Respuesta :
Answer:[tex]F_{net}=3.383\times 10^{-7}\ N[/tex]
Explanation:
Given
Mass of first object [tex]m_1=225\ kg[/tex]
Mass of second object [tex]m_2=525\ kg[/tex]
Distance between them [tex]d=3.8\ m[/tex]
[tex]m_3=61\ kg[/tex] object is placed between them
So force exerted by [tex]m_1[/tex] on [tex]m_3[/tex]
[tex]F_{13}=\frac{Gm_1m_3}{1.9^2}[/tex]
[tex]F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}[/tex]
[tex]F_{13}=2.5374141274×10^{−7}\ N[/tex]
Force exerted by [tex]m_2\ on\ m_3[/tex]
[tex]F_{23}=\frac{Gm_2m_3}{1.9^2}[/tex]
[tex]F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}[/tex]
[tex]F_{23}=5.920632964\times 10^{-7}\ N[/tex]
So net force on [tex]m_3[/tex] is
[tex]F_{net}=F_{23}-F_{13}[/tex]
[tex]F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}[/tex]
[tex]F_{net}=3.383\times 10^{-7}\ N[/tex]
i.e. net force is towards [tex]m_2[/tex]
(b)For net force to be zero on [tex]m_3[/tex], suppose
So force exerted by [tex]m_1[/tex] and [tex]m_2[/tex] must be equal
[tex]F_{13}=F_{23}[/tex]
[tex]\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}[/tex]
[tex]\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}[/tex]
[tex]\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}[/tex]
[tex]\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}[/tex]
[tex]\Rightarrow 3.8-x=1.52752x[/tex]
[tex]\Rightarrow 3.8=2.52x[/tex]
[tex]\Rightarrow x=1.507\ m[/tex]
