Question

a) What is the total force on \(m_{1}\) due to \(m_{2}, m_{3},\) and \(m_{4}\) if all four masses are located at the corners of a square of side \(a\) ? Let \(m_{1}=m_{2}=m_{3}=m_{4}\) b) Sketch all the forces acting on \(m_{1}\).

Short answer

Answer: The total gravitational force acting on mass \(m_{1}\) is given by: \(F_{total} = \sqrt{((F_{12} + F_{14})^{2} + F_{13})^2)}\) where \(F_{12}, F_{13}\), and \(F_{14}\) are the gravitational forces acting between \(m_{1}\) and \(m_{2}, m_{3},\) and \(m_{4}\), respectively.

Step-by-step solution

Understand the forces between masses

According to Newton's law of universal gravitation, the force \(F\) between two masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by: \(F = G \frac{m_{1}m_{2}}{r^{2}}\) where \(G\) is the gravitational constant.

Find the forces between the masses

Let's find the gravitational forces acting on \(m_{1}\) due to each of the other three masses. Force between \(m_{1}\) and \(m_{2}\): \(F_{12} = G \frac{m_{1}m_{2}}{a^2}\) Force between \(m_{1}\) and \(m_{3}\): Since \(m_{3}\) is located at the opposite corner of the square, the distance between \(m_{1}\) and \(m_{3}\) is the diagonal of the square, which is equal to \(a\sqrt{2}\). Therefore, we have: \(F_{13} = G \frac{m_{1}m_{3}}{(a\sqrt{2})^{2}}\) Force between \(m_{1}\) and \(m_{4}\): \(F_{14} = G \frac{m_{1}m_{4}}{a^2}\)

Find the vector components of the forces

The force between \(m_{1}\) and \(m_{3}\) acts along the diagonal whereas the other forces act along the sides of the square. Now, we will find the horizontal and vertical components of the forces \(F_{12}\), \(F_{13}\), and \(F_{14}\). Horizontal component of \(F_{12}\) is zero. Vertical component of \(F_{12}\) is \(F_{12}\). Horizontal component of \(F_{13}\) is \(F_{13}\cos{45^{\circ}}= \frac{F_{13}}{\sqrt{2}}\). Vertical component of \(F_{13}\) is \(F_{13}\sin{45^{\circ}}= \frac{F_{13}}{\sqrt{2}}\). Horizontal component of \(F_{14}\) is \(F_{14}\). Vertical component of \(F_{14}\) is zero.

Calculate the total force

To find the total gravitational force acting on \(m_{1}\), we need to sum up the horizontal and vertical components of individual forces. Total horizontal component = Horizontal component of \(F_{12}\) + Horizontal component of \(F_{13}\) + Horizontal component of \(F_{14}\) Total vertical component = Vertical component of \(F_{12}\) + Vertical component of \(F_{13}\) + Vertical component of \(F_{14}\) Now, we can find the total force \(F_{total}\) by adding the total horizontal and vertical components, which form a right-angled triangle: \(F_{total} = \sqrt{(Total\; horizontal\; component)^{2} + (Total\; vertical\; component)^{2}}\)

Plug in the values and compute the total force

For the given problem, plug in the values for \(F_{12}\), \(F_{13}\), and \(F_{14}\), and then compute the total force. \(F_{total} = \sqrt{((F_{12} + F_{14})^{2} + F_{13})^2)}\)

Sketch the forces acting on \(m_{1}\)

This is a graphical representation of the problem, showing the vector components of the forces: 1. Draw a square with side \(a\) and place the masses at its corners (\(m_{1}\) at the bottom-left corner). 2. Draw an arrow representing \(F_{12}\) pointing straight up from \(m_{1}\). 3. Draw an arrow representing \(F_{14}\) pointing straight right from \(m_{1}\). 4. Draw an arrow representing \(F_{13}\), pointing along the diagonal from \(m_{1}\) to \(m_{3}\). You can represent the horizontal and vertical components for this force by drawing dashed lines parallel to the other two forces. 5. Combine the three forces by drawing the resulting vector formed by the sum of the horizontal and vertical components of all the forces (\(F_{total}\)). This will be the total force acting on \(m_{1}\).