A meditation on Newton’s second Law

Newton’s second law is often the key to unlocking answers to so many everyday questions. Why does a car drift? How can we jump? Why are sprinters so muscular? Should you pick the light or heavy dumbbell if you’re trying to build strength? The truth is, Newton’s second law is hiding in plain sight, shaping everything from sports to cars to your gym routine. In this article, we’ll explore how this one elegant principle helps explain so much of the physical world around us.

Don’t worry ! this isn’t a technical deep dive. We’ll stick to basic undergraduate-level physics and keep the math simple (or even skip it where possible). My goal? To show you just how powerful (and surprisingly everywhere) Newton’s second law really is

Before starting, we need to clarify prerequisites :

When an object is at rest, it has no reason to move, and when an object is moving, it has no reason to stop or change direction. That’s what we call inertia.

So, to make an object change direction or move, we need to apply forces to it.

For example, to make an object turn to the right, we need to apply a force toward the center of the rotation. This force creates an acceleration vector toward the center, which will always have this value:

where v is the velocity of the object and r is the radius of the curve

Let’s begin :

The second Law of Newton is :

It is simpler to read it that way :

F is the sum of all external forces applied to the object at an instant t. m is its mass, and from this, we can obtain the acceleration vector of the object at that instant.

How does this explain how a car turns?

For a car to turn left, we need a force applied to the vehicle directed toward the left (or more precisely, pointing toward the center of the turn we want to make). What creates that force? The answer is road friction.

To get the intuition for why road friction creates that centripetal force (the force toward the center of rotation), imagine a car on a perfectly smooth sheet of ice with completely smooth tires. If we push the car forward, you can imagine that the car won’t be able to stop, right? And even if we turn the steering wheel in any direction, the car won’t change direction …it will continue straight until it hits the ice wall. This is because regardless of which way the wheels are pointed, there are still four contact points with the ground, and all four contact points will keep sliding.

movement of the car on ice

I hope you got the intuition, now, let’s understand what happens on the road

illustration of the tire from above, before and after we turn the streeing wheel to the left

When we turn the steering wheel to the left, the part of the tire in contact with the ground stays in place due to the car’s weight, while the rest of the tire becomes deformed. Since the tire is elastic, it tries to return to its original shape.

As soon as the front part of the tire touches the ground, it applies the force illustrated above to the ground. By Newton’s third law, the ground applies an equal and opposite force back to the tire: this is the centripetal force required for the car to turn

If the force generated by this friction is not sufficient to turn the car (for example, if the tires do not have adequate grip or the car is moving too fast), then the car will drift.

the centripetal force will make that velocity vector tangential to the turn

The actual forces acting on the car when taking a corner and holding the steering wheel to the left are: gravity, the normal force, and the friction force. So we can write Newton’s second law and determine the maximum speed the car can take without drifting.

Gravity and the normal force cancel out, so:

The acceleration vector will be in the same direction as the friction force. Remember what we said at the beginning: acceleration toward the center is always v²/r.

And what is the magnitude of the friction force? Well, it is proportional to the normal force. I won’t explain why in detail, but I’ll give you a simple intuition and a quick explanation of friction force: When you put a heavy box on a table and start applying a small force to push that box, it won’t move. But gradually, by pushing harder and harder, the box will eventually start moving. What prevents the box from moving when you apply a small force? It’s static friction.

So the box will start moving only when you apply a horizontal force greater than a certain threshold, and that threshold is the maximum static friction force.

A higher box weight (so a higher normal force) increases the pressure on these contact points until it reaches some stabilization, and so it makes them flatten:

due to box weight, the contact points will flatten

This flattening effect will increase the contact zone between the box and the table, and so we will have more atomic interactions between the atoms of the box and the atoms of the table. And it is these atomic interactions that create the static force. So the greater the normal force, the more contact zone we have, the more interactions we have, the more friction we have. (Of course, molecular interactions forces depend on the two materials.)

Back to our car drifting problem: we can write the car equation as:

and the normal force equals mg in this case, so the maximum velocity before drifting will be:

Let’s think a bit..what is the acceleration of an object? It is the change of velocity of that object over a period of time, right? So it is ∆v/∆t, which means Newton’s second law can be written as:

Note that ∆v/∆t is the average acceleration between two instants.

Now, imagine a person is preparing to jump vertically. The condition that confirms the person is jumping is: v_takeoff > 0 as soon as the toes leave the ground

A silly question might be: why don’t we jump when we extend our knees slowly, but if we do the same movement faster, we do jump?

so let’s try to find what make v_takeoff>0.

When we contract our calves, bend our knees, and extend them to jump, we apply a force to the ground, so the ground will apply a force back on us

When toes leave the ground, the normal force disappears

To compute v_takeoff, let’s apply Newton’s second law at the exact instant of toe-off:

N(t) — mg = m*(dv/dt) (we can’t use Δt because we want an instantaneous value, not a large interval)

We can rewrite this as:

(N(t)-mg)dt = m * dv

This applies to a particular instant, but we can sum the equalities corresponding to all instants from the moment we start to extend the knee until the moment of takeoff:

we can consider that v0 = 0

The left integral is the sum of all reaction forces at every instant. If we divide it by Δt, it gives us the average reaction force during that period

so to have v_off > 0 we need to have:

Therefore, during the entire jumping movement, we need to apply a force that is on average greater than our weight. Note that N = 0 means that we are already in the air.

Another way to look at this equation is to write it like this:

So to jump, we just need to push harder than our weight for some duration.

When we extend the knees slowly, we actually don’t apply sufficient force onto the ground, because the normal force will depend on how fast we can lift our center of mass during the push phase (use Newton’s second law to see this).

The equation we’ve seen above is called the Impulse-Momentum Theorem, and it’s simply Newton’s Second Law with some mathematical manipulations

Running is similar to jumping, it involves the same knee extension movement, but instead of pushing downward, we push the ground backward, essentially jumping horizontally.

So if you’re in a sprint and want to run faster than others, at the starting blocks when the signal goes off, you need to achieve the best acceleration with your first leg push. This increases the average ground reaction force and gives you the optimal initial velocity.

Newton’s Second Law, from the “go” to toes-off:

So if you weigh 75kg and want to achieve a speed of 5 m/s at the start in less than 0.5 seconds, you need an average reaction force of:

1050 N

this is equivalent to pushing…

We divide by mg to get the equivalent in body weight (because we compare the force of the push with the force of the weight)

…1.42 times your body weight in less than 0.5 seconds.

Another thing: during 100m races, sprinters swing their arms very fast to provide their upper body with the stability needed and thus be able to properly push off the ground with their legs.

A sprinter approximately swings their arm from back to front in around 0.15 seconds.

The weight of a human arm is approximately 10 kg

the force applied by the shoulder is not brief , it is continous. So we will need the average Force.

We can model this arm swing as follows :

We will calculate the average force during the most difficult part of the movement, so when we go from angle 0° to angle 50° in about 0.06 seconds.

From what we saw previously, we only worked on examples where forces remain fixed relative to the reference frame. But in our case, as the arm swings, it’s as if the muscles are applying a tangential force to the arm, so the coordinates of the vector will change over time.

To overcome this difficulty, we can set our y-axis along the bar and the x-axis perpendicular to it. That way, F will only have an x-component.

The direction of the reaction force is not fixed. It’s a “reaction force” that responds to all the other forces and the motion of the bar. It generally has both a horizontal and a vertical component.

What about the acceleration? Since this involves rotation, how can we represent it? You know that the length of an arc equals the radius multiplied by the angle, right? Therefore, the velocity along an arc is the angular velocity multiplied by the radius. The same principle applies to acceleration.

so in the newton law, we can plug in the angle.

this way we’ll have:

and the forces acting on the system are F_shoulder(t), weight, and the reaction force from the pivot (shoulder)

We will focus our analysis on the moment when the athlete initiates movement at the “go” position, as this is when maximum force is required.

Since F_shoulder is zero along the y-axis, we will analyze only the x-axis to derive the expression for F_shoulder.

r is the distance between the pivot and the center of mass

we will us a maths trick to get rid of that R_pivot, because we don’t know that Force.

Let’s divide the bar into numerous infinitesimal segments.

for a segment i :

alpha is the angular acceleration

These forces are: gravitational force, shoulder force (if the shoulder muscle is in contact with that segment), and internal forces from adjacent segments (internal tensions and compressions).

we can multiply that equation by r_i (r_i is the distance from the pivot to the segment i):

the thing is that the pivot reaction force is only valid on the segment having r_i = 0 and is zero elsewhere.

so let’s sum all segments together :

using the definition of center of mass…

To transform the right side of the equation, we can replace m_i with Mass of the whole arm/length of whole arm, and integrate ri² from 0 to Length_of_arm. This gives:

this is called the moment of inertia

We are studying the movement when the arm starts aligned with the torso, so at that moment, θ = 0, so:

We want to achieve the velocity of 14.54 rad/s in less than 1 second, so alpha =14.54 rad.s^-2, m =10kg, L=0.5m, rf = 0.05m

So at the start, the shoulder should produce a force of:

This is almost the same amount of force required to lift a 20kg barbell over the shoulder.

Sprinters swing their arms a lot (20+ times) during a race. So yes, upper body strength is needed to win a race. But as you noticed, we need less force to swing the arm (from torso to front) in the middle of the race than at the start because the arm already has initial speed.

It’s like when we do push-ups. If you stop the movement in the middle of the push, it will feel much harder to push the remaining distance.”

Let’s compare the push phase of a normal push-up vs. the push phase when we stop in the middle.

When you stop the movement in the middle of the push, your actual velocity is 0, so you need an acceleration. So, according to Newton’s second law, the force required to push you up is:

Whereas in the normal push-up, in the middle of the push, the force required is:

As we’ve seen, Newton’s second law adequately explains most everyday mechanical phenomena. The key is to first establish an accurate model before proceeding with calculations.

While multiple approaches exist, some prefer energy-based methods over force-based analysis.

When should I use energy-based methods versus force-based analysis?

Both energy and force approaches are indeed valid for analyzing physical systems, and choosing between them often depends on which makes the problem easier to solve.

For example,if you try to understand why a tennis ball bounces to a height less than the height from which it was dropped

If you think in terms of energy :

When a ball hits a surface, some energy is transformed into sound energy, some is transformed into thermal energy from the friction created, and some becomes elastic potential energy resulting from the deformation of the ball when it collides. This elastic potential energy is why the ball is able to bounce, or rebound. After the ball rebounds, the elastic potential energy is transformed into kinetic energy, but it will never possess as much kinetic energy as during its original fall because some of the original kinetic energy has been transformed into sound, friction, and deformation of the ball.

from https://www.youtube.com/watch?v=1yT0hxplVBg

If you think in terms of forces :

During the first fall, the only force applied on the ball is the gravity.

At the moment of ground contact, the ball exerts a brief but significant force against the surface, generating acoustic energy through molecular vibrations. As the ball undergoes compression, it continues applying downward force while experiencing internal stress distribution throughout its rubber structure(a material with imperfect elastic properties). Upon reaching maximum deformation, the forces responsible for arresting the ball’s downward motion simultaneously contribute to its structural distortion..

In returning to its original shape, the ball continues to exert a force on the ground, and the ground continues to exert an upward force on the ball, accelerating it until it leaves the ground with some upward velocity.

But inside the ball, layers of material tug on each other, slip slightly, and rub at the microscopic level. These resisting forces during motion convert some organized movement into random molecular vibrations (which produces heat). If there’s any sliding at the contact patch, surface friction adds even more heating.

Because some of the push during the “unsquish” is spent shaking air and rubbing inside, the ball can’t push itself off the ground as strongly as it did while stopping. The rebound push is weaker and slightly delayed, so the ball leaves more slowly and doesn’t reach its original height.

Upon closer inspection, by manipulating Newton’s second law, we can see how energy and force converge in a single equation:

we suppose the mass constant

T(v) is the kinectic energy : 1/2 m v²

Newton’s second law is universal, but for complicated systems it can be cumbersome : vector equations, constraints, and moving reference frames make life harder. That’s where Lagrangian mechanics comes into play. It is a different approach where we don’t use cartesian coordinates (x,y,z) like in Newton’s mechanics, but we choose a coordinate for each degree of freedom of the system.

Hamiltonian mechanics presents another powerful perspective, among others. Each framework unveils unique insights, and while physics is infinitely vast, capable of absorbing a lifetime of exploration, what matters most is to savor the journey of learning and stay endlessly curious.