1
$\begingroup$

I'm taking a module in physics, and I'm quite confused with applying Newton's third law of motion. I know in the following examples we are usually just interested in the forces acting on the book and the car, but to understand it I wanted to take a look at the big picture, so could you let me know if I have the basic idea.

Say you have a book resting on a table which is resting on the floor. You end up with four pairs of forces.

  1. Force due to gravity on the book, equal and opposite one on the earth.

  2. A contact force on the table equal to the force due to gravity on the book, and a normal reaction force on the book equal and opposite to the contact force.

... then you have two more pairs: one between the desk and the earth and one between the floor and the desk. The force of the floor on the desk is equal and opposite to the contact force of the book on the table and the weight of the table.

so thats 1 problem. I'm not sure if I have the right idea, but hopefully one of you can help me out. And with the car:

1: tire pushes back on road, road pushes forward on car 2.friction pushes back on the car and forward on the road, easily overcome.

and I guess air resistance and other forces eventually bring the car to a contstant velocity.

$\endgroup$

closed as unclear what you're asking by Aaron Stevens, ZeroTheHero, Buzz, Mozibur Ullah, John Rennie Jan 22 at 10:13

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ A contact force on the table equal to the force due to gravity on the book Be careful. This is using Newton's second law with no acceleration of the book. Newton's third law just tells us the force the book exerts on the table is equal and opposite to the force that the table exerts on the book. Newton's third law does not tell is this force is equal to the weight of the book, and this is not the case if the table and book were accelerating upwards, for example. $\endgroup$ – Aaron Stevens Jan 22 at 0:22
  • $\begingroup$ Thanks, are we talking about the contact force on the table acting downwards due to the weight of the book? $\endgroup$ – Carlos Bacca Jan 22 at 0:35
  • $\begingroup$ I am talking about the force between the book and the table. Newton's third law says nothing about what goes into those forces. For a book and table not accelerating then using Newton's second law we know that the contact force is equal in magnitude to the weight. If the book and table were accelerating upwards, then the contact force will be dependent on more than this. $\endgroup$ – Aaron Stevens Jan 22 at 0:59
  • $\begingroup$ If you are focusing on just N3L pairs then only focus on one interaction between only two bodies. As soon as you say "contact force between the book and table due to the weight (gravity)" you are invoking more than just N3L $\endgroup$ – Aaron Stevens Jan 22 at 1:00
  • $\begingroup$ Thanks, so have we got then if we are looking at forces on the table. a force from the book which is equal and opposite to the normal reaction force on the book by the table, and the weight of the table, and a normal reaction force o the table from the ground. $\endgroup$ – Carlos Bacca Jan 22 at 1:12
0
$\begingroup$

Newton's third law can be used to state that "The force due to the Earth on the table is equal and opposite to the force due to the table on the Earth"

But! The statement that the force is equal to the contact force of the table on the Earth plus the weight of the book is not necessarily true. In the above question, assuming that neither of the table or the book have an acceleration in the vertical direction, then it is true.

Now coming to the "car" problem, (Its a nice one)

Consider a road with no friction, and the wheel rotates clockwise. Now look at the bottom most point of the wheel (The point which is in contact with the road)(In my diagram it is point 'P'). The point appears to move backwards.

enter image description here

Now, just add in some friction.

Now friction has just one protocol "OPPOSE RELATIVE MOTION!"

So in other words, when you add in friction between the tire and the road, friction sees that the point which is experiencing friction (Point P) is moving backwards, So it gives a force in the forward direction. Now you can extend this argument by saying that "Using Newton's third law, since the forces will act in opposite pairs, The friction on the ground due to the tire is in the backward direction". And that's completely true! (( Ofcourse we dont see the ground moving back :p ))

So to summarize, Friction acts in the forward direction for the tire.

$\endgroup$
  • $\begingroup$ Ah yes thankyou friction does act forward on the tyre, not like if you were pushing a block along the floorwhere it would act backwards... also for the first problem when we talk about contact forces is this line of reasoning correct. earth pulls down on table with book but it does not accelrate so the floor must exert a force on the table, equal and opposite. The reaction to this force would be the force of the table on the floor, which is equal and opposite im treating the book and table as one object. $\endgroup$ – Carlos Bacca Jan 22 at 12:00
  • $\begingroup$ Yes! Precisely! "Earth pulls down the object (treating the table and book as one 'object'). But the object isn't accelerating (in the vertical direction). And hence there must be an equal and opposite force acting on the object which is infact the normal reaction due to the Earth" $\endgroup$ – The Jade Emperor Jan 22 at 15:24

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