Three balls, with masses of m, 2m, and 3m, are placed at the corners of a square

Question

Three balls, with masses of m, 2m, and 3m, are placed at the corners of a square measuring L on each side, as shown in the figure. The value of m is 2.90 kg, and L = 70.0 cm. Assume this set of three balls is not interacting with anything else in the universe. Note that, on WebAssign, a number like 0.0000456 can be entered as 4.56e-5. What is the magnitude of the net gravitational force acting on the ball of mass 2m? What is the magnitude of the net gravitational force acting on the ball of mass 3m?

Explanation / Answer

a) force on 2m due to m is F1 = G*2m*m/L^2 = (6.67*10^-11*2*2.9^2)/(0.7^2) = 2.28e-9 N

Force on 2m due to 3m is F2 = G*3m*2m/L^2 = (6.67*10^-11*6*2.9^2)/(0.7^2) = 6.86e-9 N

Net force is Fnet = sqrt(F1^2+F2^2) = sqrt(6.86^2+2.28^2)*10^-9 = 7.23e-9 N

B) Force on 3m due to m is F1 = G*3m*m/(2*L^2) = (6.67*10^-11*3*2.9^2)/(2*0.7^2) =1.717e-9 N

Force on 3m due to 2m is F2 = G*3m*2m/(L^2) = (6.67*10^-11*6*2.9^2)/0.7^2 = 6.86e-9 N

Net force is Fnet = Sqrt(F1^2+F2^2+2*F1*F2*cos(45))

Fnet = Sqrt(1.717^2+6.86^2+(2*1.717*6.86*cos(45)))*10^-9 = 8.16e-9 N

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