91Ó°ÊÓ

From figure, masses of three spheres at corner of the triangle =m

Mass of the central sphere =M

This problem is based on Newton’s gravitational law. We can find the net gravitational force by using vector addition of forces.

Formula:

Force of Gravitation,$F=\frac{G{m}_1{m}_2}{r^2}$ (i)

Using equation (i), Force due to mass m on particlecan be written by:

$F_m=\frac{Gm{m}_4}{r^{2^{}}}$

Along x axis, the form of the force can be written as:

$F_{mx}=\frac{Gm{m}_4}{r^{2^{}}}\cos \left(\mathrm{θ}\right)$ (ii)

Similarly, along y axis, the force can be written as:

$$\begin{gathered} F_{my}=\frac{Gm{m}_4}{r^{2^{}}}\sin \left(\mathrm{θ}\right) \\ =\frac{Gm{m}_4}{r^{2^{}}}\cos \left(\mathrm{θ}\right) \end{gathered}$$ (iii)

Using equation (i), Force due to masscan be written as:

$F_M=\frac{GM{m}_4}{r^{2^{}}}$

Using equation (ii), the force due to mass M along x-axis can be written as:

$F_{Mx}=0$

Similarly, the force due to mass M along y-axis can be written as:

$F_{My}=\frac{GM{m}_4}{r^{2^{}}}$ (iv)

As net force on the central sphere due to the three masses m, m, M is zero. (Given) $$\begin{gathered} \underset{}{∑F_y=0} \\ {F}_{My}-2F_{my}=0 \end{gathered}$$

From equation (iii) and (iv), we get

$\frac{GM{m}_4}{r^{2^{}}}-2\frac{Gm{m}_4}{r^{2^{}}}\sin \left(\mathrm{θ}\right)=0$

Hence,

$M=2m\sin \left(30\right)$

Therefore, the mass M is

As we double mass, it will double all the three forces. Direction of forces will not change here. So, the net force will be zero.