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

a) The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is
`4.3390* 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.

Step-by-step 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=(3.30 m)/(2)=1.65 m`

`F=G* (M_1* M_3)/(r^2)`

`F=6.674* 10−11 m^3/kg s^2(245 kg* 59.0 kg)/((1.65 m)^2)`

`F=3.5435* 10^(-7) N`

Force on
`M_3` due to
`M_2`

Distance between
`M_2 \&M_3,r'=(3.30 m)/(2)=1.65 m`

`F'=G* (M_1* M_3)/(r^2)`

`F'=6.674* 10−11 m^3/kg s^2(545 kg* 59.0 kg)/((1.65 m)^2)`

`F'=7.8825* 10^(-7) N`

Net force on
`M_3`:

`F'+ F=7.8825* 10^(-7) N+(-3.5435* 10^(-7) N)`

`=4.3390* 10^(-7) N`

The magnitude of the net gravitational force exerted by these objects on a 59.0 kg is
`4.3390* 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* (M_1* 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* (M_2* M_3)/(r'^2)`

Net force equal to zero, then :

`f=f'`

`G* (M_1* M_3)/(r^2)=G* (M_2* M_3)/(r'^2)`

`(245 kg* 59.0 kg)/(r^2)=(545 kg* 59.0 kg)/(r'^2)`

`r=1.4914* 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.