This may be a stupid question, but it is something I always wondered.
Torque is effective due to the mass of the lever having a force applied to it, right? Is the length of the lever being used as a stand-in for the mass being affected (a longer lever would necessarily have more mass)? If the lever had no mass, would there be no torque? If the lever did not have a uniform mass distribution, would the difference in applied torque differ based on where on the lever you applied the force (is the derivative of torque with respect to mass not constant for a non-uniform mass distribution)?
Nope, the mass of the lever doesn't play in to the torque that is applied at all. All that matters is that the force be transmitted via atom-to-atom motion. It's the rigidity of the wrench that mediates that, not its mass.
Of course, to create rigid objects, practically speaking they need to be made of something that will have mass. So the rigidity and the mass are related in a very loose sense. In any case, from a Newtonian physics perspective, you'll see none of those terms in there - neither a "rigidity" nor a mass. The torque is simply the length of the wrench multiplied by the magnitude of the force.
In a more detailed analysis, you might consider the flexure of the wrench by analyzing the stress and strain inside the wrench. That would no longer treat the wrench as a perfect idealized body that is completely rigid, but rather a body that can stretch based on the internal compressive or tensile forces that arise inside of it. Sometimes we don't think of metals as being stretchy, but with enough force, they're not so different from a rubber band.
With all that said, once you consider the _dynamics_ of the situation - how the forces applied give rise to motion - then mass does come into play. If you apply torque to a wheel, that will cause an angular acceleration proportional to the mass of the wheel. If you were using a wrench to spin a gear, then the mass of the gear decides how fast it will start to spin. Also, the mass of the wrench matters here, since presumably it will be spinning too: some of your effort has to go into angularly rotating the wrench. So it would have a "parasitic" effect on how fast you could get your gear spinning.
The mass of the lever just adds or subtracts from any other external force(s) applied to the lever - depending on the direction of the external force relative to the direction of gravity.
So a heavy wrench is helpful when pushing down, while loosening a rusty nut. But a light carbon fibre wrench would be better when pulling up.
How much torque gravity adds or subtracts in that case is also dependent on the distribution of mass along the length of the lever.
And to add to the fun, the mass of the leaver does add momentum into the equation, which comes into play as extra work required when changing speed of the rotation (e.g. starting, stopping the turn). And how much momentum also depends on the distribution of mass along the length of the lever.
No, the mass of the object that has a force applied to it has absolutely nothing to do with the torque.
Here is how Wikipedia defines the torque caused by a force acting on an object with a rotation axis: Torque is the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation.
So the force generated by the torque is completely unaffected by the mass of the lever? Then, why does applying the force on a longer portion of the lever create more torque? I had thought it was because there is more mass acting on the point of rotation (longer lever = more mass).
> So the force generated by the torque is completely unaffected by the mass of the lever?
Yes, that's right.
> Then, why does applying the force on a longer portion of the lever create more torque?
Most of the answers to this question reduce, upon examination, to "that's how we define torque". We define the torque of 100 newtons at a lever distance of one meter as the product of 100 newtons and a meter, which we can call 100 newton-meters, which is equal to 1000 newtons at a lever distance of 0.1 meters.
But that doesn't really answer the question, which becomes, why is torque defined in this way an interesting thing to think about? And the answer is that if the lever is a rigid body free to rotate around a fulcrum, then 100 newtons at one meter in one direction will make it start to rotate, while 1000 newtons at 0.1 meters in the opposite direction will precisely cancel that "moment", as we call it, and there will be no tendency to start rotating. It's about what forces are needed to cancel each other.
Well, but, why should that be? Why does it take exactly 1000 newtons and not, say, 316.2 newtons? And I don't think I have a really good answer for that question. In the case of an elastic solid body it falls out of Hooke's law and the geometry of the situation, which you can reduce to two long, skinny triangles sharing a common side bisected by the fulcrum. But it seems to be much more general than that.
> I had thought it was because there is more mass acting on the point of rotation (longer lever = more mass).
Nope. You can try using a pair of scissors or a folding ladder as a lever, or pull in different directions on the end of a fixed-geometry lever. The lever's mass doesn't change, but the leverage certainly does.
A generalization which applies to levers, pulleys, and hydraulics is mechanical advantage while conserving energy. You have a system with input work and output work (energy) that are the same, ignoring frictional losses.
Recall that work is force over distance. The mechanical system relates the input and output distances by a scalar coefficient. Since the working distances are related by a ratio, the working forces are related by the reciprocal of that ratio.
You can find the lever and fulcrum ratio with simple geometry. The input and output lever segments are radii, and the travel is distance along two arcs. Since the arc length is directly proportional to radius, the ratio of lever radii translates directly to the same ratio of arc lengths, and the reciprocal ratio is the force multiplier. Your 10:1 lever sweeps 10:1 arc lengths and balances with 1:10 opposing forces.
Yes, that's an excellent point, but I think the lever law is more general than that. For example, it continues to apply when the lever in question is stationary, even though no value of the forces involved would violate conservation. In fact, it holds to higher precision in that situation because your measurements aren't confounded by vibration and accelerating masses.
Maybe you can derive it from some kind of generalization of Hooke’s Law to cover nonlinear stress–strain relationships, elastic hysteresis, anisotropy, viscoelastic behavior, and so on, but it's not obvious to me what that would be. Also, I feel like the concept of angular moments acting to produce angular acceleration is simpler and more general than all that stuff, but I'm not sure if conservation of energy and geometry alone are sufficient to derive it.
Torque is nothing more than "spinny" force. For example, sometimes you will see the term "generalized force" to mean both force and torque, because it doesn't really matter in some contexts. For example, if I have a robot arm that has some linear joints (like that of a 3D printer) and rotational joints (like that of an arm), you can talk about the generalized forces of each joint. Some of those generalized forces are linear (and people call that "force"), and some of them are rotational (and people call that "torque").
They are exactly analogous to (linear) velocity and _angular_ velocity.
When you talk about force _generating_ a torque, the only thing I can understand is how the linear velocity at a point on a disk (say you blow across its surface) _generates_ an angular velocity.
It's the distance that matters. If you push a car it will speed up. If you push it twice as far it will speed up more.
If you apply the same force using a longer lever then the end of the lever moves further (it follows a bigger circle) so you are applying the same force over a bigger distance.
Torque is effective due to the mass of the lever having a force applied to it, right? Is the length of the lever being used as a stand-in for the mass being affected (a longer lever would necessarily have more mass)? If the lever had no mass, would there be no torque? If the lever did not have a uniform mass distribution, would the difference in applied torque differ based on where on the lever you applied the force (is the derivative of torque with respect to mass not constant for a non-uniform mass distribution)?