FIGURE P13.35 shows three masses. What are the magnitude and the direction of the net gravitational force on (a) the 20.0 kg mass and (b) the 5.0 kg mass? Give the direction as an angle cw or ccw from the y-axis.

The given diagram is

Gravitational force,

$$\begin{gathered} \text{F=G}\frac{{\text{m}}_1{\text{m}}_2}{{\text{r}}^2} \\ \text{where} \\ {\text{G=Gravitationalconstant=6.67×10}}^{-11}{\text{N·m}}^2{\text{/kg}}^2 \\ {\text{m}}_1{\text{,m}}_2\text{aremassesofdifferentobjects} \\ \text{risdistancebetweentwoobjects} \end{gathered}$$

Gravitational force on 20 kg mass due to 5kg mass is

$$\begin{gathered} {\text{F}}_1{\text{=6.67×10}}^{-11}\text{·}\frac{20\times 5}{0.2^2} \\ {\text{=1.67×10}}^{-7}\text{N} \end{gathered}$$

This force is in vertical direction.

Gravitational force on 20 kg mass due to 5kg mass is

$$\begin{gathered} {\text{F}}_2{\text{=6.67×10}}^{-11}\text{·}\frac{20\times 10}{0.2^2} \\ {\text{=1.34×10}}^{-6}\text{N} \end{gathered}$$

This force is in horizontal direction.

Net force is

$$\begin{gathered} \text{F=}\sqrt{{{\text{F}}_1}^2{{\text{+F}}_2}^2} \\ \text{=}\sqrt{\left(1.67\times {10}^{-7}\right)\text{+}\left(1.34\times {10}^{-6}\right)}\text{N} \\ =1.35\times {10}^{-6}\text{N} \end{gathered}$$

Direction of net force from horizontal x-axis is

$$\begin{gathered} \theta ={\tan }^{-1}\frac{{\text{F}}_1}{{\text{F}}_2} \\ ={\tan }^{-1}\left(\frac{1.67\times {10}^{-7}}{1.34\times {10}^{-6}}\right) \\ =7.1° \end{gathered}$$

Now direction of net force from vertical y-axis is

$$\begin{gathered} \left(90-7.1\right)° \\ =82.9° \end{gathered}$$

The diagram is

Gravitational Force is

$$\begin{gathered} \text{F=G}\frac{{\text{m}}_1{\text{m}}_2}{{\text{r}}^2} \\ \text{where} \\ {\text{G=Gravitationalconstant=6.67×10}}^{-11}{\text{N·m}}^2{\text{/kg}}^2 \\ {\text{m}}_1{\text{,m}}_2\text{aremassesofdifferentobjects} \\ \text{risdistancebetweentwoobjects} \end{gathered}$$

Gravitational force on 20 kg mass due to 5kg mass is

$$\begin{gathered} {\text{F}}_1{\text{=6.67×10}}^{-11}\text{·}\frac{5\times 20}{0.2^2} \\ {\text{=1.67×10}}^{-7}\text{N} \end{gathered}$$

This force is in vertical direction.

Perpendicular distance between 5 kg mass and 10 kg mass is$\sqrt{5^2+{10}^2}=22.37cm$

Gravitational force on 20 kg mass due to 5kg mass is

$$\begin{gathered} {\text{F}}_1{\text{=6.67×10}}^{-11}\text{·}\frac{5\times 10}{0.{22}^2} \\ {\text{=6.9×10}}^{-8}\text{N} \end{gathered}$$

The direction of force from horizontal direction.

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{20}{10}\right) \\ =63.43° \end{gathered}$$

The horizontal component of $F_2$is

localid="1648404758938" $$\begin{gathered} F_H=F_2\cos \theta \\ =6.9\times {10}^{-8}·\cos 63.43 \\ =3.0\times {10}^{-8}N \end{gathered}$$

The vertical component of $F_2$is

role="math" localid="1648405091261" $$\begin{gathered} F_V=F_2\sin \theta \\ =6.9\times {10}^{-8}·\sin 63.43 \\ =6.17\times {10}^{-8}N \end{gathered}$$

Now, total force in vertical direction is

$$\begin{gathered} F_{v^{\text{'}}}=F_1+F_v \\ =1.67\times {10}^{-7}+6.17\times {10}^{-8} \\ =2.287\times {10}^{-7}N \end{gathered}$$

Net force is

$$\begin{gathered} F=\sqrt{{F_H}^2+{F_{V\text{'}}}^2} \\ =\sqrt{{\left(3\times {10}^{-8}\right)}^2+{\left(2.287\times {10}^{-7}\right)}^2}N \\ =2.3\times {10}^{-7}N \end{gathered}$$

Direction of net force from horizontal x-axis is

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{F_{V\text{'}}}{F_H}\right) \\ ={\tan }^{-1}\left(\frac{2.287\times {10}^{-7}}{3\times {10}^{-8}}\right) \\ =82.53° \end{gathered}$$

Now direction of net force from vertical y-axis is

$\left(90-82.53\right)°=7.46°$