Question

Four masses are arranged as shown in the figure below. Determine the x and y components of the gravitational force on the mass (m). m=100 gr, x_0=15 cm, y_0=10 cm.

Answer 1

Please refer to diagram.

Now Force on m due to 2m is given by

$F_{B} = \frac{Gm_{1}m_{2}}{AB^{2}}=\frac{Gm\times 2m}{0.15^{2}}= 88.89Gm^{2}$ along + X axis

$F_{C} = \frac{Gm_{1}m_{3}}{AC^{2}}=\frac{Gm\times 3m}{0.18^{2}}= 92.59Gm^{2}$ Along AC at an angle 33.69⁰ with + X axis

$F_{D} = \frac{Gm_{1}m_{4}}{AD^{2}}=\frac{Gm\times 4m}{0.10^{2}}= 400Gm^{2}$ Along Y axis.

Now F_c will be resolved in components forms and added to respective x or y axis component to get F_x  and F_y Components of Net force on mass m placed at point A

$F_{x}= F_{B}+F_{C}\cos \theta$

$F_{x}= 88.89Gm^{2}+(92.59Gm^{2})\cos 33.69^{0} = 165.93 Gm^{2}$

$F_{y}= F_{D}+F_{C}\sin \theta$

$F_{y}= 400Gm^{2}+(92.59Gm^{2})\sin 33.69^{0} = 451.36 Gm^{2}$

$F_{x}= 165.93 Gm^{2} = 165.93\times 6.67\times 10^{-11}\times (0.100)^{2}= 11.07\times 10^{-11}N$

$F_{y}= 451.36 Gm^{2} = 451.36\times 6.67\times 10^{-11}\times (0.100)^{2}= 30.10\times 10^{-11}N$

So Net Force on m will be

$\vec{F_{net}}= (11.07\times 10^{-11}N)\hat{i}+(30.10\times 10^{-11}N)\hat{j}$