Question: Four uniform spheres, with masses m_A 40 kg ,m_B = 35 kg , m_C = 200 kg , and m_D = 50 kg , have (x, y) coordinates of(0,50 cm), (0,0) ,(-80 cm,0) , and (40 cm,0) , respectively. In unit-vector notation, what is the net gravitational force on sphere Bdue to the other spheres?

The mass of the spherewith coordinates is$m_A=40kg,\left(0,50cm\right)$

The mass of the spherewith coordinates is$m_B=35kg,\left(0,0cm\right)$

The mass of the spherewith coordinates is$m_c=200kg,\left(-80cm,0\right)$

The mass of the sphere with coordinates is $m_D=50kg,\left(40cm,0\right)$

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

We can use the concept of Newton’s law of gravitation and principle of superposition of forces to find the net force on the object.

Formula:

$$\begin{gathered} \overrightarrow{F}=\frac{GMm}{r^2}\hat{r} \\ \overrightarrow{F}=\sum _{i=1}^n\overrightarrow{F} \end{gathered}$$

Where,G is the gravitational constant ($6.67\times {10}^{-11} \frac{N.m^2}{k{g}^2}$ )

The force acting on the ball B due to ball A is ${\overrightarrow{F}}_{BA}$

The force acting on the ball B due to ball D is ${\overrightarrow{F}}_{BD}$

The force acting on the ball B due to ball C is ${\overrightarrow{F}}_{BC}$

According to Newton’s law of gravitation, the force of attraction is

$$\begin{gathered} \overrightarrow{F}=\frac{GMm}{r^2}\hat{r} \\ \overrightarrow{{\mathbf{F}}_{\mathbf{B}\mathbf{A}}}\mathbf{=}\overrightarrow{{\mathbf{F}}_{\mathbf{BD}}}\mathbf{+}\overrightarrow{{\mathbf{F}}_{\mathbf{BC}}} \\ {\stackrel{\rightharpoonup }{F}}_{net}=Gm\left[\left(\underset{n=1}{\overset{3}{\sum \frac{m_n{x}_n}{{r^3}_n}}}\right)\hat{i}+\left(\underset{n=1}{\overset{3}{\sum \frac{m_n{x}_n}{{r^3}_n}}}\right)\hat{j}\right] \\ {\stackrel{\rightharpoonup }{F}}_{net}=\left(6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(35\text{kg}\right)\left[\left(\frac{40\text{kg}\times 0}{{\left(0.50\text{m}\right)}^3}\right)-\left(\frac{200\text{kg}\times \left(-0.80\text{m}\right)}{{\left(-0.80\text{m}\right)}^3}\right)+\left(\frac{50\text{kg}\times \left(0.40\text{m}\right)}{{\left(-0.40\text{m}\right)}^3}\right)\right]\hat{i}+ \\ \left(6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(35\text{kg}\right)\left[\left(\frac{40\text{kg}\times 0.50\text{m}}{{\left(0.50\text{m}\right)}^3}\right)+\left(\frac{200\text{kg}\times \left(0\text{m}\right)}{{\left(-0.80\text{m}\right)}^3}\right)+\left(\frac{50kg\times \left(0\text{m}\right)}{{\left(0.40\text{m}\right)}^3}\right)\right]\hat{j} \\ =\left(3.7\times {10}^{-7}\text{N}\right)\hat{j} \end{gathered}$$

..The net gravitational force on sphere due to the other sphere is$\overrightarrow{F}=\left(3.7\times {10}^{-7}N\right)\hat{j}$