A 0.500 m long wrench is applied to a nut with a force of 80.0 N. Due to the cramped space, the force is exerted upward at a 60.0-degree angle with respect to a line from the bolt through the end of the wrench. What is the torque applied to the nut?

 To calculate the torque τ applied to the nut by the wrench, we use the formula: \[ τ = r \cdot F \cdot \sin(\theta) \]

where: - \( τ \) is the torque. - \( r \) is the lever arm, the distance from the axis of rotation to the point where the force is applied (0.500 m in this case). - \( F \) is the magnitude of the force applied (80.0 N). - \( \theta \) is the angle between the force vector and the lever arm (60.0 degrees here).

Now, we will substitute the given values into this formula:

\[ τ = 0.500 \, \text{m} \times 80.0 \, \text{N} \times \sin(60.0^\circ) \]

First, we need to find the value of \( \sin(60.0^\circ) \). The sine of a 60-degree angle is \(\frac{\sqrt{3}}{2}\). So, we compute:

\[ τ = 0.500 \, \text{m} \times 80.0 \, \text{N} \times \frac{\sqrt{3}}{2} \]

Now, we can calculate the numerical value of the torque:

\[ τ = 0.500 \, \text{m} \times 80.0 \, \text{N} \times 0.866 \] \[ τ = 40.0 \, \text{N} \cdot \text{m} \times 0.866 \] \[ τ = 34.64 \, \text{N} \cdot \text{m} \]

Therefore, the torque applied to the nut is 34.64 Newton meters.