 2(3Gm²/a²)² x 1/2] = 3Gm² / a²

Since,

Force in OC = 3Gm² / a² which is equal and opposite to F which cancel each other.

Net gravitational force = 0

Hope it helps!

Answer 2

Answer:

above answer is perfect