Suppose I am pushing a box on table with $10$ N force due to friction it is not moving and if i applied $20$ N the box started accelerating. Then, how does the box apply $20$ N force on me?Well actually,I really know fundamental laws are certainly true.But from where the opposing force on us is coming?
[Physics] Is Newton’s third law really correct?
forcesfree-body-diagramnewtonian-mechanics
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I think it's a great question, and enjoyed it very much when I grappled with it myself.
Here's a picture of some of the forces in this scenario.$^\dagger$ The ones that are the same colour as each other are pairs of equal magnitude, opposite direction forces from Newton's third law. (W and R are of equal magnitude in opposite directions, but they're acting on the same object - that's Newton's first law in action.)
While $F_{matchbox}$ does press back on my finger with an equal magnitude to $F_{finger}$, it's no match for $F_{muscles}$ (even though I've not been to the gym in years).
At the matchbox, the forward force from my finger overcomes the friction force from the table. Each object has an imbalance of forces giving rise to acceleration leftwards.
The point of the diagram is to make clear that the third law makes matched pairs of forces that act on different objects. Equilibrium from Newton's first or second law is about the resultant force at a single object.
$\dagger$ (Sorry that the finger doesn't actually touch the matchbox in the diagram. If it had, I wouldn't have had space for the important safety notice on the matches. I wouldn't want any children to be harmed because of a misplaced force arrow. Come to think of it, the dagger on this footnote looks a bit sharp.)
Yes, the normal force applied over an tiny amount of time changes the speed (momentum) of the objects. The exact magnitude of the force depends on the time it takes for the impact to happen.
A force $F$ applied for a time period $\Delta t$ has impulse of $ J = F \Delta t $. The impulse is equal to the change of linear momentum in an opposing fashion (3rd law)
$$ \begin{aligned} m_1 \Delta v_1 & = J \\ m_2 \Delta v_2 & = -J \end{aligned} $$
The impact is characterized by a certain coefficient of restitution $\epsilon$ definiting the final relative velocities $v_1^F$ and $v_2^F$ in terms of the initial velocities $v_1^I$ and $v_2^I$ as
$$ \begin{aligned} (v_2^F-v_1^F) & = -\epsilon (v_2^I-v_1^I) \\ (v_2^I + \Delta v_2-v_1^I - \Delta v_1) & = -\epsilon (v_2^I-v_1^I) \\ ( \Delta v_2 - \Delta v_1) & = -(\epsilon+1) (v_2^I-v_1^I) \\ \frac{-J}{m_2} - \frac{J}{m_1} & = -(\epsilon+1) (v_2^I-v_1^I) \end{aligned}$$
$$ J = \frac{(1+\epsilon) v_{rel} } { \frac{1}{m_2} + \frac{1}{m_1} } $$
where $v_{rel} = v_2^I-v_1^I$ is the impact relative velocity. Once $J$ is known, $\Delta v_1$ and $\Delta v_2$ are calculated, as well as the force $F = \frac{J}{\Delta t}$.
Best Answer
The difficulty you are having is quite common when first learning Newtons laws because you think Newton’s third law means the forces always cancel each other and nothing should accelerate. They don’t. You have to look at the net force acting on each body individually and apply Newton's second law to each individually.
When you were pushing on the box, say to the right, with a force of 10 N, an equal static friction force of 10 N was acting on the box to the left in opposition to your force. The net force on the box was zero and therefore it did not accelerate.
Per Newton's third law the box was exerting an equal and opposite force of 10 N on you to the left. You did not accelerate because of an equal and opposite static friction force between your feet and the ground that acted to the right. Therefore you do not accelerate.
Now when you were pushing with a 20 N force to the right the box accelerated because your force apparently exceeded the maximum static friction force between the box and ground acting to the left. The box is now sliding and it is the kinetic friction force, which is generally less than the static friction force, between the box and the ground that is now opposing your force. But since the box is accelerating your 20 N force to the right was greater than the kinetic friction force to the left, meaning there was a net force to the right, $F_{net}$, causing the box to accelerate per Newton's second law. Then, per Newton's second, the acceleration of the box to the right is $a=\frac{F_{net}}{m}$, where $m$ is the mass of the box.
Now the box also exerts an equal and opposite 20 N force on you per Newton's third law. What keeps you from accelerating? It is the static friction force between your feet and the ground acting to the right that is equal to the force the box exerts on you. The net force on you is zero as long as you don't push too hard that the maximum static friction force between your feet and the ground is exceeded, in which case you will start slipping.
Hope this helps.