0
$\begingroup$

My question is regarding a specific case displaying Newton's third law.

In the diagram below, a man is shown exerting a force on the wall, which in turn causes an equal and opposite reaction force on the man. I understand that the reason the man doesn't fly backwards is due to friction with the ground - the man exerts an action force on the ground which causes a reaction force back on the man. Hence, there is a 0 net force acting on the man as the original reaction force from the wall cancels out the reaction force from the ground.

enter image description here

I am not confused about how objects move in general with respect to the third law since i understand that the action and reaction forces act on two separate objects. However, the logic behind this specific example would suggest that nothing on a plane would be able to move when a force is applied, since the friction from the ground would result in a 0 net force on the object.

For example, in the diagram shown below, a hand is exerting an action force on the brick. This in turn causes the brick to exert an action force on the floor which causes a reaction force on the brick. The action force on the brick and the reaction force on the brick exerted by the floor cancel each other out resulting in a 0 net force on the brick. Hence, the brick (or any other object on a plane like this) should not move and accelerate no matter how large a force is supplied.

enter image description here

However, we are clearly able to push objects like bricks across a surface like a table in real life. I don't understand how this could be achieved as my understanding tells me that you cannot ever overcome the frictional force of the floor.

The original YouTube video where i got my diagrams from: https://www.youtube.com/watch?v=91QYouih4bQ

$\endgroup$
  • 1
    $\begingroup$ What makes you think that the reaction force from the ground has to match the action force from the hand? $\endgroup$ – Chet Miller Dec 25 '18 at 13:24
  • $\begingroup$ The action force on the brick from the hand causes an action force on the ground. Then, the ground will exert an opposite and equal reaction force back on the brick. $\endgroup$ – weary27 Dec 25 '18 at 13:37
2
$\begingroup$

You are assuming the force applied to the block and the friction force must be equal in magnitude, but this is not always true. Adopting the typical simple model of friction, all you have to do is apply a large enough force to overcome static friction between the block and the ground in order for the block to move.

Action-reaction force pairs are always equal in magnitude and opposite in direction, but there isn't any "law" in general relating forces in different pairs. There is no reason to assume that the force you apply is always equal and opposite to the friction force.

If this is too much, instead think of a block on a frictionless surface with two people pushing on either end. Of course the block will push back on each person with the same magnitude of force each person exerts on the block, but the direction of the block's acceleration is just determined by whoever chooses to push more (or if they push with the same force then the block won't move). There isn't any law that relates the two applied forces.

$\endgroup$
  • $\begingroup$ Does that mean that in the diagram where the man was pushing on the wall, the two red reaction forces don't exactly cancel each other out? $\endgroup$ – weary27 Dec 25 '18 at 14:25
  • $\begingroup$ @weary27 They can cancel out, but they don't have to. If he isn't moving, then the forces obviously cancel out. But they don't have to cancel out, in which case he would start to slip backwards. $\endgroup$ – Aaron Stevens Dec 25 '18 at 14:26
  • $\begingroup$ So every material has a different static friction value which needs to be overcome in order to make an object move across the surface? Ice for example would have a lower frictional value than rubber? $\endgroup$ – weary27 Dec 25 '18 at 14:35
  • $\begingroup$ @weary27 Yes that is correct $\endgroup$ – Aaron Stevens Dec 25 '18 at 14:53

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