7) A light (massless) rod of length L = 1.5 m is freely pivoted at one end. Three forces act as shownin the figure below. The force F3 acts at the midpoint. What is the torque due to each force? TakeF1 = 6.90 N, Fz = 4.00 N, F3 = 2.00 N, 0 = 20°, and a = 30°Look at picture draw picture and show all work looking for equations **Problem Description:**A light (massless) rod of length \( L = 1.5 \, \text{m} \) is freely pivoted at one end. Three forces act as shown in the figure below. The force \( \vec{F}_3 \) acts at the midpoint. What is the torque due to each force? The given values are:- \( F_1 = 6.90 \, \text{N} \)- \( F_2 = 4.00 \, \text{N} \)- \( F_3 = 2.00 \, \text{N} \)- \( \theta = 20^\circ \)- \( \alpha = 30^\circ \)**Diagram Explanation:**- The rod is pivoted vertically on the left.- Force \( F_1 \) is applied at an angle \(\alpha\) from the rod.- Force \( F_2 \) acts horizontally at the end of the rod.- Force \( F_3 \) is applied vertically downward at the midpoint of the rod.- The lengths and angles are labeled as follows: \( F_2 \) acts at the full length \( L = 1.5 \, \text{m} \), and \( F_3 \) acts at \( \frac{L}{2} = 0.75 \, \text{m} \).**Torque Calculation:**Torque \(\tau\) due to a force is calculated using the formula:\[\tau = r \cdot F \cdot \sin(\theta)\]Where:- \( r \) is the lever arm length from the pivot.- \( F \) is the force magnitude.- \( \theta \) is the angle between the force and lever arm.Apply this formula for each force:1. **Torque due to \( F_1 \):** \[ \tau_1 = L \cdot F_1 \cdot \sin(\alpha) \]2. **Torque due to \( F_2 \):** \[ \tau_2 = L \cdot F_2 \cdot \sin(\theta) \]3. **Torque due to \( F_3 \):** \[ \tau_3 = \frac{L}{2} \cd

Answered: Image /qna-images/answer/ce7aba24-e8e4-45af-829d-bc626702b1e7.jpgWe know torque produced depends on the magnitude of the force and the perpendicular distance between the point about which torque is calculated and the point of application of force.