Question

(hrw8c13p85) The masses and coordinates of three spheres are as follows: 16 kg, x = 1.50 m, y = 1.50 m; 40 kg, x = -2.50 m, y = -3.00 m; 56 kg, x = 0.00 m, y= -0.50 m. What is the magnitude of the gravitational force on a 22 kg sphere located at the origin due to the other spheres?

Answer 1

m1 = 16 kg  

m2 = 40 kg

m3 = 56 kg

m4 = 22 kg

r14 = 1.5i + 1.5 j    lr14l = sqrt(1.5^2+1.5^2) = 2.12 m

r24 = -2.5i - 3j    lr24l = sqrt(2.5^2+3^2) = 3.9 m

r34 = -0.5j         lr34l = 0.5 m

F14 = G*m1*m4*r14/lr14l^3

F14 = (6.67*10^-11*16*22/2.12^3)*(1.5i + 1.5 j )

F14 = 2.46*10^-9*(1.5i + 1.5 j )

F14 = 3.7*10^-9i + 3.7*10^-9 j

F24 = G*m2*m4*r14/lr14l^3

F24 = (6.67*10^-11*40*22/3.9^3)*(-2.5i - 3j)

F24 = 0.98*10^-9*(-2.5i - 3j)

F24 = -2.5*10^-9i - 2.97*10^-9 j

-------------------------

F34 = G*m3*m4*r14/lr14l^3

F34 = (6.67*10^-11*56*22/0.5^3)*( - 0.5j)

F34 = 6.57*10^-7*(-0.5j)

F34 = -3.3*10^-7j

net force on 22 kg

F = F14 + F24 + F34

F = 1.2*10^-9 i - 3.3*10^-7 j N

magnitue of force F = 3.3*10^-7 N