
#19
Dec2112, 03:55 AM

P: 5,462

Please note that you cannot apply the equations and arguments of equilibrium to an object that is not in equilibrium.
An object that is accelerating is not in equilibrium. In particular when an object is not in equilibrium there is never any need to 'balance' anything. This is a very common error. Hope this general comment helps. 



#20
Dec2112, 02:55 PM

P: 197





#21
Dec2512, 05:24 AM

P: 342





#22
Dec2512, 06:10 PM

P: 440

Here is a little exercise:
When we apply a force of 1 N to the end of a lever, the lever exerts a force of 10 N on a mass near the fulcrum. The force 10 N accelerates the mass. Now we add a second identical mass to the end of the lever, I mean that end where the force is applied. Question: how does the force exerted on the mass near the fulcrum change, when the force applied to the end of the lever is still 1 N? My answer: New force is about one hundreth of the old force. 



#23
Dec2612, 01:29 PM

P: 3,552





#24
Dec2612, 04:20 PM

P: 16

Ok I'll try my hand at this. First a little elaboration on internal forces.
Imagine two situations: you're out in the middle of space and in the first case you see a rigid rod rotating around and around with no external forces or acceleration. In the other case you see a rope floating around in space and you give it a little spin, but because it's not rigid, it bends around a bit. What's the difference between these two cases? It's the intramolecular forces keeping the rod together and keeping the rope together. In general you would need quantum mechanics to understand these forces and explain why one material is rigid and another is not. No one before the 20th century understood them, so you wouldn't expect that a first year classical physics class would be able to explain them. Instead we simply assume that whatever forces are needed to keep the rod rigid are at work when talking about a lever. If they weren't then we'd be talking about a rope and not a rigid lever and we'd need to use other equations. Now let's look at your case. As others have alluded to, you're trying to use a static equation for a dynamic situation. If you apply a torque M to the lever on one side, then it will take a force F=M/l to balance the torque so the lever won't rotate. So far so good. But then you wanted to start talking about acceleration of a point on the lever. That requires a bit more work. If you simply apply a torque M to one end of the lever, then the angular acceleration is M/i, where i is the moment of inertia of the lever system. So the linear acceleration of a test mass placed on the other side a distance of r from the fulcrum is rM/i. (Keeping in mind that the moment of inertia is for the lever AND the test mass). Assuming that the test mass doesn't affect the moment of inertia much, then the acceleration is indeed proportional to the distance r. In the other extreme when the lever is massless, then the size of the test mass matters. If the test mass has mass m, then the moment of inertia is m*(r squared) and the linear acceleration is M/(rm). In this case as you slide the mass closer to the fulcrum, the acceleration increases. That's simply because as you move it closer to the fulcrum it gets easier to turn the lever and if you keep the torque constant then the lever will accelerate more. Finally when the test mass is sitting right on the fulcrum then the tiniest torque will cause an infinite angular acceleration and the lever will just spin around and around because the lever system has zero moment of inertia. 



#25
Dec2612, 04:33 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

It might be better to look at energy rather than force or acceleration. There is a 'conservation of energy' law, but no 'conservation of force'. If we have an object we want to move "x" m from the fulcrum, and apply force f Newtons at a point "y" m from the fulcrum, moving the lever through an angle [itex]\theta[/itex] radians, we have applied the force, f, through distance [itex]x\theta[/itex] meters and so have done [itex]fx\theta[/itex] Joules of work. That must be the same energy applied at the other end. If we call the force applied there "F", have moved the lever through [itex]y\theta[/itex] meters and so we must have [itex]Fy\theta= fx\theta[/itex] and so we must have [itex]F/f= x/y[/itex].



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