Answer:
(a) [tex]F_{net}[/tex] = 4.19 x [tex]10^{-5}[/tex] N, and its direction is towards [tex]m_{2}[/tex].
(b) It must be placed inside a hollow shell.
Explanation:
Let, [tex]m_{1}[/tex] = 165 kg, [tex]m_{2}[/tex] = 465 kg, [tex]m_{3}[/tex] = 60 kg, and the distance between [tex]m_{1}[/tex] and [tex]m_{2}[/tex] is 0.340 m.
(a) Since [tex]m_{3}[/tex] is placed midway between [tex]m_{1}[/tex] and [tex]m_{2}[/tex], then its distance to both masses is 0.170 m.
From the Newton's law of universal gravitation,
F = [tex]\frac{Gm_{1}m_{2} }{r^{2} }[/tex]
Where all variables have their usual meaning.
Then,
a. [tex]F_{net}[/tex] = [tex]F_{23}[/tex] - [tex]F_{13}[/tex]
[tex]F_{13}[/tex] = [tex]\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }[/tex]
= 2.25 x [tex]10^{-5}[/tex] N
[tex]F_{23}[/tex] = [tex]\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }[/tex]
= 6.44 x [tex]10^{-5}[/tex] N
∴ [tex]F_{net}[/tex] = = 6.44 x [tex]10^{-5}[/tex] - 2.25 x [tex]10^{-5}[/tex]
= 4.19 x [tex]10^{-5}[/tex] N
The net force exerted by the two masses on the 60 kg object is 4.19 x [tex]10^{-5}[/tex] N.
(ii) /[tex]F_{net}[/tex]/ = /[tex]F_{23}[/tex]/ - /[tex]F_{13}[/tex]/
= 6.44 x [tex]10^{-5}[/tex] - 2.25 x [tex]10^{-5}[/tex]
= 4.19 x [tex]10^{-5}[/tex] N
(iii) The direction of the net force is to the right i.e towards [tex]m_{2}[/tex].
(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.
