Question

49
in the below figure, where should the doorknob be located to make the door easier to open?

figure: door with hinges at top, labeled axis of rotation, height h, doorknob on the door

multiple choice
○ along the rotational axis.
○ above the rotational axis.
○ along the lower edge.
○ closer to the rotational axis.

Explanation:

To open a door (rotating about an axis), torque \( \tau = rF\sin\theta \) is needed. A larger \( r \) (distance from axis) gives more torque for the same force. If the axis is horizontal (top of the door), the lower edge is farthest from the axis, so locating the doorknob there maximizes \( r \), making it easier to open.

Answer:

None of the provided options (along the rotational axis, above the rotational axis, along the lower edge, closer to the rotational axis) are correct. The correct location should be as far as possible from the rotational axis (e.g., on the edge opposite to the hinges, at a suitable height like the middle vertically) because torque \( \tau = rF\sin\theta \), and a larger \( r \) (distance from axis) gives more torque for the same force, making it easier to open. But if we have to choose from the given (maybe a typo in options, assuming "along the lower edge" was misphrased or there's a mistake, but re - evaluating the options:

Wait, maybe the original options had a mistake. The correct physical principle is that the doorknob should be far from the axis. But among the given, if we assume a misprint and "along the lower edge" is meant to be "far from the axis" (but no). Wait, maybe the options were miswritten. The correct answer based on torque (rotational force) is that the doorknob should be located as far as possible from the rotational axis (hinges). But since the options don't have that, there might be an error. However, if we consider the options again:

• Along the rotational axis: Torque would be zero ( \( r = 0 \) ), hard to open.
• Above the rotational axis: \( r \) is small, not best.
• Along the lower edge: This is far from the rotational axis (hinges are at the top, lower edge is far in terms of vertical? No, horizontal distance from hinges (axis) is what matters. Wait, the door's axis is the hinges (vertical axis? No, in the diagram, the axis is horizontal at the top). So the door rotates around a horizontal axis at the top. So the distance from the axis (top) vertically. Wait, no, a door rotates around a vertical axis (hinges on the side). Oh! Maybe the diagram has a horizontal axis by mistake. If it's a normal door, axis is vertical (hinges on the side). Then the doorknob should be far from the vertical axis (hinges), i.e., on the opposite edge. But in the given options, if we assume the axis is vertical (hinges on the left, axis is the left edge), then "along the lower edge" - no, the horizontal distance from the axis (left edge) is what matters. The farther from the left edge (axis), the more torque. So the doorknob should be on the right edge (far from axis). But the options don't have that. There's a mistake in the options. But if we have to choose from the given, maybe the intended answer is "Along the lower edge" (but that's incorrect). Wait, maybe the axis is horizontal (like a trapdoor), rotating around the top. Then the distance from the axis (top) is the vertical distance. So to get more torque, we need a larger vertical distance, so lower edge (since \( r \) is the distance from axis, so lower edge is farther from the top axis). So in that case, "Along the lower edge" would be correct, as \( r \) is larger, so torque \( \tau = rF \) (since \( \theta = 90^{\circ} \), \( \sin\theta = 1 \)) is larger, making it easier to open.

So the answer is: Along the lower edge.