Question

A
Question 7 - Rank by force Three balls, with masses of 4m, 2m, and m, are equally spaced along a line. The spacing between neighboring balls is r. We can arrange the balls in three different ways, as shown in the figure. In each case, the balls are in an isolated region of space very far from anything else. 2m 4m 4m 2m

B

Answer 1

a) I hope you will understand the meaning if I write, for example in the first case, for the mass 4m, the force due to 2m and m will be

$F_4 = \frac{4\times2}{1^{2}} +\frac{4\times1}{2^{2}} = 9$

(If you write as per equation,there will be a Gmm/r² outside the above expression always. We only need to calculate using the coefficient of mass and distance)

$F_2 = -\frac{4\times2}{1^{2}} +\frac{2\times1}{1^{2}} = -6$

$F_1 = -\frac{4\times1}{2^{2}} -\frac{2\times1}{1^{2}} = -3$

So the net force will be 9-6-3=0.

This is to show that the net force of the system will always be zero.

NB: In the first part the question is not given properly. If it was for the system or for any of the particular mass you want the comparison is not specified. That is why I showed you the procedures step by step so that now you will be in a position to solve it your self even if it is for a single mass. Hope you got my point.❤

b) In case ll the net force on the mass 4m will be,

$F_4 = \frac{Gmm}{r^{2}}(-\frac{4\times2}{1^{2}} +\frac{4\times1}{1^{2}}) = -4\frac{Gm^{2}}{r^{2}}$

$F_4 = -4\times \frac{6.674\times 10^{-11}\times3^{2}}{0.7^{2}} =- 4.9\times 10^{-9}N =-4.9e(-9)N$

Hope the answer is clear. All the best for your studies ❤.

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