Question

What are the magnitude and direction of the net gravitational

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

Short answer

The magnitude and direction of the net gravitational force on 20kg mass are$6\times {10}^{-7}N$and$90°$respectively.

Step-by-step solution

Step 1. Given information

The given diagram is

Step 2. Formula used

Gravitational Force is

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

Step 3. Calculation

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

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

Step 4. magnitude and direction of force

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.