Two uniform spheres, each of mass 0.26 kg , are fixed at points A and B (Fig. E13.5).Find the magnitude and direction of the initial acceleration of a uniform sphere with mass 0.01 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.

Mass of the uniform spheres, M = 0.26 kg

Mass of the sphere being pulled, m = 0.01 kg

Distance between each of the uniform spheres and the small sphere,

$\begin{array}{l}r=10\mathrm{cm}\\ =10.\left(1\mathrm{cm}\times \frac{1m}{100\mathrm{cm}}\right)\\ =0.1\mathrm{mm}\end{array}$

Distance between the two uniform spheres is, AB = 16 cm

The gravitational force between two masses${\mathit{m}}_{\mathbf{1}}$and ${\mathit{m}}_{\mathbf{2}}$at a distance from each other is

$\mathit{F}\mathbf{=}\mathit{G}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$ ..... (1)

The force is attractive and is directed along the line joining the two masses. Here,Gis the gravitational constant having value

Here,

$G=6.67\times {10}^{-11}N·m^2/k{g}^2$

From equation (I), the force on m by each of the spheres of mass M is.

$F=G\frac{mM}{r^2}$

Each uniform sphere tries to pull m towards itself.

The horizontal components of these two force are oppositely directed and cancel each other. The vertical components add up. The net force on is thus,

role="math" localid="1668059782497" $F_{net}=2F\cos \theta$

Here, $\theta$ is the angle between AP or BP and the vertical axis. Name the mid point of AB as C. Then from triangle ACP,

$\begin{array}{r}\cos \theta =\frac{CP}{AP}\\ =\frac{6\mathrm{cm}}{10\mathrm{cm}}\\ =0.6\\ F_{\text{net}}=2F\times 0.6\\ =1.2F\\ =1.2G\frac{mM}{r^2}\end{array}$

The acceleration of m is thus

$\begin{array}{l}a=\frac{F_{net}}{m}\\ =1.2G\frac{M}{r^2}\end{array}$

Substitute the values to get

$\begin{array}{r}a=1.2\times 6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\times \frac{0.26\mathrm{kg}}{(0.1\mathrm{m}{)}^2}\\ =208\times {10}^{-11}\times \left(1\mathrm{N}\times \frac{1\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)\cdot \left(1{\mathrm{m}}^2\right)\cdot \left(\frac{1}{1{\mathrm{kg}}^2}\right)\cdot \left(1\mathrm{kg}\right)\cdot \left(\frac{1}{1{\mathrm{m}}^2}\right)\\ =2.08\times {10}^{-9}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the required acceleration is $2.08\times {10}^{-9}\mathrm{m}/{\mathrm{s}}^2$.