Question

Crmm a 2 2

For each problem, find the magnitude and show the direction of the gravitational force exerted on the black circle. All objects has a mass of 4kg

Answer 1

In general, the gravitational force between exerted by mass 1 on mass 0 can be expressed as

$\:$

$\vec{F}_{g}=-\frac{Gm_{0}m_{1}}{r_{01}^{3}} \, \vec{r}_{01}$

$\:$

where r₀₁ is the distance between masses labeled as 0 and 1 and $\vec{r}_{01}$ is a vector poiting from point 0 to point 1. I'll assume for each part that every tile measures 1 m, you can replace 1 m for whatever unit used for length.

$\:$

Part a)

The net force exerted on mass m₀ is

$\:$

$\vec{F}_{net0}=\vec{F}_{01}+\vec{F}_{02}$

$\Rightarrow \: \: \: \vec{F}_{net0}=-\frac{Gm_{0}m_{1}}{r_{01}^{3}} \, \vec{r}_{01}-\frac{Gm_{0}m_{2}}{r_{02}^{3}} \, \vec{r}_{02}$

$\Rightarrow \: \: \: \vec{F}_{net0}=-\frac{Gm_{0}m_{1}}{r_{01}^{3}} \, (\vec{r}_{1}-\vec{r}_{0})-\frac{Gm_{0}m_{2}}{r_{02}^{3}} \, (\vec{r}_{2}-\vec{r}_{0})$

$\:$

since

$\:$

$m_{0}=m_{1}\equiv m=4 \, kg$

$\:$

we have

$\:$

$\vec{F}_{net0}=-Gm^{2}\left [\frac{\vec{r}_{1}-\vec{r}_{0}}{r_{01}^{3}} \, +\frac{\vec{r}_{2}-\vec{r}_{0}}{r_{02}^{3}} \right ]$

$\vec{F}_{net0}=-(6.67\times 10^{-11} \, N\cdot m^{2}/kg^{2})(4kg)^{2}\left [\frac{4m\, \hat{i}-(-4m\, \hat{i})}{(8m)^{3}} \, +\frac{(4m\, \hat{i }-2m\, \hat{j})-(-4m\, \hat{i})}{\left [ (-4m-4m)^{2}+(-2m)^{2} \right ]^{3/2}} \right ]$

$\vec{F}_{net0}=-(106.7\times 10^{-11} \, N)\left (\frac{\hat{i}}{64} +\frac{8\, \hat{i }-2\hat{j} }{561} \right )$

$\:$

or

$\:$

${\color{Blue} \vec{F}_{net0}=(-3.16\times 10^{-11} \, \hat{i}+3.80\times 10^{-12} \, \hat{j}) \, N}$

$\:$

Part b)

m3

The net force exerted on mass m₀ is

$\:$

$\vec{F}_{net0}=\vec{F}_{01}+\vec{F}_{02}+\vec{F}_{03}$

$\vec{F}_{net0}=-Gm^{2}\left [\frac{\vec{r}_{1}-\vec{r}_{0}}{r_{01}^{3}} \, +\frac{\vec{r}_{2}-\vec{r}_{0}}{r_{02}^{3}}+\frac{\vec{r}_{3}-\vec{r}_{0}}{r_{03}^{3}} \right ]$

$\vec{F}_{net0}=-(6.67\times 10^{-11} \, N)(4)^{2}\left [ \frac{\hat{i}}{64}+\frac{\hat{j}}{64}+ \frac{-4\, \hat{i}-4\, \hat{j}}{(32)^{3/2}} \right ]$

$\:$

or

$\:$

${\color{Blue} \vec{F}_{net0}=(6.94\times 10^{-12} \, \hat{i}+6.94\times 10^{-12}\, \hat{j}) \, N}$

$\:$

Part c)

The net force exerted on mass m₀ is

$\:$

$\vec{F}_{net0}=\vec{F}_{01}+\vec{F}_{02}+\vec{F}_{03}+\vec{F}_{04}$

$\vec{F}_{net0}=-Gm^{2}\left [\frac{\vec{r}_{1}-\vec{r}_{0}}{r_{01}^{3}} \, +\frac{\vec{r}_{2}-\vec{r}_{0}}{r_{02}^{3}}+\frac{\vec{r}_{3}-\vec{r}_{0}}{r_{03}^{3}}+\frac{\vec{r}_{4}-\vec{r}_{0}}{r_{04}^{3}} \right ]$

$\vec{F}_{net0}=-(6.67\times 10^{-11} \, N)(4)^{2}\left [ \frac{4 \, \hat{j}}{32^{3/2}}+\frac{4 \, \hat{j}}{32^{3/2}}+\frac{8 \, \hat{j}}{8^{3}} -\frac{9 \, \hat{j}}{9^{3}} \right ]$

$\:$

or

$\:$$\:$

${\color{Blue} \vec{F}_{net0}=-5.07\times 10^{-11} \,\hat{j} \, N}$