Question 5 A coordinate system (in meters) is constructed on the surface of a pool table, and three objects are placed on the table as follows: a 2 kg object at the origin of the coordinate system, a 3 kg object at \( (0,2) \), and a 4 kg object at \( (4,0) \). a) Calculate the magnitude of the resultant gravitational force exerted by the other two objects on the object at the origin. b) Calculate the direction/angle of the resultant force

Let the gravitational force between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) be given by \[ F = G\frac{m_1 m_2}{r^2}, \] where \( G \) is the gravitational constant. The 2 kg object is located at the origin, the 3 kg object at \( (0,2) \), and the 4 kg object at \( (4,0) \). 1. **Force from the 3 kg Object at \( (0,2) \):** - Distance from the origin is \[ r_1 = 2 \, \text{m}. \] - The force magnitude is \[ F_1 = G\frac{2 \times 3}{(2)^2} = G\frac{6}{4} = \frac{3G}{2}. \] - The direction is along the positive \( y \)-axis (upward), so its vector form is \[ \vec{F}_1 = \left(0, \frac{3G}{2}\right). \] 2. **Force from the 4 kg Object at \( (4,0) \):** - Distance from the origin is \[ r_2 = 4 \, \text{m}. \] - The force magnitude is \[ F_2 = G\frac{2 \times 4}{(4)^2} = G\frac{8}{16} = \frac{G}{2}. \] - The direction is along the positive \( x \)-axis. In vector form, \[ \vec{F}_2 = \left(\frac{G}{2}, 0\right). \] 3. **Resultant Force on the Object at the Origin:** - The net force vector is the sum of the two force vectors: \[ \vec{F} = \vec{F}_1 + \vec{F}_2 = \left(\frac{G}{2}, \frac{3G}{2}\right). \] - (a) The magnitude of the resultant force is \[ |\vec{F}| = \sqrt{\left(\frac{G}{2}\right)^2 + \left(\frac{3G}{2}\right)^2} = \frac{G}{2}\sqrt{1 + 9} = \frac{G}{2}\sqrt{10}. \] - (b) The direction (angle \( \theta \) measured from the positive \( x \)-axis) is given by \[ \theta = \tan^{-1}\left(\frac{\frac{3G}{2}}{\frac{G}{2}}\right) = \tan^{-1}(3). \] In approximate numerical form, \[ \theta \approx 71.6^\circ. \] Thus, the answers are: (a) The magnitude of the resultant force is \[ \frac{G}{2}\sqrt{10}. \] (b) The direction of the force is \[ \theta = \tan^{-1}(3) \approx 71.6^\circ. \]