How Is Torque Related to Frictional Force?

PJani
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I am trying to figure out how exactly Angular friction ("torque") is connected with friction force.

Is there any connection between torque friction and force friction?
 
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Well torque is related to force by T=rxF
 
Yes i know for that relation but what if i have box on the ground which fully touches ground with all bottom area. Which i spin and i know that static friction is k_s and kinetic is k_k. How can i figure out when box will stop rotating?

Box has some moment of inertia and some mass...

I know that formula is.
[tex]\omega = 0[/tex]

[tex]\omega = \omega_0 + \alpha t[/tex]

[tex]\tau = I \alpha[/tex]

If i toss around elements
[tex]\frac{(\omega - \omega_0)}{\alpha} = t[/tex]

[tex]\alpha = \frac{\tau}{I}[/tex]

and if i join equations together
[tex]\frac{(0 - \omega_0) I}{\tau} = t[/tex]

then what the heck is torque(T)? If i insert [tex]\tau = r \times F[/tex] Then i have missing variable r.
 
r would be the distance from the center to where the force is applied.
 
Yes i know, that why i can't get it because box spins on ground and it has full contact on ground whole area is in contact. That why i can't figure out what i could do with r. because r is scalar and not contact area!
 
PJani said:
Yes i know, that why i can't get it because box spins on ground and it has full contact on ground whole area is in contact. That why i can't figure out what i could do with r. because r is scalar and not contact area!

Friction is independent of area and given by Ffriction= μN where N is the normal reaction.
 
I can use [tex]\tau = r \times F[/tex] for point contacts with ground and then calculate produced torque. Which gives me no problem.

But what if i spin box around ground normal. how can i get then torque which will stop the box from spining on ground?
 
Multiply the frictional force by the distance from the center of rotation.
 

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