Five balls, two of mass m and three of mass 2m, are arranged as shown in the fig
ID: 1478824 • Letter: F
Question
Five balls, two of mass m and three of mass 2m, are arranged as shown in the figure. The value of m is 2.20 kg and the value of r is 65.0 cm. Note that, in this question, you are only asked to find the magnitude of the net force in each case, but you should also think about the direction of the net force. What is the magnitude of the net gravitational force on the ball of mass m that is located at the origin? This is the net force because of the other four balls, which are all located a distance r from the origin. 7.64e-10 N What is the magnitude of the net gravitational force on the other ball of mass m, located on the positive y-axis?Explanation / Answer
Left and right balls of mass 2m will have gravitational force in y axis only and there x component if force cancels each other.
net force due to this 2 masses are = G*m*2m / d^2 * cos 45 + G*m*2m / d^2 cos 45
d = r*sqrt(2)
net force due to this 2 masses are =2* G*m*2m / d^2 * cos 45
=2* G*m*2m / (r*sqrt(2))^2 * cos 45
=2* G*m*2m / 2*r^2 * cos 45
=2* G*m*2m / {2*sqrt(2)r^2}
=2* (6.673*10^-11)*2.2*2*2.2 / {2*sqrt(2)*0.65^2}
=2* (6.673*10^-11)*2.2*2*2.2 / {2*sqrt(2)*0.65^2}
= 1.081*10^-9 N
force due to lower 2m mass = G*m*2m / (2r)^2
= (6.673*10^-11)*2.2 * 2*2.2 / (2*0.65)^2
= 3.822*10^-10 N
force due to lower m mass = G*m*m / (r)^2
= (6.673*10^-11)*2.2 * 2.2 / (0.65)^2
= 7.644*10^-10 N
total force = 1.081*10^-9 N + 3.822*10^-10 N + 7.644*10^-10 N
= 2.23*10^-9 N
Answer: 2.23*10^-9 N
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