How Is Torque Related to Frictional Force?

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Torque is related to frictional force through the equation T = r x F, where r is the distance from the center of rotation to the point of force application. In the case of a box on the ground, the static and kinetic friction coefficients (k_s and k_k) determine when the box will stop rotating. The moment of inertia and angular acceleration are also crucial, as they relate to the torque applied to the box. The frictional force is independent of the contact area and is calculated using F_friction = μN, where N is the normal force. To determine the torque that stops the box from spinning, the frictional force must be multiplied by the distance from the center of rotation.
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.
\omega = 0

\omega = \omega_0 + \alpha t

\tau = I \alpha

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

\alpha = \frac{\tau}{I}

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

then what the heck is torque(T)? If i insert \tau = r \times F 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 \tau = r \times F 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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