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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. Provide your answer in Newtons (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? Provide your answer in meters (m) from the 545 kg mass toward the 245 kg mass.

Answer :

Answer:

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

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= [tex]M_1=245 kg[/tex]

Mass of an object-2= [tex]M_2=545 kg[/tex]

Distance between [tex]M_1 \&M_2[/tex] = 3.30 m

Mass of an object-3 =[tex]M_3=59.0 kg[/tex]

If [tex]M_3[/tex] placed midway between them [tex]M_1 \&M_2[/tex].

Force on [tex]M_3 [/tex] due to [tex]M_1[/tex]

Distance between [tex]M_1 \&M_3,r=\frac{3.30 m}{2}=1.65 m[/tex]

[tex]F=G\times \frac{M_1\times M_3}{r^2}[/tex]

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

[tex]F=3.5435\times 10^{-7} N[/tex]

Force on [tex]M_3 [/tex] due to [tex]M_2[/tex]

Distance between [tex]M_2 \&M_3,r'=\frac{3.30 m}{2}=1.65 m[/tex]

[tex]F'=G\times \frac{M_1\times M_3}{r^2}[/tex]

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

[tex]F'=7.8825\times 10^{-7} N[/tex]

Net force on [tex]M_3[/tex]:

[tex]F'+ F=7.8825\times 10^{-7} N+(-3.5435\times 10^{-7} N)[/tex]

[tex]=4.3390\times 10^{-7} N[/tex]

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

b) Distance between [tex]M_1 \&M_2[/tex] = r = 3.30 m

Let the distanced between [tex]M_1 \&M_3[/tex] be r

Force experienced by [tex]M_3[/tex] due to [tex]M_1[/tex]:

[tex]f=G\times \frac{M_1\times M_3}{r^2}[/tex]

Let the distanced between [tex]M_2 \&M_3[/tex] be r'

Force experienced by [tex]M_3[/tex] due to [tex]M_2[/tex]:

[tex]f'=G\times \frac{M_2\times M_3}{r'^2}[/tex]

Net force equal to zero, then :

[tex]f=f'[/tex]

[tex]G\times \frac{M_1\times M_3}{r^2}=G\times \frac{M_2\times M_3}{r'^2}[/tex]

[tex]\frac{245 kg\times 59.0 kg}{r^2}=\frac{545 kg\times 59.0 kg}{r'^2}[/tex]

[tex]r=1.4914\times r'[/tex]

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.

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