When do friction forces stop acting?

AI Thread Summary
Friction forces act to resist motion, adjusting based on the applied force up to their maximum limit. When an object is pushed with a force less than the maximum static friction, the static friction matches the applied force, resulting in a net force of zero and preventing movement. As the object slows down due to kinetic friction, the net force can become negative if the applied force is insufficient to overcome friction. Once the object stops, static friction takes over, equalizing with any remaining applied force to maintain a net force of zero. Understanding these dynamics clarifies how friction operates in both static and kinetic scenarios.
Starby
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Another classic case of "I understand it in reality, but on paper it just makes no sense and I'm confused".

There is an object sliding north on a surface where the normal is 10N. The co-efficient for kinetic friction for the surface is 0.5. Simple math will tell me that the force of friction acting in the opposite direction (south) will be 0.5*10N = 5N. There's also the co-efficient for static friction for when the object isn't moving. Let's say it has the value of 0.6. So I need 6N to overcome static friction and make the object even budge. Everything makes perfect sense so far.

But what happens when a force applied isn't strong enough? Would there be a net force in some direction even though the object isn't moving? What happens in the moments when the object stops on its own (read below)?

t0 - Object has force 10N north. It slides across the surface.
t1 - The object is now moving very slowly due to friction, it's just about to stop.
t2 - The force North acting on the object is maybe 4N now. That isn't enough to continue, so the object stops. But what happens to the net force? Is it reduced to 0? If so, what makes it reduce to 0?

If I apply a weaker force than the static force the object doesn't move. What is resisting my force. For the cases where I push with a size not large enough, does friction resist with an equal force in the opposite direction?
 
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Starby said:
For the cases where I push with a size not large enough, does friction resist with an equal force in the opposite direction?
Exactly. Realize that ##\mu_s N## is the maximum value of static friction. Static friction will adjust to prevent slipping between surfaces (up to its limit); it can vary from 0 to the max value, as needed.
 
Starby said:
t2 - The force North acting on the object is maybe 4N now. That isn't enough to continue, so the object stops. But what happens to the net force? Is it reduced to 0? If so, what makes it reduce to 0?

The object will not stop immediately when the force drops to 4N. Remember that the net force only causes a deceleration and Force does not equal to motion.

But back to your question about friction: say the pushing force experienced by the object is 4N (at t2). The net force is now -1N. By Newton's second law, the object slows down and comes to a stop eventually. During this time, assuming the force remains at 4N, the kinetic friction is still 5N in the opposite direction up till the point where the object comes to a complete stop, then static friction kicks in, taking the value of 4N so that the net force on the object is now 0N and the object is unable to move further.

Starby said:
If I apply a weaker force than the static force the object doesn't move. What is resisting my force. For the cases where I push with a size not large enough, does friction resist with an equal force in the opposite direction?

Yes, if you apply a weaker force than the maximum static friction, then the magnitude of the static friction is simply the same magnitude as the applied force, such that the net force on the object is zero and the object remains at rest.
 
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