In opening a door, a 39.5 N force is applied as shown in the figure below. Find the torque produced by this force with respect to point O. (Let r = 0.875 m and ? = 62.0.)Magnitude _____Direction _____ The image depicts a lever arm setup in a two-dimensional coordinate system, with a horizontal x-axis and a vertical y-axis. The lever extends from point O towards the right, ending at a point where a force is applied. Key components in the diagram:1. **Coordinate System**: The x and y axes are shown with the origin labeled as O. 2. **Lever Arm**: The horizontal rod extends from point O to the right. The lever arm's position is indicated by a distance labeled \( r \) from point O to the point where the force is applied. 3. **Force Application**: A force \( \mathbf{F} \) is applied at the end of the lever arm. The force is represented by a red arrow pointing downward and outward at an angle. This angle is labeled as \( \theta \).4. **Angle \( \theta \)**: This is the angle between the lever arm (horizontal) and the direction of the force \( \mathbf{F} \).This setup is typically used to study torque and rotational motion in physics. The torque (\( \tau \)) produced by the force can be calculated using the equation \( \tau = r \cdot F \cdot \sin(\theta) \), where \( r \) is the distance from the pivot point (O) to the point of force application, \( F \) is the magnitude of the force applied, and \( \theta \) is the angle between the lever arm and the force vector.

Force F = 39.5 N , r = 0.875 m Angle = 62 ° torque is a dot product, so you multiply the radius vector and force vector tail to tail. since Cos component cannot provide torque as it acts along the pivot . Torque = r× F Torque = rF Sinθ Torque = 0.875 ×39.5 Sin 62 Torque = 30.51 Nm Torque = 30.51 Nm.