In this example, we explore how to calculate the torque produced by a force acting at different angles. Imagine you're fishing, holding a 3-meter-long fishing pole at an angle of 50 degrees above the x-axis. The pole has a mass of 2 kilograms, and a fish pulls on it with a force of 40 newtons directed at 20 degrees below the x-axis. The goal is to determine the torque about the axis of rotation at your hands.

To find the torque, we use the formula:

\(\tau = F \cdot r \cdot \sin(\theta)\)

where \(\tau\) is the torque, \(F\) is the force applied, \(r\) is the distance from the axis of rotation to the point of force application, and \(\theta\) is the angle between the force vector and the position vector.

First, we identify the position vector \(r\), which is simply the length of the fishing pole, measuring 3 meters. Next, we need to determine the angle \(\theta\). The angle between the force and the position vector is not directly one of the angles given (50 degrees or -20 degrees). Instead, we find the angle by visualizing both vectors originating from the same point.

By adjusting the vectors, we see that the angle between the position vector \(r\) and the force vector \(F\) is actually 70 degrees, calculated by adding the 50 degrees (the angle of the pole) and 20 degrees (the angle of the force below the x-axis).

Now, substituting the values into the torque equation:

\(\tau = 40 \, \text{N} \cdot 3 \, \text{m} \cdot \sin(70^\circ)\)

Calculating this gives a torque of approximately 113 newton-meters. This example illustrates the importance of correctly identifying the angle between the force and the position vector when calculating torque, especially in scenarios involving multiple angles.