1. The problem involves calculating the force needed to loosen a bolt using a wrench. 2. We know that torque ($\tau$) is given by the formula $$\tau = r \times F \times \sin(\theta)$$ where $r$ is the distance from the pivot point to the point of force application, $F$ is the force applied, and $\theta$ is the angle between the force and the lever arm. 3. Here, $r = 0.20$ m, the force perpendicular to the wrench is $F_\perp = 50$ N, and the angle for the new force $\theta = 30^\circ$ from perpendicular. 4. First, calculate the torque when the force is perpendicular (angle $= 90^\circ$): $$\tau = 0.20 \times 50 \times \sin(90^\circ) = 0.20 \times 50 \times 1 = 10~\text{Nm}$$ 5. This torque must be the same when the force is applied at a $30^\circ$ angle from perpendicular in order to loosen the bolt. 6. Let $F$ be the new force applied at $30^\circ$ from perpendicular. The angle between the force and lever arm is therefore $90^\circ - 30^\circ = 60^\circ$ because $0^\circ$ would be along the wrench. 7. So, we use torque formula: $$10 = 0.20 \times F \times \sin(60^\circ)$$ 8. Solve for $F$: $$F = \frac{10}{0.20 \times \sin(60^\circ)} = \frac{10}{0.20 \times \frac{\sqrt{3}}{2}} = \frac{10}{0.20 \times 0.866} = \frac{10}{0.1732} \approx 57.7\,\text{N}$$ 9. Therefore, if the force is applied at a 30-degree angle from perpendicular, it would take approximately 57.7 newtons to loosen the bolt.