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Could anyone correct my explanation for how things move despite the reaction-action force? (I've just started learning this topic so I might be wrong)

Explanation: Let's take two scenarios: (I'm going to ignore the - sign mostly)

  1. A book on a table -> the book has a weight of 4 N. According to Newton's third law, the table will apply an equal reaction force of 4 N. This prevents the book from passing the table and hitting the floor. This is also because the table is solid. Hence, the book will stay still, with a net force of 0 because of the reaction force.
  2. A person pushing a box -> the person applies an action force of 10 N to a box and the box will apply an equal reaction force to the hand. However, let's say that this force was not enough to push the box. So, let's say in general the box needed 20 N to move. The 10 N (action-reaction force) applied to the box would only allow the hand to stay still on it (remember the table-book scenario), as it is less than 20 N, so it will not move. So like the book on the table, nothing will actually move because the force is not great enough to move the box.

I think it would be important to note that the actual external force applied to the box is the one that actually MOVES the box. If you think about it, even if you applied 20 N to the box, there will be a reaction force of -20 N, giving a "net force" of 0 (this would not make sense right? Like how can anything move then???). So that's why the actual external force applied to the box would be considered the net force and not the reaction force + external force. The internal force isn't considered while doing vector addition or free-body diagrams. (Please correct me if this is wrong)

Also: According to Newton's second law, Net force = mass (of the receiving object)* acceleration. So the greater the mass, the less the acceleration will be, so even though you apply the same net force to two objects of different masses, the heavier one will have less acceleration. If the acceleration is almost nothing, it's probably because the mass of the receiving object is significantly greater than the object applying the force, or the net force (on the lighter object) is almost nothing (so no force is being applied).

Qmechanic
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Erina Kim
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2 Answers2

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You have a few misconceptions or misunderstandings ...

the actual external force applied to the box is the one that actually MOVES the box

You don't need a force to move the box - if there were no external forces at all acting on the box, then it could still be moving in a straight line at constant speed (Newton's 1st Law). You do need an external force to accelerate the box. But if there is more than one force acting on the box (in this case the pushing force and friction) then we cannot say that one force accelerates the box and the other does not. We can only say that the box will accelerate (or decelerate) if the net vector sum of all external forces is not zero.

even if you applied 20 N to the box, there will be a reaction force of -20 N, giving a "net force" of 0

This incorrect, although it is a common misunderstanding of Newton's 3rd Law. You apply a force of 20 N to the box, and the box applies a force of 20N to you. We cannot net these two action-reaction forces to make zero because they are applied to two different objects. If friction is sufficiently strong (or the box is nailed to the floor) then the net force on the box is zero, but not because of the action-reaction pair - it is because the force of friction that the ground exerts on the box acts in the opposite direction to your push, and the net value of your push (+20N) and friction (-20N) is zero. If the maximum value of friction acting on the box is only 10N then the net force acting on the box is 20N (your push) - 10N (friction) which is 10N, and so the box will accelerate.

The internal force isn't considered while doing vector addition or free-body diagrams

This would be correct if there was an internal force. An internal force is a force applied by one part of an object on another part of the same object. So when considering the motion of the object as a whole, we can in this case net off the action and reaction parts of an internal force because they are both applied to the same object. But in your example (pushing a box) there are no internal forces - your push on the box (action) and the box's push on you (reaction) are both external forces, as are friction acting on the box from the ground, and friction acting on you from the ground. The reason why the reaction force is not shown in a free body diagram of the box is that it is not acting on the box - it is acting on you. The external forces acting on the box are your push and friction.

If you drew a free body diagram of you and the box considered as one object then your push becomes an internal force. Now your push on the box and the box's push on you can be netted to zero and need not be shown in the free body diagram. The external forces acting on the combined you+box object are the friction that the ground applies to you and the friction that the ground applies to the box. In order to accelerate the box, the friction that the ground applies to you must be greater than the friction that the ground applies to the box.

gandalf61
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Newton's 3rd law states, as you say, that a force always is met by a reaction force, identical in magnitude.

But these two forces do not act on the same object. When two objects collide, then one object produces a force on the other, and the other produces the reaction force on the first.

Thus each of them experiences a force which is not cancelled out. This force (along with all other forces that may be acting as well in your scenario) can then be analysed in Newton's second law for each object individually which will tell how each object will individually move (accelerate) do to this force.

In your book example, where the book is lying stationary on the table, then the book is exerting a downwards pushing force (a normal force) on the table, and the table is responding with a reaction force (a normal force) on the book. When you then set up Newton's 2nd law for, say, the book then you only include one of these normal forces -- but you also include gravity as always, which in this case causes the book to be stationary.

Steeven
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