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Suppose there is a crate in contact with the ground when an external force directed to the right is applied to it. I understand that if this external force does not exceed the maximum static frictional force: 1. the crate will not move and 2. the static frictional force will equal the external force being applied.

However, lets say that the external force is greater than the maximum static frictional force and the force is applied at a time t = 0s. It is clear now that at time t = 0s the crate will have some acceleration to the right.

My confusion is that since we know the crate will move, if a free body diagram of the forces acting on the crate at time t = 0 was drawn would the forces acting in the horizontal direction be:

  1. the max static friction force acting to the left and the external force to the right or
  2. since we know the crate will move, the kinetic friction force to the left and the external force to the right.

My reasoning is that since at time t = 0s, the crate is still at rest, we cannot use kinetic friction since the surfaces (ground and crate) are not sliding over one another. Using this reasoning the forces acting on the crate at t = 0 must be the max static friction force and the external force. Is this correct?

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Note that the concept of static and dynamic friction is an empirical approximation. In general, the way this is usually presented would lead one to assume that there is some discontinuous change in the forces as you go from $\vec{v} = 0$ to $\vec{v}\neq 0$. But really, the friction forces just describe the electromagnetic interactions between the ground and the crate, and there is no sudden discontinuous change as the object starts moving.

As such, in the regime of extremely small velocities I would expect this approximation to be invalid. Indeed, imagine we're moving at a velocity of 1cm per year - would you consider this to be a "static" or "dynamic" situation? Which force of friction is relevant?

Therefore, as with all empirical laws, it is important to not overstretch their realm of applicability. In general, once we have some appreciable speed, we can use dynamic friction to a good approximation, whereas at zero or negligible velocities we can use static friction.

What happens the crossover between these regimes is the hard part, where you generically need to look at the actual physical interactions to get the behavior right.

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My reasoning is that since at time t = 0s, the crate is still at rest, we cannot use kinetic friction since the surfaces (ground and crate) are not sliding over one another. Using this reasoning the forces acting on the crate at t = 0 must be the max static friction force and the external force. Is this correct?

It depends on the exact conditions at time t=0.

Up until the maximum static friction force the static friction force will match the applied force for a net force of zero and the crate will not move. When the applied force equals the maximum static friction force, motion of the crate is "impending". As soon as the applied force exceeds the maximum static friction force, the crate moves and the friction force changes from static friction to kinetic friction. Since the coefficient of kinetic friction is generally less than the coefficient of static friction, the kinetic friction force opposing the applied force will be less than the maximum static friction force before relative motion. You can see this progression in the friction plot of the following link:

http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

So, if t=0 corresponds to the time before the static friction has changed to kinetic friction, the crate will remain stationary. If t=0 corresponds to the time when static friction has changed to kinetic friction, and if the applied force is not changed (still has a magnitude equal to the the maximum static friction force before relative motion occurs) then there will be a net force on the crate of

$$F_{net}=F-\mu_{k}mg=\mu_{s}mg-\mu_{k}mg$$

where $\mu_{s}$ and $\mu_{k}$ are the coefficients of static and kinetic friction, respectively, and where $\mu_{k}<\mu_{s}$.

and an acceleration of the crate of

$$a=\frac{\mu_{s}mg-\mu_{k}mg}{m}$$

$$a=(\mu_{s}-\mu_{k})g$$

Hope this helps.

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