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Old Oct 26th 2012, 10:35 AM   #1
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torque,rotational motion of an object

We all know that
T = Fd
Where F = applied force,T=torque about the pivot.
Mathematically,to produce the same size of torque,if F increases,d decreases.
If F decreases,d increases.
But,let's think about what's happening at the atomic level.
Please see the attached.
Suppose A is the pivot.
In picture 1 and 2,you are applying a force at different position.
By the formula,to produce the same size of clockwise torque about A,
In fact the force is applied on the whole object,
i.e.When you apply a force on E in figure 2,the force is actually acting on the atoms(or group of atoms) at E.By repulsive forces between atoms,force is "transmitted" to B,
so the object rotates.
In figure 1,you apply force at B,when the atoms at B move,they pull the other atoms forwards,so the whole object rotates about A.
(These are what I thought about what's happening in atoms when an object rotate.)
However,this analogy cannot explain why as d decreases F increases.
1.Can someone give me an explanation of the equation in terms of movement of atoms and molecules?
i.e. Why d decreases then F increases (for the same torque)

2.And why the atoms at pivot point don't perform rotational and translational motion?
Thx a lot.

I have been solving this for one week though I can't figure it out....
Please forgive for my foolishness.
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torque,rotational motion of an object-img_5071.jpg   torque,rotational motion of an object-img_5072.jpg   torque,rotational motion of an object-img_5073.jpg  
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Old Oct 29th 2012, 08:08 AM   #2
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The key is to recognize that molecules transmit shear forces as well as moments upon their neighbors. Imagine a force F pressing on a molecule - that force must be counteracted by the next molecule. But that force also creates a moment (or torque) of Fd, where d is the physical size of the molecule, and that moment must also be resisted by teh next molecule. So the second molecule experiences a force F at its top edge plus a moment Fd. This must be resisted at its bottom edge by a force F plus a moment of Fd + Fd = 2Fd. Which means the next molecule below that experiences a force F and moment 2Fd at its top edge, which must be resisted at its bottom edge by a force F plus a moment 3Fd. And so on... do this for the full length D of the object and you find that the resisting moment at the base is FD, where D is the length from the base to the point where the force is applied. And yes, the pivot also experience a translational force F as well.
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