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I might be asking a very elementary question here. How do we identify where static friction is acting or kinetic friction is acting?

For example, Consider this case:

enter image description here

Now, here we are talking about "sliding" a body along the ground with the least possible force.. If we are talking about sliding , then that would mean it will involve relative motion between the ground and the body.. causing kinetic friction. As $\mu_s$=$\mu_k$ , it does not cause much of a problem. However in the solution , it was given as follows :

When the block is about to start sliding, the frictional force acting on block reaches to its limiting value f =$μ_s$N

Why do we not directly take the case of sliding friction here? I mean , if $\mu_s$ would not have been given equal as $\mu_k$, then why would i go with limiting case instead of directly considering the case where the object is sliding?

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    $\begingroup$ The reason they set static and kinetic friction the same in the problem, is precisely that they don't want to have to state the problem carefully, in terms of whether and when the block is moving or not (because it doesn't matter.) If the two coefficients were different, then presumably they would have stated the problem more carefully and the question wouldn't come up. $\endgroup$ Commented Jan 28 at 18:00
  • $\begingroup$ Static friction acts whenever two surfaces are not moving relative to each other. When they are rubbing past on and other, you have kinetic friction, which is significantly lower. For instance, a sliding box experiences kinetic friction, and a ball rolling without slipping experiences static friction. $\endgroup$
    – Cat Bisque
    Commented Jan 28 at 21:38
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    $\begingroup$ It is odd to set the coefficient of static friction equal to that of kinetic friction. I believe they are trying to make it so you don't have to consider whether the block was initially moving or not. The solution addresses the case where the block started at rest, and to get it moving you have to overcome the static friction. I believe they use the word "limiting" because the friction experienced by the block decreases in proportion to the normal force (N). Since the pulling force has an upward component, it reduces the normal force: mg - P*sin(45) = N, where g = 9.8 m/s^2 $\endgroup$
    – Cat Bisque
    Commented Jan 28 at 21:52

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In general, static friction is greater than kinetic friction. That can be easily shown by placing an object on a flat surface (ceramic surface), and gradually increasing the angle of incline of the surface. A moment comes when the object starts moving downwards.

If kinetic friction were equal to the static one, we would see velocities as small as we wished, by inclining the surface very slowly. But experience tells us that when the static friction is overcome, the acceleration $a = mg\ \sin(\theta) - f_a$ jumps from zero to a finite value well above zero.

So, $f_a$ just after the object starts to move is smaller than it was before.

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    $\begingroup$ Nice explanation, I would also include this chart from wiki. $\endgroup$ Commented Jan 27 at 23:10
  • $\begingroup$ So like , how do we identify when static force is acting and when kinetic friction is acting? The textbooks describe this in term of "Relative sliding" between the two surfaces.... Here in this question as well , there must be a relative sliding right ( as we are sliding an object along the ground) , so why do we consider static friction to be acting here? Why are we considering the first case onlyy? Rest of the answer was perfect, thanks. $\endgroup$
    – Adhway
    Commented Jan 28 at 5:02
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In general, the type of friction to calculate with can be determined by assessing which of the following situations the physics problem is about:

  • The objects with friction are not sliding, and will continue not sliding.
  • The objects with friction are initially not sliding, but transition into sliding.
  • The objects with friction are already sliding.

The first and last of those situations are obvious - sliding uses kinetic friction, and not sliding uses static friction. The middle one, with a transition into sliding, is trickier to understand, but once you do understand it is equally simple.

When a physics problem is about the transition from not sliding to sliding, it uses static friction. This is because there is only one question to be asked about such a situation: What force is needed to cause the transition to occur? Or, practically equivalently, is a given force sufficient to cause the transition or not? All other questions about friction (at the level of analysis depth that static and kinetic friction are relevant to) fall into one or the other of the other two situations.

For any force too small to cause a transition to sliding, the objects will continue not sliding, so obviously static friction applies. Therefore, any force less than static friction will not cause a transition to sliding. Accordingly, calculating the necessary force to cause the transition uses static friction.

For your example problem, it is actually two problems combined.

The first one, "Find the least pulling force which [...] will slide", is a textbook example of a transition problem. It is telling you to find how much force is necessary to cause the transition from not sliding to sliding. It therefore uses static friction.

The other one, "find the resulting acceleration", is about sliding. The setup of the problem includes a transition, but this part of it is about what happens after the transition. It instructs you to find acceleration, and non-zero acceleration of a non rolling object relative to a surface it has friction with can only happen with sliding. It therefore uses kinetic friction.

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  • $\begingroup$ Thank you!! It answers my question! $\endgroup$
    – Adhway
    Commented Jan 29 at 5:41
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Your question is more involved with language and terminology as you seem to be new to these phrases like "just after" , " just before" etc.

In mathematics, it's senseless to say something like " a real number just greater than 4" because such a real number does not exist. But in physics , authors usually don't care about these formal things. The concept of limits is not taken with much caution here.

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  • $\begingroup$ Um yeah I am having problem with these things only....... $\endgroup$
    – Adhway
    Commented Jan 28 at 15:08

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