# Applying force in zero seconds? How is this possible?

$$F\,\Delta t=m\,\Delta v$$, is the equation of impulse. Let's say that I push a wall with my hands, we can have $$0$$ change in velocity; therefore, $$F\,\Delta t=0$$, since I'm pushing the wall there is force, the only way the equation equals zero is that $$\Delta t$$ equals zero, applying force in zero seconds? How is this be possible?

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Mar 28, 2020 at 22:25

Beware of blindly applying equations.

In thinking about the velocity of the wall, then for $$F\Delta t=m\Delta v$$, $$F$$ is the net force acting on the wall. If $$\Delta v=0$$, then we know that $$F=0$$ (and the same is true vice versa).

Note that this is just Newton's second law: $$F=m\frac{\Delta v}{\Delta t}=ma$$. If $$a=0$$ then we can conclude that the net force $$F$$ must be $$0$$.

Also, if we really did have $$\Delta t=0$$, then of course $$\Delta v=0$$ since nothing can change over no time.

• Comments are not for extended discussion; this conversation has been moved to chat. Apr 1, 2020 at 1:01

the only way the equation equals zero is that Δ𝑡 equals zero, applying force in zero seconds?

No. The force $$F$$ is the net force acting on the wall, not the force you apply to the wall. So the net force can be zero.

You apply a force $$F$$ to the wall. The floor that holds the wall in place applies an equal and opposite force of $$F$$ on the wall, for a net force on the wall of zero and no acceleration of the wall per Newton's second law $$F_\text{net}=ma$$.

At the same time you apply a force $$F$$ to the wall it applies an equal and opposite force $$F$$ on you per Newton's third law. But you do not accelerate because the static friction force between your feet and the floor is equal and opposite to the force the wall exerts on you, for a net force on you of zero.

Hope this helps.

• I find this answer confusing. It uses the symbol $F$ for too many different things. It uses the contradictory phrase "equal and opposite" (vectors in different directions are not equal!), which should be banned in serious physics discussions. Even worse, the second paragraph uses this phrase in a situation that is not described by Newton's third law, which can only create confusion for readers not familiar with that law. Dec 15, 2020 at 12:15
• @BrianDrake Sorry you are unhappy with my answer, but I think you did not read it properly. By "equal" I meant equal in magnitude. I know what vectors are. The second paragraph describes a situation covered by N2 as it says, not N3. The first sentence of the third paragraph describes N3. Dec 15, 2020 at 12:31
• I said that it was confusing, not that I couldn't understand it. I know what you meant. But will the target audience of your post (presumably a beginner at physics) know what you meant? Dec 15, 2020 at 12:41