The magnitude of the torque is $$ \tau = rF\sin\theta $$. We want to maximize $$ r \sin\theta $$. In case A, the distance to a corner is $$ r = \sqrt{(L/2)^2 + (L/2)^2} = L/\sqrt{2} $$, and the force is perpendicular ($$ \sin\theta=1 $$), so $$ \tau_A = (L/\sqrt{2})F \approx 0.707LF $$. In case B, the distance is $$ r=L/2 $$ and the force is perpendicular, so $$ \tau_B = (L/2)F = 0.5LF $$. In case C, the distance is $$ r = L/\sqrt{2} $$ but the angle is not 90 degrees; the torque is $$ LF/2 = 0.5LF $$. In case D, the force points toward the pivot, so $$ \theta=180^\circ $$ and the torque is zero. Comparing the magnitudes, case A produces the greatest torque.

The original torque magnitude is given by $$ \tau = rF $$, since the force is applied perpendicularly. The new torque, $$ \tau' $$, is calculated with the new force $$F' = 2F$$ and new distance $$r' = r/2$$. Thus, $$ \tau' = r'F' = (r/2)(2F) = rF $$. The new torque is equal to the original torque, $$ \tau' = \tau $$.

Torque is given by the expression $$ \vec{\tau} = \vec{r} \times \vec{F} $$, and its magnitude is $$ \tau = rF\sin\theta $$. To maximize the torque for a given force magnitude $$F$$, the distance from the pivot (the bolt), $$r$$, must be maximized, and the angle $$ \theta $$ between the position vector $$ \vec{r} $$ and the force vector $$ \vec{F} $$ must be $$90^\circ$$ (so $$ \sin\theta = 1 $$). Therefore, the force should be applied as far from the bolt as possible and perpendicular to the handle.

Torque is calculated as the cross product $$ \vec{\tau} = \vec{r} \times \vec{F} $$. Substituting the given vectors: $$ \vec{\tau} = (r_0 \hat{j}) \times(F_0 \hat{k}) = r_0 F_0 (\hat{j} \times \hat{k}) $$. Using the right-hand rule for unit vectors, $$ \hat{j} \times \hat{k} = \hat{i} $$. Therefore, the torque vector is in the positive x-direction ($$+\hat{i}$$).

In static equilibrium problems, it is often advantageous to choose a pivot point where one or more unknown forces are acting. This is because the lever arm for those forces is zero, and thus they produce zero torque, simplifying the torque equation. At the point of contact with the floor, both the normal force from the floor and the static friction force act. By choosing this point as the pivot, these two unknown forces are eliminated from the torque equation, making it easier to solve for other unknowns.

The position vector is $$ \vec{r} = r\hat{i} $$ (positive x-axis) and the force vector is $$ \vec{F} = F\hat{j} $$ (positive y-direction). The torque is $$ \vec{\tau} = \vec{r} \times \vec{F} = (r\hat{i}) \times(F\hat{j}) = rF (\hat{i} \times \hat{j}) $$. By the right-hand rule for cross products, $$ \hat{i} \times \hat{j} = \hat{k} $$. Therefore, the torque vector is in the positive z-direction.

The original torque is $$ \vec{\tau}_O = \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 $$. The new position vectors relative to $$ \vec{r}_P $$ are $$ \vec{r}'_1 = \vec{r}_1 - \vec{r}_P $$ and $$ \vec{r}'_2 = \vec{r}_2 - \vec{r}_P $$. The new torque is $$ \vec{\tau}_P = (\vec{r}_1 - \vec{r}_P) \times \vec{F}_1 + (\vec{r}_2 - \vec{r}_P) \times \vec{F}_2 = (\vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2) - (\vec{r}_P \times \vec{F}_1 + \vec{r}_P \times \vec{F}_2) = \vec{\tau}_O - \vec{r}_P \times(\vec{F}_1 + \vec{F}_2) $$. The torque is independent of the pivot only when the net force $$(\vec{F}_1 + \vec{F}_2)$$ is zero.

The magnitude of the cross product is defined as $$ |\vec{r} \times \vec{F}| = rF\sin\theta $$, where $$ \theta $$ is the angle between the vectors. This can be grouped as $$ r(F\sin\theta) $$. The term $$ F\sin\theta $$ is the magnitude of the component of the force vector $$ \vec{F} $$ that is perpendicular to the position vector $$ \vec{r} $$. Thus, the torque magnitude is the full distance multiplied by the perpendicular component of the force.

For rotational equilibrium, the sum of the torques must be zero. The block creates a counter-clockwise torque of $$ (L/2)mg $$. The applied force F creates a clockwise torque of $$ rF $$, where r is the distance from the pivot. To balance, $$ rF = (L/2)mg $$, so $$ F = (L/2r)mg $$. To minimize the required force $$F$$, the distance $$r$$ must be maximized. The maximum possible value for $$r$$ is $$L/2$$, which occurs when the force is applied at the far right end.

This question tests AP Physics C: Mechanics skills, specifically understanding torque and rotational dynamics with focus on comparing torques from forces at different angles. For maximum torque, the force should be perpendicular to the lever arm. F₁ acts perpendicular to the door, producing τ₁ = (1.00 m)(10 N)sin(90°) = 10 N·m. F₂ acts at 30° to the door, producing τ₂ = (1.00 m)(10 N)sin(30°) = 5 N·m. Choice A is correct because the perpendicular force F₁ produces greater torque (10 N·m) than the angled force F₂ (5 N·m). Choice C might result from not considering the angle effect on torque magnitude. To help students: Demonstrate that torque is maximized when force is perpendicular to the lever arm. Use visual aids to show how the effective force component decreases as the angle deviates from 90°.