Question

A 295-kg object and a 595-kg object are separated by 4.10 m.

(a) Find the magnitude of the net gravitational force exerted by these objects on a 63.0-kg object placed midway between them. Incorrect: Your answer is incorrect. Draw a picture of the three objects, including the forces exerted on the third. N
(b) At what position (other than an infinitely remote one) can the 63.0-kg object be placed so as to experience a net force of zero from the other two objects? m from the 595 kg mass toward the 295 kg mass

Answer 1

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Step-by-step explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

`F_(13)=(Gm_1m_3)/(r^2)`

by putting the values

`F_(13)=(Gm_1m_3)/(r^2)`

`F_(13)=(6.67* 10^(-11)* 295* 63 )/(2.05^2)`

F₁₃=2.94 x 10⁻⁷ N

The force between the mass m₂ and m₃

by putting the values

`F_(23)=(Gm_2m_3)/(r^2)`

`F_(23)=(6.67* 10^(-11)* 595* 63 )/(2.05^2)`

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

`F_(23)=(Gm_2m_3)/(x^2)`

`F_(13)=(Gm_1m_3)/((4.1-x)^2)`

Form above two equation

`(Gm_1m_3)/((4.1-x)^2)=(Gm_2m_3)/(x^2)`

`(m_1)/((4.1-x)^2)=(m_2)/(x^2)`

`(295)/((4.1-x)^2)=(595)/(x^2)`

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m