# Physics

posted by on .

Three masses that are 10 kg each are located at the corners of an equilateral triangle, with side length of 0.55m. What is the magnitude of the total force acting on one mass due to the other two masses?

The answer is supposed to be 3.82e-08 N. Any help is appreciated. I'm not sure where to even start this question.

• Physics - ,

Let the mass m3 be in the origin of the coordinate system, m2 be on x-axis (separated by 0.55 m from origin), and m1 is above x-axis. m1=m2=m3=m=10 kg, a=0.55 m.
The gravitational constant is
G =6.67•10^-11 N•m²/kg²,

F2 =F2x=G•m2•m3/a²,
F1x = (G•m1•m3/a²)•cos60º,
F1y=(G•m1•m3/a²)•sin60º,
F(net)x= F2x+F1x= G•m2•m3/a² +(G•m1•m3/a²)•cos60º=1.5•G•m2•m3/a²,
F(net)y = F1y=(G•m1•m3/a²)•sin60º= 0.866• G•m2•m3/a²,
F(net) =sqrt[(F(net)x)²+( F(net)y)²] = G•m²/a² •sqrt(1.5²+0.866²)=
=1.73•6.67•10^-11•100/0.55²=3.82•10^-8 N.

### Answer This Question

 First Name: School Subject: Answer:

### Related Questions

More Related Questions

Post a New Question