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Suppose you put your hands on a wall and push it. If neither the wall or you accelerate, how could one calculate the force?

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    $\begingroup$ No, I don't think so that we can calculate the force exerted by us simply with equations and we don't accelerate back because there is enough friction force to stop us, but if we do the same with socks or slippery shoes on then we will definitely move back. However, we can calculate force by attaching a spring to the wall with high spring constant and after applying force we can calculate the force applied by F=kx , x is the displacement of the spring by our force. $\endgroup$
    – Hitman A7
    Dec 14 '20 at 3:10
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    $\begingroup$ Not sure if you're looking for a practical answer but hold a bathroom scale against the wall and push. Do this while wearing socks, bar feet, tennis shoes, soccer cleats, etc... and call it a day. I do understand that this simply measures the force instead of calculating it so I don't know if this answers your question. $\endgroup$
    – MonkeyZeus
    Dec 14 '20 at 17:18
  • $\begingroup$ "how could one calculate the force" Define "the force". There are many forces involved here, as suggested by free-body-diagram. Define "calculate". Do you mean "calculate" or "measure"? $\endgroup$ Dec 15 '20 at 12:00
  • $\begingroup$ You could tape a scale on the wall and push it instead $\endgroup$ Dec 15 '20 at 14:18
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This very common misconception about Newton's third law.

The sum of these two forces are zero that is why nothing is moving

That is wrong. The two forces acts on different body.

You do get pushed by the same amount you pushed the wall, but you don't move horizontally because of friction present between ground and your feet.

We can't directly calculate the exact force exterted by you but we can tell by looking that is less than maximum value of static friction, if you are not accelerating.


The two forces here means:

  1. The force by which you push the wall.
  2. The force by which the wall push you back.

These two forces are equal in magnitude and opposite in direction. They also form an action reaction pair.

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    $\begingroup$ you mean the friction between feet and floor, isnt it? $\endgroup$
    – lalala
    Dec 14 '20 at 12:23
  • $\begingroup$ @lalalaL yes. There is no relative motion btw the wall and the hands (assuming you are keeping the hands still and not trying to slide up and down the wall). $\endgroup$ Dec 14 '20 at 18:28
  • $\begingroup$ You can prove this if you put on some socks and stand on slippery tiles. Under conditions where there's a relatively low amount of friction between your feet and the floor, if you're strong enough you can push yourself away from the wall. Eventually you'll end up too far away to apply enough force against the wall. $\endgroup$
    – Kayndarr
    Dec 15 '20 at 7:51
  • $\begingroup$ It looks like the answer you refer to has been deleted, and therefore this answer no longer makes sense. Please either include the relevant context from the other answer, or don't refer to it at all. $\endgroup$ Dec 15 '20 at 23:22
  • $\begingroup$ @BrianDrake Yes it has been deleted. Thank you for pointing it out. I will edit my answer accordingly. $\endgroup$ Dec 15 '20 at 23:45
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I believe that can we can surely measure the force exerrted on the wall by you when you are not slipping to a close accuracy(and complete by taking an ideal case)

1.For this the best suited is the parallel combination of springs sandwiched between two wood sheets enter image description here

2.So just take that apparatus and put it between your hands and the contact of the wall and just calculate the compression in the springs and using spring force formula. $$\vec{F}= -K\vec{X}$$ where K is the net spring constant. And cheers we are done with calculating force exertedby us on the wall

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    $\begingroup$ "That apparatus" has a name: It's called a spring scale. $\endgroup$ Dec 14 '20 at 21:43
  • $\begingroup$ In principle if I knew the properties of the material in the wall, or the compressibility of the flesh on my palms, we could do this without adding springs, but same idea of course. $\endgroup$
    – kaylimekay
    Dec 15 '20 at 3:41
  • $\begingroup$ @kaylimekay but that would be very difficult because that constant of compressibility can change with time and even it shows very small compressions leading to low precision $\endgroup$
    – Anonymous
    Dec 15 '20 at 3:58
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This situation is known in engineering as a "statically indeterminate system". It literally you cannot determine the forces using a force balance alone. You also need some form of flexibility on the part to decide how load is shared between the contacts.

You can try it yourself, by pushing on the wall, you can shift the pushing from one arm to the other, or you can increase and decrease the force your feet apply (with equal and opposite force through both hands).

If you treat yourself as a rigid body, there is no way to discern all the forces in a statically indeterminate system.

But if you were to model on the bone flexibility, muscle tension and kinematic joint in your body you could estimate the forces given certain (physiological) assumptions.

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One thing to keep in mind when doing force problems is that acceleration is net force divided by mass. An object can still be experiencing forces without accelerating, as long as there are other forces. You are pushing on the wall, so that is a force on the wall. The wall is presumably attached to the floor, so the floor will exert a force that opposes your force. There is also a reaction force from the wall to you, and that force is resisted by the friction between you and the floor. Since the floor is exerting a force on you, you are exerting a reaction force on it.

So if label the direction you're pushing on the wall as being "positive", there is a positive force from you to the wall, a negative force from the wall to you, a positive force from the wall to the floor, a negative force from the floor to the wall, a positive force from the floor to you, and a negative force from you to the floor. Each body has one negative force and one positive force acting on it, and they're all the same magnitude, so none of them experience acceleration.

As far as knowing what that magnitude is, there would have to be some empirical measurement.

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