Question

Three spheres are arranged in the xy plane as in the figure below. The first sphere, of mass

m₁ = 15.6 kg,

is located at the origin; the second sphere, of mass

m₂ = 4.47 kg

is located at (−6.00, 0.00) m; and the third sphere, of mass

m₃ = 2.00 kg

is located at (0.00, 5.00) m.Assuming an isolated system, what is the net gravitational force on the sphere located at the origin?

Answer 1

Given masses are

m1 = 15.6 kg at origin

m2 = 4.47 kg at (-6.00,0.00)m

m3 = 2.00 kg at (0.00,5.00)m

the net gravitational force on m1 by m2.m3 , F = ?

here m2 attracts m1 in -ve x direction and m3 attracts m1 along +y direction

writing the forces

F12 = F12x i + F12y j

F13 = F13x i + F13y j

here F12y,F13x are zero because sin 180 =0 , cos 90 =0 so

F12 = F12x i + 0 j

F13 = 0 i + F13y j

the net force on m1 is  

F = F12xi + F13y j

we know that the force between two masses separated by a distance r is

F = G*m1*m2/d^2

same way here

F12x = Gm1*m2/r12^2 cos (180) i ; r12 = 6 m

F13y = Gm1*m3/r13^2 sin (90) i ; r13 = 5 m

F = G*m1(-m2/r12^2+m3/r13^2)

substituting the values  

F12x = 6.67*10^-11*15.6*4.47 cos(180) /(-6)^2 = -1.29275*10^-10 N

F13y = 6.67*10^-11*15.6*2 sin(90) /(5)^2= 8.32416*10^-11 N

the net force is  

F = -1.29275*10^-10 N i +8.32416*10^-11 N j

the magnitude is  

F = sqrt(F12x^2+F13y^2)

F = sqrt((-1.29275*10^-10)^2+(8.32416*10^-11)^2) N

F = 1.53757*10^-10 N

the net gravitational force on the sphere located at the origin(m1) is 1.53757*10^-10 N