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Q42.

Expert-verifiedFound in: Page 134

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 ${F}_{0}=1.1\times {10}^{-8}\mathrm{N}$ and $\varphi =45\xb0$ .

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

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

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

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 $\theta =45\xb0$.

On plugging the values in the above relation.

$\begin{array}{c}{F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}+\left(G\frac{{m}^{2}}{{\left(\sqrt{2}d\right)}^{2}}\right)\mathrm{cos}45\xb0\\ {F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}+\left(G\frac{{m}^{2}}{{\left(\sqrt{2}d\right)}^{2}}\right)\frac{1}{\sqrt{2}}\\ {F}_{\mathrm{x}}=G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)\end{array}$

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

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.

$\begin{array}{l}{F}_{0}=\sqrt{2}\left[G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)\right]\\ {F}_{0}=\sqrt{2}\left[\left(6.67\times {10}^{-11}\mathrm{N}\xb7{\mathrm{m}}^{2}/{\mathrm{kg}}^{2}\right)\frac{{\left(7.5\mathrm{kg}\right)}^{2}}{{\left(0.8\mathrm{m}\right)}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)\right]\\ {F}_{0}=1.1\times {10}^{-8}\mathrm{N}\end{array}$

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

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.

$\begin{array}{c}\mathrm{tan}\varphi =\left(\frac{G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)}{G\frac{{m}^{2}}{{d}^{2}}\left(1+\frac{1}{2\sqrt{2}}\right)}\right)\\ \varphi =45\xb0\end{array}$

Thus, $\varphi =45\xb0$ is the required direction of the net force.

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