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I know that the magnitude of a torque increases with the distance from the axis of rotation.

What is the reason of that : how can a force exert a greater torque when the force is applied at a greater distance from the axis of rotation?

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    $\begingroup$ Because torque is $\vec \tau = \vec r \times \vec F$? $\endgroup$ – ACuriousMind Nov 20 '14 at 15:14
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A net force on an object likes to act through the centre of mass of that object; in this case, the middle of the door. Apply a force to the centre of mass and it will accelerate uniformly because there is an even amount of matter on all opposing sides that can resist the motion with an even amount of inertia. If a force is applied off the centre of mass, it will cause rotations because there is now one side that contributes more inertia (and so is more resistive to change in motion) than its opposing side. The opposing side, therefore, accelerates faster and the result is rotation. However, you still only have one applied force and the total acceleration applied to the centre of mass must obey Newton's Second Law. So if one side is accelerating faster than the centre of mass, the other side must accelerate slower (or even backwards sometimes) to compensate. That is for a free object (not fixed by a hinge or anything).

For the door, it's very similar except that the hinge end cannot move. If you apply force at the hinge, it won't rotate because all of that force will be opposed by the wall to prevent the hinge from moving, so the door won't move. As the applied force moves away from the hinge, the process I described in the above paragraph kicks in. The more of the door is on the same side of the force as the hinge, the more inertia that side produces and the slower it accelerates. Again, because the hinge can't move, it provides an opposing force. But if that side accelerates slower, then the opposing force decreases and your net force increases (that means the door becomes easier to accelerate). As you move to the other end of the door, the hinge can actually gain the desire to accelerate backwards (like I described above), so the opposing force would push back in the same direction as the applied force.

I want to clarify something in case it was confusing. If you're pushing on the same side of the centre of mass of the door as the hinge, the door wants to rotate the other way (as if the hinge was on the other side). But the hinge is fixed and the net force you apply means the centre of mass has to go in the direction you push. So the hinge provides a counter-force large enough to prevent that end from accelerating but not large enough to stop the door from moving. This makes the free end accelerate faster to compensate, thus rotation occurs.

But what it comes down to is that the further from the hinge you apply a force, the less inertia will be on the side of the force opposite the hinge and the faster it will be able to accelerate and the less opposing force the hinge has to provide to cancel the acceleration on its end. This means your forces are more effective (less opposed) when further from the fixed point.

Hope this wasn't too hard to follow

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  • $\begingroup$ Can you tell me what would be the direction of torque when we would pull to open a door? $\endgroup$ – pcforgeek Nov 21 '14 at 1:06
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    $\begingroup$ Jim -- are you thinking about a spring hinge or something? The OP just means a completely ordinary swinging hinge. (The only reason it is colloquially "easier" to push further out, is leverage.) $\endgroup$ – Fattie Nov 24 '14 at 10:40
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    $\begingroup$ @JoeBlow I'm not thinking of a spring hinge; I'm trying to explain this without equations and the easiest way to do that is in the frame of reference of the CoM of the door. In that frame, this is how leverage works $\endgroup$ – Jim Nov 24 '14 at 14:06
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One other way of looking at it is that the total amount of work needed for opening the door is constant. The work done by a force is the force multiplied by the path length: $W = F\cdot s$. If the force changes along the path (or is not parallel to it), this expression generalises to an integral: $$W = \int_{s1}^{s2}\vec{F}\cdot \vec{ds}$$

From this, you can see that if you apply the force over a longer path, you need less force to perform the same amount of work. This is exactly what you're doing when you're pushing the door at a point further away from the hinge. The arc your hand makes becomes larger (see image), so you don't need to apply as much force to open the door.

Arc length difference

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Because you have to push it further.

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  • $\begingroup$ Force doesn't depend on the distance you need to push. You're thinking of work. For example, I could apply a force of 10 N on the door very close to the hinge, or very close to the edge. When I apply it close to the hinge, the door will not open very fast, which is not the case when I apply the force close to the edge. However, in both cases, the force I've applied is the same. $\endgroup$ – Pranav Hosangadi Nov 24 '14 at 7:55
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    $\begingroup$ Well no: of course, you can open a door with any force, at any point. You can use a tiny force near or far, or a big force near or far. What the OP "means" is how can you open it at the same speed with less force further out. Answer, you have to push for longer. (TBC your comment could, simply, exactly apply to the OP's question.) $\endgroup$ – Fattie Nov 24 '14 at 10:38

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