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If I neglect air resistance is it possible that I can walk with constant velocity? Suppose that I take equal steps of equal length every second and therefore my velocity doesn't change. That means acceleration of my body is $0$.

But there is a net friction force on me which is making me walk. Now according to me there are no other forces acting. Shouldn't that mean that I should be accelerating but I am not accelerating?

I am talking about my motion after a time $t$ from my start of the motion.

Edit: why doesn't my velocity of centre of mass change ?

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  • $\begingroup$ Hi Ohw, and welcome to Physics Stack Exchange! I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Jun 25 '20 at 19:40
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Mathematical analysis

The net momentum imparted by the friction force on your body is zero. Mathematically,

$$\Delta \mathbf p =\int_0^T \mathbf F(t)\: \mathrm dt =0\tag{1}$$

where $\mathbf F(t)$ is the friction force acting on you at any time $t$. Now, since you have the same velocity at every instant, thus the integral in equation $(1)$ should evaluate to $0$ for any possible time $T$. This is only possible when the integrand itself is $0$, that is

$$\mathbf F(t)=0\qquad\forall \:\:t\in(0,T)$$

But we know that this isn't practically true, or physically correct. Here comes the physical analysis to help us out.

Physical analysis

Physically (or practically), we know that we do experience a force of friction while walking. Then this must also imply that we must not have a constant velocity. And the truth is that we don't have a constant velocity. Our velocity isn't constant and wobbles/swings/oscillates. If you graphed velocity versus time, you'd get something similar to this:

graph

In the above graph, the $x$-axis corresponds to time, and the $y$-axis corresponds to the velocity.
Note: This graph is just a guess/approximation to how a real velocity varies. I have only included it to give you a feel about the velocity dependence of time. If a real person's gait is measured, it's extremely likely that it might not be this uniform and monotonic and might not even have the same shape. Though, the above example will suffice for our further analysis.

Now, the acceleration is defined as the time derivative of velocity,

$$\mathbf a=\frac{\mathrm d\mathbf v}{\mathrm dt}$$

Since we are dealing with one dimensional motion, we can drop off the vectors (and the vector notation) to get

$$a=\frac{\mathrm dv}{\mathrm dt}$$

So, we can see that acceleration is just the slope of the velocity-time graph. In the above example, the curve does have a non-zero slope (and thus a non-zero acceleration) at many points. This also implies non-zero net force at those points. And what could be this net force? You got that, it's friction. Friction acts at you all along the time.

Now, secondly, notice that the graph has regions with both, positive and negative slope. This implies that friction also acts in both, the positive and negative direction, or in our context, the forward and the backward directions. So, we now know that friction doesn't always push you forward, rather, it also pushes you backwards. And, now we can modify our mathematical expression $(1)$ to fit our new model:

$$\Delta \mathbf p =\int_0^{nt'} \mathbf F(t)\: \mathrm dt =0\tag{2}$$

where $t'$ is the time period of the oscillations and $n$ is a natural number. Before we move any further, I would like to clarify that equation $(2)$ only holds if the motion is perfectly periodic (which it isn't in reality), but it's a good approximation and can get us a bit further.

Now, suddenly the question arises, how did I get to equation $(2)$? Well, it's simple. The left hand side of the eauation $(1)$ denotes the change in momentum from time $t=0$ to $t=T$. The momentum of any body is given as $\mathbf p=m\mathbf v$ (where $m$ is the mass of the body). Now since the velocity $v$ stays the same after time intervals of $t'$ (because of the periodic variation of velocity with time), thus the change in momentum between this time interval vanishes and thus we get $\Delta \mathbf p=0$. And there we are, with our corrected version of equation $(1)$, equation $(2)$.

Note: The value of the integral,

$$\int_0^T \mathbf F(t)\: \mathrm dt =0$$

for any $T\neq t'$, is not equal to zero. The value is a finite, but non zero quantity.

But how does it happen?

Most of the details of how and why it happens, lie in biophysics. I am not going to explain it in great detail, however, the image below aptly describes the process in a visual way:

image

Image source

Summary

This shows that the friction does change the momentum, but it keeps on compensating for that change in momentum. In other words, friction, alternately, accelerates you and then decelerates. This goes on and on, until you decide to stop :-)

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But there is a net friction force on me which is making me walk. Now according to me there are no other forces acting. Shouldn't that mean that I should be accelerating but I am not accelerating?

As @FakeMod pointed out its answer, you are alternatively accelerating and decelerating in such a way that your overall average velocity is constant.

Consider when you begin at rest. You exert a force backwards on one foot and the ground exerts and equal and opposite friction force on your foot forwards per Newton's third law. At that initial stride, before your lead foot contacts the ground, the static friction force on your rear foot is the only external force acting on you (except air resistance) and does so in the forward direction, causing you to accelerate from rest. Without static friction, your foot would slip.

Next, your forward foot makes contact with the ground at an angle such it exerts a forward force on the ground resulting in the ground exerting an equal and opposite static friction force backwards causing you to decelerate. But before it brings you back to rest you take the next step with your back foot.

Imagine that if after the initial acceleration there was no friction acting on either foot. You would then slide on the surface at the constant velocity attained by the initial acceleration, in the absence of air friction.

Bottom line: Following the initial accelerating step, the acceleration and deceleration associated with each complete stride causes your velocity to rise and fall around a constant velocity.

Hope this helps.

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When we walk, the friction with the ground is important not only to move ahead.

When people fall down because of a wet floor, they can fall forwards or backwards. Friction is essential to avoid our foot ahead to slip. The friction force is then backward.

When we walk at a constant speed, (negleting air drag), friction forces acting in our feet should be balanced forwards and backwards. The net friction force is zero for any cycle of some steps.

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