Applications of Newton's Laws

Friction is a contact force between two surfaces that resists their relative motion or tendency to slide. Far from being a nuisance, friction is what keeps tires on roads, allows you to walk, and lets nails stay in wood. Understanding it quantitatively is essential for solving a vast range of Newton's-law problems.

Static and Kinetic Friction

Two distinct types of friction act depending on whether surfaces are sliding relative to each other.

Static friction $f_s$ acts when surfaces are not sliding. It is a reactive force that adjusts in magnitude to exactly cancel any applied force component that would otherwise cause sliding — up to a maximum limit. This is why a book on a tilted desk stays put: static friction matches and opposes the gravitational component trying to slide it. Once you push hard enough to exceed that maximum, the book moves.

Kinetic friction $f_k$ acts when surfaces are sliding past each other. Unlike static friction, $f_k$ has a fixed magnitude independent of sliding speed or the size of the applied force. It is always directed opposite to the velocity of sliding.

The dimensionless constants $\mu_s$ (coefficient of static friction) and $\mu_k$ (coefficient of kinetic friction) depend on the pair of surfaces in contact. Crucially, $\mu_k < \mu_s$ always — it takes more force to start something sliding than to keep it sliding. This is why a heavy couch requires a bigger initial push than the force needed to keep sliding it across the floor.

Microscopic Origin: Cold Welding

Even polished surfaces are microscopically rough, touching only at tiny raised bumps called asperities. At these contact points, atoms from each surface are so close that they bond together — a process called cold welding. Static friction is the force required to break these microscopic bonds. Pressing harder (larger $N$) creates more and stronger bonds, which is why $f_{s,\max} \propto N$. The total contact area does not appear because doubling the area also halves the pressure per asperity — the two effects cancel exactly, giving the remarkably simple formula $f_{s,\max} = \mu_s N$.

Finding the Normal Force

The most common friction mistake is assuming $N = mg$. This is only true on a horizontal surface with no vertical applied forces. In general, find $N$ from Newton's second law perpendicular to the surface. On an incline at angle $\theta$, the perpendicular equilibrium gives $N = mg\cos\theta$ — less than $mg$. If someone pushes the object down into the surface, $N$ increases and so does friction. If they pull it upward, $N$ decreases.

When an object moves through a fluid (liquid or gas), the fluid exerts a drag force that opposes the motion. This is the force that prevents falling objects from accelerating forever, sets the top speed of vehicles, and is why opening a parachute saves your life. Unlike kinetic friction between solid surfaces, drag depends strongly on speed.

The Drag Equation

For an object moving through a fluid at speed $v$, experiment and theory give:

where $C$ is the dimensionless drag coefficient (depends on the shape of the object — a sphere: $C \approx 0.47$; a person spread-eagle: $C \approx 1.0$; a streamlined car: $C \approx 0.3$); $\rho$ is the fluid density (air at sea level: $\rho \approx 1.21\,\text{kg/m}^3$); and $A$ is the cross-sectional area perpendicular to the velocity. The $v^2$ dependence is crucial: doubling your speed quadruples the drag force.

Terminal Speed

Consider an object dropped from rest and falling through air. At $t = 0$, $v = 0$ so $D = 0$ and the object accelerates downward at $g$. As $v$ increases, drag $D = \frac{1}{2}C\rho A v^2$ grows, reducing the net downward force and therefore the acceleration. Eventually $D$ equals the gravitational force $F_g = mg$ and the net force reaches zero — the object then falls at constant terminal speed $v_t$.

A heavier object (larger $mg$) reaches a higher terminal speed. A larger or flatter object (larger $A$) reaches a lower terminal speed — this is precisely why a parachute works. Skydivers control their terminal speed by changing body orientation: spread-eagle gives $v_t \approx 55\,\text{m/s}$; head-down gives $v_t \approx 90\,\text{m/s}$.

An object moving at constant speed $v$ around a circle of radius $R$ is not in equilibrium — it is continuously changing direction, so it has an acceleration. That acceleration points toward the center of the circle and has magnitude:

Applying Newton's Second Law in the centripetal (toward-center) direction, the net force must equal $ma_c$:

Car on a Flat Circular Curve

A car of mass $m$ rounds a horizontal circular curve of radius $R$ at speed $v$. On a flat road, the only horizontal force is static friction between the tires and road surface. Static friction must provide the centripetal force:

The car can navigate the curve without skidding only when $v^2 \leq \mu_s g R$. Notice that mass cancels — the maximum safe speed is independent of the car's weight! A lighter car and a heavier truck have the same maximum cornering speed on the same road. Greater radius $R$ or higher friction coefficient $\mu_s$ both allow faster cornering.

The Banked Curve

If the roadway is banked (tilted inward) at angle $\theta$, the horizontal component of the normal force also contributes to the centripetal force. At the ideal banking angle, the normal force alone provides all the centripetal force needed — no friction required at all. Setting up the equations (with the road surface tilted by $\theta$):

Dividing these equations eliminates $N$ and $m$:

Highway engineers use this formula to design banked curves for the posted speed. At the design speed, a car needs zero friction — so vehicles can navigate the curve safely even on ice. NASCAR tracks are banked up to 33°, allowing cars to corner at over 200 km/h.

Vertical Circular Motion: Top of a Loop

For an object at the top of a vertical circular loop of radius $R$, both the normal force $N$ and weight $mg$ point downward (toward the center). Applying Newton's second law centripetally:

The normal force decreases as speed decreases. When $N = 0$ the track exerts no force on the object — the object is momentarily in free fall while still moving in a circle. This gives the minimum speed at the top to maintain contact with the track:

Below this speed, the required centripetal force exceeds $mg$ alone, which would require $N < 0$ — impossible for a normal force. The object leaves the track. Above this speed, the rider feels a normal force pressing them into the seat (toward the center from above), giving a sensation of "heaviness" even at the top of the loop.

These three circular-motion scenarios share the same approach: (1) identify the physical force(s) that point toward the center, (2) set their centripetal component equal to $mv^2/R$, and (3) write a second equation for the perpendicular direction if needed. The centripetal direction is always the key equation.