Five balls, two of mass m and three of mass 2m, are arranged as shown in the fig
ID: 1480541 • 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 3.20 kg and the value of r is 55.0 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 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 four balls, which are all located a distance r from the origin. What is the magnitude of the net gravitational force on the other ball of mass m, located on the positive y-axis?Explanation / Answer
a)
net force by two mass "2m" on left and right of mass at origin "m" cancel out being equal and opposite.
force along Y-direction:
Force by mass "m" = F1 = Gm(m) /r2 = Gm2 /r2 in upward direction
Force by mass "2m" = F2 = Gm(2m) /r2 = 2 Gm2 /r2 in down direction
net force = Fnet = F2 - F1 = 2 Gm2 /r2 - Gm2 /r2 = Gm2 /r2 = (6.67 x 10-11)(3.20)2 / (0.55)2 = 2.3 x 10-9 N
b)
force by each mass "2m" on X-axis on mass "m" is given as
F = 2 Gm2 /(sqrt(2)r)2 = Gm2/r2
net force on mass "m" by two mass "2m" on X-axis
F' = sqrt (F2 + F2) = F sqrt(2) = sqrt(2) Gm2 /r2
force by mass "m" on mass "m" on Y-axis
F1 = Gm2/r2
force by mass "2m"
F2 = 2Gm2/(2r)2 = Gm2/2r2
net force = F' - F1 - F2 = sqrt(2) Gm2 /r2 - Gm2/r2 - Gm2/2r2 = - 0.086 Gm2/r2 in down direction
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