Question

An equilateral triangle 12.0 m on a side has a m1 = 10.00 kg mass at one corner, a m2 = 65.00 kg mass at another corner, and a m3 = 125.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 10.00 kg mass. a) Newtons b) Degrees counterclockwise from the positive x-axis.

Answer 1

The law of gravitation states the the force between any two masses m1 and m2, seperated by distance r, is

F = G*m1*m2 / r^2 , where G = 6.67*10^(-11) N*(m^2) / (Kg^2) is the gravitational constant.

Assuming that m1 lies at the top corner of the triangle and that the side connecting m2 and m3 is parallel to the x-axis, there are two forces acting on m1

The force between m1 and m2 :
F(1,2) = G*m1*m2 / r^2 = 6.67*10^(-11)*10.00*65.00 / 12.0^2 =3.010*10^(-10) N
at a 60 degrees angle

The force between m1 and m3 :
F(1,3) = G*m1*m3 / r^2 = 6.67*10^(-11)*10.00*125.00 / 6.0^2 = 5.78*10^(-10) N
at a 120 degrees angle.

Analyzing the two force vectors into two components, one parallel to the x-axis and the other parallel to the y axis we get

Fy(1,2) = F(1,2)sin(60) = 3.010*10^(-10)*0.87 = 2.61*10^(-10) N
Fx(1,2) = F(1,2)cos(60) = 3.010*10^(-10)*0.50 = 1.50*10^(-10) N

Fy(1,3) = F(1,3)sin(120) = 5.78*10^(-10)*0.87 = 5.02*10^(-10) N
Fx(1,3) = F(1,3)cos(120) = 5.78*10^(-10)*(-0.5) = -2.89*10^(-10) N

So, the net force on the y-axis is Fy = Fy(1,2)+Fy(1,3) = 7.63*10^(-10) N
and on the x-axis Fx = Fx(1,2)+Fx(1,3) = -1.39*10^(-10) N

The magnitute of the net force is F = sqrt(Fx^2 + Fy^2) =7.75*10^(-10) N