Question

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objects on a 60.0 kg object placed midway between them. magnitude N direction ---Select--- (b) At what position (other than infinitely remote ones) can the 60.0 kg object be placed so as to experience a net force of zero

Answer 1

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$.

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

     = 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

     = 6.44 x $10^{-5}$ N

∴ $F_{net}$ =  = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

              = 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$  N.

(ii) /$F_{net}$/ = /$F_{23}$/ - /$F_{13}$/

              = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

              = 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.