IP Three 700kg masses are at the corners of an equilateral t

IP Three 7.00-kg masses are at the corners of an equilateral triangle and located in space far from any other masses.

IP Three 7.00-kg masses are at the corners of an equilateral triangle and located in space far from any other masses.

Part A

If the sides of the triangle are 1.90 m long, find the magnitude of the net force exerted on each of the three masses.

Part B

How does your answer to part A change if the sides of the triangle are doubled in length?

Solution

Fg = Gm1m2 / d^2

where G is universal gravitational constant equal to 6.67 x 10^-11 Nm^2/kg^2

Force on one mass due to mass at one corner,

F1 = (6.67 x 10^-11 x 7 x 7) / (1.90)^2

F1 = 9.1 x 10^-10 N

Force on one mass due to mass at another corner,

F2 = (6.67 x 10^-11 x 7 x 7) / (1.90)^2

F2 = 9.1 x 10^-10 N

As the triangle is equilateral, the angle between any two sides is 60 degree

The two forces F1 and F2 are inclined at 60 degree and magnitude of F1= magnitude of F2 = (9.1 x 10^-10N) = F

Resultant of two equal forces of magnitude F each inclined at 60 degrees = sqrt(3)*F = 1.732

F = 1.732 x 9.1 x 10^-10 = 1.576 x 10^-9 N

(a)

The magnitude of the net force exerted on each of the three masses is 1.576 x 10^-9 N

(b)

if the sides of the triangle are doubled in length, the net force will be (1/4) of original.

The net force after sides are doubled will be (1/4) of 1.576 x 10^-9 N

The net force after sides are doubled will be 3.94 x 10^-10 N