What are the magnitude and direction of the net gravitational

force on the 20.0 kg mass in FIGURE P13.36?

The given diagram is

Gravitational Force is

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

Distance between 20kg mass and 10kg mass is

$$\begin{gathered} d=\sqrt{{20}^2+{10}^2} \\ =22.36 \end{gathered}$$

Gravitational force on 20kg mass due to 10 kg mass is

$$\begin{gathered} F=6.17\times {10}^{-11}·\frac{20\times 10}{22.{36}^2} \\ =3.14\times {10}^{-7}N \end{gathered}$$

The direction of force from vertical direction is

$$\begin{gathered} \theta ={\tan }^{-1}\frac{5}{20} \\ =14° \end{gathered}$$

The horizontal component of the force is

$$\begin{gathered} F_H=F_1\sin \theta \\ =3.14\times {10}^{-7}·\sin 14° \\ =0.76\times {10}^{-7}N \end{gathered}$$

Total horizontal force is

$F_{H^{\text{'}}}=F_H-F_H=0N$

The vertical component of the force is

$$\begin{gathered} F_V=F_1\cos \theta \\ =3.14\times {10}^{-7}·\cos 14° \\ =3\times {10}^{-7}N \end{gathered}$$

Total force in vertical direction is

$$\begin{gathered} F_{V^{\text{'}}}=F_V+F_V \\ =3\times {10}^{-7}+3\times {10}^{-7}N \\ =6\times {10}^{-7}N \end{gathered}$$

Net force is

$$\begin{gathered} R=\sqrt{{F_H}^2+{F_V}^2} \\ =\sqrt{0^2+{\left(6\times {10}^{-7}\right)}^2} \\ =6\times {10}^{-7}N \end{gathered}$$

Direction of force is

$$\begin{gathered} \theta ={\tan }^{-1}\frac{F_V}{F_H} \\ ={\tan }^{-1}\frac{6\times {10}^{-7}}{0} \\ =90° \end{gathered}$$

This is from horizontal x-axis.