Suppose you apply a force of 40 N to a 0.25 -meter-long wrench attached to a bolt in a direction perpendicular to the bolt. Determine the magnitude of the torque when the force is applied at an angle of \( 120^{\circ} \) to the wrench. The magnitude of the torque is (Type an exact answer.)

To determine the torque (τ), you can use the formula: \[ τ = r \times F \times \sin(θ) \] Where: - \( r \) is the length of the wrench (0.25 m), - \( F \) is the force applied (40 N), and - \( θ \) is the angle between the wrench and the force (120°). Substituting the values: \[ τ = 0.25 \, \text{m} \times 40 \, \text{N} \times \sin(120°) \] Since \( \sin(120°) = \sin(180° - 60°) = \sin(60°) = \frac{\sqrt{3}}{2} \): \[ τ = 0.25 \times 40 \times \frac{\sqrt{3}}{2} \] \[ τ = 0.25 \times 20\sqrt{3} \] \[ τ = 5\sqrt{3} \] Thus, the magnitude of the torque is \( 5\sqrt{3} \, \text{N·m} \).