1
$\begingroup$

Suppose we push a wall with some force $F$, and the friction between the wall and my hands is $F_{fr}$. If we push a wall, we see that we bounce back, which means the wall applied a force greater than what we applied on the wall as reaction. But mathematically, I think that: $$F_{net} = F-F_{fr} $$

So net force acting on the wall is obviously less than total force which we applied, so the reaction provided by the wall is also less than the force we applied. But this means the wall will move towards the direction of the applied force, which obviously does not happen.

So what is the physics behind this? Can someone explain it?

$\endgroup$
10
  • 1
    $\begingroup$ What do you mean by we bounce back? Are you considering that our body as some pivotal points, elbows and our feet on the ground? $\endgroup$
    – Alchimista
    Nov 2, 2021 at 14:01
  • 3
    $\begingroup$ Re, "friction between the wall and my hands." If the force you apply is normal (i.e., perpendicular) to the wall, then no friction is needed. You only need the wall to be made of something solid that can "push back." OTOH of course, if you push at an angle to the wall, then you need the friction to keep your hand from sliding away until you end up "pushing" on the wall with your face. $\endgroup$ Nov 2, 2021 at 15:02
  • 1
    $\begingroup$ Re, "this means the wall will move..., which obviously does not happen." But the wall, does move. Suppose you push off from a wall in your house. The wall is rigidly attached to the house. The house is attached to the Earth. When you push off, you are trying to move the entire Earth... And in theory, you will move it, but because the mass of the Earth is so much greater than your mass, the amount by which it moves will be proportionally smaller than the amount by which you move–too small a movement for instruments to measure. $\endgroup$ Nov 2, 2021 at 15:09
  • 1
    $\begingroup$ The force the wall exerts on you is not friction, but the normal (perpendicular) force in reaction to your force. $\endgroup$
    – Bob D
    Nov 2, 2021 at 15:26
  • 1
    $\begingroup$ @JavaMonke This is not true if you look at the whole system. The net force on the whole system (wall connected to Earth + you) is zero, since the force on wall from you and vice versa are the same, as Newtons axiom predicts, but you are so much smaller that only you actually will move. If an apple falls down a tree, the force from Earth applied to apple is also the same as from apple applied to Earth, but the Earth does not care, so we can describe it as if the Earth is not moving at all, and only the apple feels a force. But looking at the whole system, Newtons actio-reactio axiom applies. $\endgroup$
    – Koschi
    Nov 2, 2021 at 15:45

3 Answers 3

2
$\begingroup$

In the expression in the OP, $F$ is the force exerted by the person on the wall, but $F_{\textrm{fr}}$ (which really should be $F_{\textrm{normal}}$) is the force exerted on the person by the wall.

You should never be subtracting these.$^1$

Newton's second law says that the net force exerted on an object determines the force exerted on that object, i.e., $$ m_{\textrm{person}}\vec{a}_{\textrm{person}} = \vec{F}_{\textrm{net on person}} = \mbox{(sum of forces on that person)}\,. $$ Nowhere in the expression should the force exerted by the person on the wall show up. This is a very common mistake made by students learning physics for the first time, and it always comes into conflict with Newton's 3rd Law. That is, the confusion arises because Newton's 3rd law says that the forces exerted between the person and the wall must be equal and opposite, but then how could there be a net force? Well, again, the point is that acceleration is caused by the net force exerted on the object, and it doesn't matter what forces that object is exerting on other things.

$^1$With the caveat that we need to generalize this statement i n the following way. When computing the motion of a single particle, only consider the forces acting on that particle. When computing the motion of the center of mass of a system of particles, only consider the forces exerted by objects outside the system on objects inside the system. The forces exerted by objects inside the system "cancel" in the sense that they don't contribute to the motion of the center of mass.


How does the normal force exerted on the person by the wall arise? We imagine the wall as a very stiff spring. The person pressing into the wall compresses this spring, leading to a spring force exerted by the wall on the person. The wall being very stiff, we can't really detect this microscopic compression with our hands, because the displacement of the wall is extremely small, but it's there, and it leads to the wall pushing back on the person.


Why does the person move and not the wall?

While the forces exerted by the person and the wall on each other are equal and opposite, the results of these forces are not, and that's because the wall is much more massive than the person (in fact, the wall, being connected to the Earth, is part of the Earth and therefore much more massive). Since the acceleration is the net force divided by the mass, the wall has a much smaller acceleration.


What causes the motion of the person?

The resultant motion of the person is in fact more complicated to analyze. Imagine a spring attached to and compressed against a wall. The force exerted by the wall on the spring acts at the point of contact. We release the spring, and it expands outward away from the wall (but it is still connected to the wall). The point of contact doesn't move as the spring expands, and so, paradoxically, despite the fact that the wall is exerting a force on that point on that part of the spring, the spring experiences no acceleration. How does that not contradict Newton's 2nd law?

Well, the part of the spring connected to the wall is being pushed backward toward the wall by the next part of the spring that it's connected to, and so on down the spring. The net result is that the center of mass of the spring experiences an acceleration determined by Newton's 2nd Law in the form of $$ m_{\textrm{spring}}\vec{a}_{\textrm{spring COM}} = \vec{F}_{\textrm{net}} = \vec{F}_{\textrm{N, by wall on spring}}\,. $$ The main idea is that the inner structure of the system (the spring, or a person with muscles flexing and tendons pulling) determines the net result of the force exerted by the wall.

$\endgroup$
3
  • $\begingroup$ So can we say that the physics of our hands "bouncing back" is similar to the physics behind a bouncing ball? $\endgroup$ Nov 3, 2021 at 6:08
  • $\begingroup$ Also, I recalled a statement of the 3rd Law; "Action and reaction cannot act on the same body". So is it ever possible that both action and reaction can cancel each other out? Is that why you said "You should never be subtracting these."? $\endgroup$ Nov 3, 2021 at 6:45
  • 1
    $\begingroup$ For your first question, yes, in way, except that we are actively use our muscles to "re-expand" the ball (i.e. extend our arm, say). As for the second, I should walk that statement back a little. I'll update the post with this, but in physics we think about systems: sometimes we include multiple particles in the system, and so the forces that objects within the system exert on each other "cancel". This happens when we start considering the potential energy of a system of particles. But, when finding out the motion of a single particle, only consider the forces acting on it. $\endgroup$
    – march
    Nov 3, 2021 at 15:25
2
$\begingroup$

Suppose we push a wall with some force $F$, and the friction between the wall and my hands is $F_{fr}$.

The force you exert on the wall and the wall exerts on you has nothing to do with friction. The force the wall exerts on you is a normal (perpendicular) reaction force.

If we push a wall, we see that we bounce back, which means the wall applied a force greater than what we applied on the wall as reaction.

No it is not.

The force the wall exerts on you will always be equal and opposite to the force you exert on it, per Newton's 3rd law, regardless of any movement of you and/or the wall. Your "bouncing back" is not due to a net force, but due to torque, as discussed below.

But mathematically, I think that: $$F_{net} = F-F_{fr} $$

So net force acting on the wall is obviously less than total force which we applied, so the reaction provided by the wall is also less than the force we applied.

The net force acting on you is not the difference between the force you exert on the wall and the wall exerts on you (the action-reaction pair of forces). The net force on you is the sum of all external forces acting on you and not just the wall reaction force. The forces that act on you are the force the wall exerts on you and the opposing friction force that the ground exerts on your feet. See the left figure below. As long as the reaction force of the wall does not exceed the maximum possible static friction force between your feet and the ground, these forces will be equal for a net force of zero resulting in no translational acceleration of the center of mass (COM) of your body per Newton's 2nd law.

But as you can see from the figure on the right, these two equal and opposite parallel forces (known as a "couple") can cause rotation without translation. This is what causes your upper body to move backwards and for our weight to shift towards the back of your feet. Not being a rigid body you can also bend ("give") somewhat in response to the torque (which is not shown in the figure). As long as you do not move too far backwards so that you become unstable you will not topple. You will become unstable if your center of gravity (COG) lies outside your base (your feet) as shown in the right figure below.

Similarly for the wall, in addition to the force you exert on the wall there is the opposing force the ground exerts on the wall. If the base of wall remains fixed (imbedded in the ground), these forces are equal and opposite for a net force of zero and its COM will not accelerate. Unlike you, being imbedded in the ground the ground offers an opposing torque reaction preventing the wall from toppling over. Also being more of a rigid body than you, it will not bend too much in response to your force, making any deflection so slight as to be unnoticeable.

Hope this helps.

enter image description here

$\endgroup$
1
$\begingroup$

Your post seems to say that your concepts are tangled.
When we push a wall we bounce back because our mass is way too less than the wall.
Suppose you apply $F$ force on the wall then according to Newton's third law the wall will also apply $F$ force on you (if you applied the force perpendicular to the wall then there is no need of friction in the system) and according to Newton's second law $$F=ma$$ which means that the force is divided between mass and acceleration (in simple words mass and acceleration share the part of force according to their superiority, if the mass is greater it takes more part and acceleration gets small part and vice versa).
For example: when you jump you apply perpendicular force to the earth and it is the reaction force which lifts you but since you applied some force on earth also it must also accelerate (as I read somewhere force applied by you causes earth to shift by half the width of hydrogen atom) and this is because earth has a huge mass thus it takes more part in force than the acceleration which gets very small part of force, but since you have a small mass the acceleration is huge which is the cause of your lift.
Note : I didn't use many mathematical terms and just explained in simple words.
Hope it helps. :)
Edit:
So when you apply the force $F_y$ on wall then the wall applies the force $-F_w$ on you. So $$F_y = -F_w$$ $\therefore$ , acceleration is also opposite $$a_y = -a_w$$ or $$a_w = -a_y$$ so your velocity will increase in opposite direction and that's the cause of your bouncing back.
I don't think this need much mathematics to understand.

$\endgroup$
4
  • $\begingroup$ Oh, so since $F=ma$, my $m$ is smaller, hence my $a$ will be more, whereas the $m$ of a wall is larger, so $a$ of wall is less, almost 0, hence the wall does not move back. But how does this explain me bouncing back from the wall? Edit: Has it got something to do with elasticity of my hand? $\endgroup$ Nov 2, 2021 at 14:38
  • $\begingroup$ Acceleration is always in direction of the force so if the wall is applying force in opposite direction on you then your acceleration will also be in opposite direction. $\endgroup$ Nov 2, 2021 at 15:02
  • $\begingroup$ I think I need some mathematical expressions here... $\endgroup$ Nov 2, 2021 at 15:13
  • $\begingroup$ Got it, thanks! $\endgroup$ Nov 3, 2021 at 4:15

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