0
$\begingroup$

Suppose there is a wooden block resting on a friction less surface. You give it a little push (force, F) and then it starts moving. Since there is no friction acting on the block at its bottom surface, would you feel a counter force against F? If you feel it, does this originate due to the friction acting on particles/ molecules (because you are pushing those molecules) those make up the wooden block?

Improve this question
$\endgroup$
1
  • $\begingroup$ You will definitely feel a force. Friction cannot be its source, since its absent. Newton's third law is the source of this force. Even if the body was 100% rigid, you will still feel the force. To know more, you should read more about Newton's third law and how it works. $\endgroup$
    – Mitchell
    Sep 14, 2017 at 16:33

2 Answers 2

Reset to default
2
$\begingroup$

According to Newton's Third Law, each force has an equal and opposite reaction force. In this case, the reaction force will be the same as if you pushed the object on a surface which did have a non-zero coefficient of friction, but in this case, the object would gain velocity faster, because there was no friction force countering the force applied by your push.

The force you feel when you push on the object, or any object, the reaction force you feel is due to the object's inertia, its resistance to change in motion, as described by Newton's First Law.

The difference between an object on a surface with friction and an object on a surface without friction is simply the resultant acceleration of the object after the forces are applied, not the reaction force on your hand when you push it.

Example

If you could travel through space freely, at any velocity you like, and you flew to the moon and tried to push it, you would still feel a reaction force against your hand, because you're pushing against the surface. Is there a wall behind the moon that is preventing you from pushing it through friction-less space? No, it is simply the mass of the moon that resists the change in velocity that you are trying to impose on it.

Improve this answer
$\endgroup$
0
1
$\begingroup$

Forces do not exist to balance each other out.

Forces exist to cause acceleration.

So, yes, to accelerate something, you need a force. This is Newton's 2nd law: $\sum F=ma$.

Newton's 3rd law then says that this force, you apply on something else, is also applied on you backwards. This reaction force is still not friction, since that is still not present. It is merely the resistance of this object to start moving, which you feel pushing back.

Improve this answer
$\endgroup$
12
  • $\begingroup$ @ Steeven. Thanks for your answer. So you say that the inertial force pushes your finger back. My question is what's the origin of this inertial force? $\endgroup$
    – Kosala
    Sep 14, 2017 at 16:54
  • 1
    $\begingroup$ @Kosala The reaction force originates from the object in the same way that your push originates from you. $\endgroup$
    – Steeven
    Sep 14, 2017 at 18:55
  • 1
    $\begingroup$ @Kosala What if I throw a stone into the object? That stone exerts a force - a push - on the object at impact. Muscle fibres is the force-creating mechanism in humans, yes, but this is a special case. $\endgroup$
    – Steeven
    Sep 14, 2017 at 19:46
  • 1
    $\begingroup$ @Kosala The point is that a force originates not from a specific "thing" in the object/body, but in every single particle because that particle does not want to move. Mass is a resistance against changes in motion. This resistance causes the forces. If you wish, you can say that the presence of mass is the "origin" of these "inertia" forces. Every single particles tries not to move by hold back against whatever pushes them. This is what makes the particles in your finger push the object (they are squeezed from behind) and this is what makes the object push back. $\endgroup$
    – Steeven
    Sep 14, 2017 at 19:48
  • 1
    $\begingroup$ @Kosala you are very welcome. Note that it is important not to mix this reaction force (or normal force or "inertia" force or whichever we call it) into the calculation on the object. This force does not act on the object itself, but on you who are pushing. It's a typical pitfall and mistake when thinking of Newton's 3rd law to include the reaction force on the object itself. If you do Newton's second law on the object you are pushing, then only the pushing force is involved in this calculation, not the reaction force. $\endgroup$
    – Steeven
    Sep 14, 2017 at 22:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.