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

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

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

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

by putting the values

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

$F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}$

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

$F_{23}=\dfrac{Gm_2m_3}{r^2}$

$F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}$

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.

$F_{23}=\dfrac{Gm_2m_3}{x^2}$

$F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}$

Form above two equation

$\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}$

$\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}$

$\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}$

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m