Question

A 245-kg object and a 545-kg object are separated by 3.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 59.0-kg object placed midway between them. N (b) At what position (other than an infinitely remote one) can the 59.0-kg object be placed so as to experience a net force of zero from the other two objects? m from the 545 kg mass toward the 245 kg mass

Answer 1

Answer:

a) The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is $4.3390\times 10^{-7} N$.

b) By placing object of 59.0 kg 1.976 m from 245 kg object and 1.324 m from 545 kg object.

Explanation:

Mass of an object-1= $M_1=245 kg$

Mass of an object-2= $M_2=545 kg$

Distance between  $M_1 \&M_2$ = 3.30 m

Mass of an object-3 =$M_3=59.0 kg$

If $M_3$ placed midway between them $M_1 \&M_2$.

Force on $M_3$ due to $M_1$

Distance between  $M_1 \&M_3,r=\frac{3.30 m}{2}=1.65 m$

$F=G\times \frac{M_1\times M_3}{r^2}$

$F=6.674\times 10−11 m^3/kg s^2\frac{245 kg\times 59.0 kg}{(1.65 m)^2}$

$F=3.5435\times 10^{-7} N$

Force on $M_3$ due to $M_2$

Distance between  $M_2 \&M_3,r'=\frac{3.30 m}{2}=1.65 m$

$F'=G\times \frac{M_1\times M_3}{r^2}$

$F'=6.674\times 10−11 m^3/kg s^2\frac{545 kg\times 59.0 kg}{(1.65 m)^2}$

$F'=7.8825\times 10^{-7} N$

Net force on $M_3$:

$F'+ F=7.8825\times 10^{-7} N+(-3.5435\times 10^{-7} N)$

$=4.3390\times 10^{-7} N$

The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is $4.3390\times 10^{-7} N$.

b) Distance between  $M_1 \&M_2$ = r = 3.30 m

Let the distanced between $M_1 \&M_3$ be r

Force experienced by $M_3$ due to $M_1$:

$f=G\times \frac{M_1\times M_3}{r^2}$

Let the distanced between $M_2 \&M_3$ be r'

Force experienced by $M_3$ due to $M_2$:

$f'=G\times \frac{M_2\times M_3}{r'^2}$

Net force equal to zero, then :

$f=f'$

$G\times \frac{M_1\times M_3}{r^2}=G\times \frac{M_2\times M_3}{r'^2}$

$\frac{245 kg\times 59.0 kg}{r^2}=\frac{545 kg\times 59.0 kg}{r'^2}$

$r=1.4914\times r'$

R = r+r'

3.30 m = 1.4914\times r' + r'

r' = 1.324 m

r = 1.976 m

By placing object of 59.0 kg 1.976 m from 245 kg object and 1.324 m from 545 kg object.