Question

Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Particle 3 has mass 2.6 kg and is at the origin. What is the direction of the net gravitational force on particle 3, measured as a counterclockwise angle from the +x axis?

Answer 1

To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

$F_g = \frac{Gm_1m_2}{r^2}$

Where,

G = Gravitational Universal Force

$m_i =$ Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

$m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m$

Applying the previous equation at X-Axis,

$F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N$

Applying the previous equation at Y-Axis,

$F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N$

Therefore the angle can be calculated as,

$tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°$

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.