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Expert-verified Found in: Page 134 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # Four 7.5-kg spheres are located at the corners of a square of side 0.80 m. Calculate the magnitude and direction of the gravitational force exerted on one sphere by the other three.

The magnitude and direction of the gravitational force is ${F}_{0}=1.1×{10}^{-8}\mathrm{N}$ and $\varphi =45°$ .

See the step by step solution

## Step 1. Given Data

The mass of spheres is $m=7.5\mathrm{kg}$.

The side of the square is $d=0.8\mathrm{m}$.

## Step 2. Understanding the net force on the sphere

In this problem, the net forces on the sphere in the horizontal as well as vertical direction will be equal due to symmetry.

## Step 3. Estimating the magnitude of net force exerted on the sphere

The free body diagram of the sphere is as follows: The relation of net force in the horizontal direction is given by,

$\begin{array}{c}{F}_{\mathrm{x}}={F}_{\mathrm{R}}+F\mathrm{cos}\theta \\ {F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}+\left(G\frac{{m}^{2}}{{\left(\sqrt{2}d\right)}^{2}}\right)\mathrm{cos}\theta \end{array}$

Here, G is the gravitational constant, ${F}_{\mathrm{R}}$ is the force on the right direction, F is the force on the diagonal and $\theta$ is the angle made by each sphere whose value is $\theta =45°$.

On plugging the values in the above relation.

$\begin{array}{c}{F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}+\left(G\frac{{m}^{2}}{{\left(\sqrt{2}d\right)}^{2}}\right)\mathrm{cos}45°\\ {F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}+\left(G\frac{{m}^{2}}{{\left(\sqrt{2}d\right)}^{2}}\right)\frac{1}{\sqrt{2}}\\ {F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)\end{array}$

The force on the vertical direction will be having the same magnitude because of the symmetry.

## Step 4. Estimating the resultant of the net force exerted on the sphere

The relation of resultant of the net force given by,

$\begin{array}{l}{F}_{0}=\sqrt{{{F}_{\mathrm{x}}}^{2}+{{F}_{\mathrm{y}}}^{2}}\\ {F}_{0}=\sqrt{{{F}_{\mathrm{x}}}^{2}+{{F}_{\mathrm{x}}}^{2}}\\ {F}_{0}=\sqrt{2{{F}_{\mathrm{x}}}^{2}}\\ {F}_{0}=\sqrt{2}{F}_{\mathrm{x}}\end{array}$

On plugging the values in the above relation.

$\begin{array}{l}{F}_{0}=\sqrt{2}\left[G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)\right]\\ {F}_{0}=\sqrt{2}\left[\left(6.67×{10}^{-11}\mathrm{N}·{\mathrm{m}}^{2}/{\mathrm{kg}}^{2}\right)\frac{{\left(7.5\mathrm{kg}\right)}^{2}}{{\left(0.8\mathrm{m}\right)}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)\right]\\ {F}_{0}=1.1×{10}^{-8}\mathrm{N}\end{array}$

Thus, ${F}_{0}=1.1×{10}^{-8}\mathrm{N}$ is the required magnitude of the net force.

## Step 5. Estimating the direction of net force exerted on the sphere

The relation of direction of the net force given by,

$\mathrm{tan}\varphi =\frac{{F}_{\mathrm{y}}}{{F}_{\mathrm{x}}}$

On plugging the values in the above relation.

$\begin{array}{c}\mathrm{tan}\varphi =\left(\frac{G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)}{G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)}\right)\\ \varphi =45°\end{array}$

Thus, $\varphi =45°$ is the required direction of the net force. ### Want to see more solutions like these? 