91Ó°ÊÓ

The given data is listed below:

• The mass of first uniform sphere is${\mathrm{m}}_1=50\mathrm{kg}$.
• The mass of second uniform sphere is$m_2=80kg$.
• The mass of third uniform sphere is$m_3=0.500kg$.
• The centre is at point is x=0 and y=3m .
• The third sphere is placed at x=4 m and y=0.

In this problem, the relation of gravitational force will be used to determine the magnitude and direction of the third sphere having a mass of 0.500 kg.

The relation from Pythagoras theorem is,

The following is the schematic diagram of the three spheres.

The relation of gravitational force on sphere 1 can be written as,

$F_1=G\frac{m_1{m}_3}{r_{13}^2}$

Here,G is the gravitational constant whose value is $\left(6.67×{10}^{-11}{\mathrm{m}}^3\mathrm{kg}.{\mathrm{s}}^2\right)$.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_1=\left[\frac{\left(6.67×{10}^{-11}{\mathrm{m}}^3\mathrm{kg}.{\mathrm{s}}^2\right)×\left(60\mathrm{kg}\right)×\left(0.5\mathrm{kg}\right)}{{\left(4\mathrm{m}\right)}^2}\right] \\ {\mathrm{F}}_1=1.25×{10}^{-10}\mathrm{N} \end{gathered}$$

The relation of gravitational force on sphere 2 can be written as,

$F_2=G\frac{m_2{m}_3}{r_{23}}$

Here,G is the gravitational constant.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_2=\left[\frac{\left(6.67×{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}.{\mathrm{s}}^2\right)×\left(80\mathrm{kg}\right)×\left(0.5\mathrm{kg}\right)}{{\left(5\mathrm{m}\right)}^2}\right] \\ {\mathrm{F}}_2=1.06×{10}^{-10}\mathrm{N} \end{gathered}$$

The total force on the x direction is calculated as,

$$\begin{gathered} F_x=F_{1x}+F_{2x} \\ {F}_x=F_{1x}+\left(-F_x\cos \mathrm{θ}\right) \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} F_x=\left(-1.25×{10}^{-10}N\right)+\left(-\left(1.06×{10}^{-10}N\right)\left(\frac{4m}{5m}\right)\right) \\ {F}_x=-2.10×{10}^{-10}N \end{gathered}$$

The total force on y direction is calculated as,

$$\begin{gathered} F_y=F_{1y}+F_{2y} \\ {F}_y=0+\left(F_y\sin \mathrm{θ}\right) \end{gathered}$$

Here, $F_{1y}$is the vertical gravitational force due to sphere 1 whose value is zero.

Substitute the values in the above relation.

$$\begin{gathered} F_y=0+\left(\left(1.06×{10}^{-10}N\right)\left(\frac{3m}{5m}\right)\right) \\ {F}_y=6.36×{10}^{-11}\mathrm{N} \end{gathered}$$

The relation of resultant force can be written as,

$F_R=\sqrt{F_x^2+F_y^2}$

Substitute the values in the above relation.

$$\begin{gathered} F_R=\sqrt{{\left(-2.10×{10}^{-10}N\right)}^2+{\left(6.36×{10}^{-11}N\right)}^2} \\ {F}_R=2.19×{10}^{-10}\mathrm{N} \end{gathered}$$

The direction is calculated as,

$$\begin{gathered} \tan \mathrm{θ}=\frac{F_y}{F_x} \\ \tan \mathrm{θ}=\left(\frac{6.36×{10}^{-11}N}{-2.10×{10}^{-10}N}\right) \\ \mathrm{θ}=-16.849° \\ \mathrm{This}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}, \\ \mathrm{θ}=180°-16.849° \\ \mathrm{θ}=163.16° \\ \mathrm{Thus},\mathrm{the}\mathrm{magnitude}\mathrm{and}\mathrm{direction}\mathrm{is}2.19×{10}^{-10}\mathrm{N}\mathrm{and}163.16°. \end{gathered}$$

The schematic diagram of the forces on the sphere is shown below.

As given in the question, the net force will be zero.

The total force can be calculated as,

$$\begin{gathered} F_{net}=0 \\ {F}_1=F_2 \\ \frac{G{m}_1{m}_2}{y^2}=\frac{G{m}_3{m}_2}{{\left(3m-y\right)}^2} \\ \mathrm{Here},\mathrm{y}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{distance}\mathrm{at}\mathrm{which}\mathrm{third}\mathrm{sphere}\mathrm{is}\mathrm{located}. \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ \frac{{\mathrm{m}}_1}{{\mathrm{y}}^2}=\frac{{\mathrm{m}}_3}{{\left(3\mathrm{m}-\mathrm{y}\right)}^2} \\ \frac{60\mathrm{kg}}{{\mathrm{y}}^2}=\frac{80\mathrm{kg}}{{\left(3\mathrm{y}-\mathrm{y}\right)}^2} \\ \left(3\mathrm{m}-\mathrm{y}\right)\sqrt{\left(60\mathrm{kg}\right)}=\sqrt{\left(80\mathrm{kg}\right)}\mathrm{y} \\ \mathrm{y}=1.39\mathrm{m} \\ \mathrm{Thus},\mathrm{the}\mathrm{third}\mathrm{sphere}\mathrm{will}\mathrm{be}\mathrm{placed}\mathrm{at}1.39\mathrm{m}. \end{gathered}$$