Please help with both parts. Thanks! Three balls, with masses or m, 2m, and 3m,

Question

Please help with both parts. Thanks!

Three balls, with masses or m, 2m, and 3m, are placed at the corners of a square measuring L on each side, as shown in the figure. The value or m is 2.80 kg, and L = 50.0 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. (a) What is the magnitude of the net gravitational force acting on the ball of mass 2m? N (b) What is the magnitude of the net gravitational force acting on the ball of mass 3m? N

Explanation / Answer

Force on 2m due to m

F1x = -G*m*2m/L^2 = -2*G*m^2/L^2 = -2*6.67*10^-11*2.8^2/0.5^2 = -4.18*10^-9 N

F1y = 0

Force on 2m due to 3m

F2y = -G*3m*2m/L^2 = -6*G*m^2/L^2 = -6*6.67*10^-11*2.8^2/0.5^2 = -12.6*10^-9 N

F2x = 0

Fx = F1x + F2x

Fy = F1y + F2y

net force on 2m = sqrt(Fx^2+Fy^2)

net force on 2m = 13.3e-9 N <<<<<=============ANSWER

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(b)

Force on 3m due to m

F1x = -G*3m*m*cos45/2L^2 = -3*G*m^2*cos45/2L^2 = -3*6.67*10^-11*2.8^2*cos45/(2*0.5^2) = -2.22*10^-9 N

F1y = G*3m*m*sin45/2L^2 = 3*G*m^2*sin45/2L^2 = 3*6.67*10^-11*2.8^2*sin45/(2*0.5^2) = 2.22*10^-9 N

Force on 3m due to 2m

F2y = G*2m*3m/L^2 = 6*G*m^2/L^2 = 6*6.67*10^-11*2.8^2/0.5^2 = 12.6*10^-9 N

F2x = 0

Fx = F1x + F2x = -2.22*10^-9 N

Fy = F1y + F2y = ( 2.22*10^-9) + (12.6*10^-9) = 1.48*10^-9 N

net force on 3m = sqrt(Fx^2+Fy^2)

net force on 3m = 2.67e-9 N <<<<<=============ANSWER

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