Force and Motion — II
Three forces that shape everyday motion — friction, drag, and the pull toward the center
- The difference between static and kinetic friction, and how each relates to the normal force.
- The friction laws \(f_{s,\max} = \mu _{s}F_{N}\) and \(f_{k} = \mu _{k}F_{N}\), and what the coefficients mean.
- The drag force on a body moving through air and the terminal speed it sets.
- That uniform circular motion needs a centripetal force — and that this is a role, not a new kind of force.
- How to handle flat and banked turns and vertical loops with free-body diagrams.
What Is Physics?
This chapter zooms in on three forces that govern ordinary motion: friction, drag, and centripetal force. An engineer prepping a car for the Indianapolis 500 must respect all three — friction on the tires for grip out of the pits and the corners, drag from the rushing air to keep fuel use down, and enough centripetal force in the turns to keep the car off the wall. We begin with friction.
Friction
Friction is everywhere, and we depend on it utterly: without it we could not walk, hold a pencil, drive a nail, or tie a knot — yet about 20% of a car's fuel goes just to fighting it. Consider pushing a heavy crate. Push gently and it stays put; a frictional force exactly matches your push. Push harder and it still holds — the frictional force grows to match. Push hard enough and the crate suddenly breaks free and slides. So friction comes in two regimes:
Acts while the surfaces are not sliding. It adjusts itself to balance whatever you apply, up to a maximum value. Below that maximum, the body stays at rest.
Acts once the surfaces are sliding. It has a single value — usually smaller than the static maximum — which is why a crate lurches forward at breakaway and you must ease off to keep it moving steadily.
Microscopically, friction is the vector sum of countless tiny cold-welds where the high points of two surfaces touch. The true contact area is far smaller than it looks — perhaps by a factor of 10⁴. Sliding continually tears and re-forms these welds, and the stick-then-slip can sing as a violin string or squeal as skidding tires.
Properties of Friction
For a dry, unlubricated body pressed against a surface, experiment gives three rules:
(1) If the body is still, static friction equals the parallel applied force, opposing it. (2) Its maximum is \(f_{s,\max} = \mu _{s}F_{N}\); exceed it and sliding begins. (3) Once sliding, friction drops to the roughly constant \(f_{k} = \mu _{k}F_{N}\).
The Drag Force & Terminal Speed
A fluid is anything that flows — a gas or a liquid. When a body and a fluid move relative to each other, the fluid exerts a drag force \(D\) opposing the relative motion. For a blunt body in air, fast enough to leave turbulent swirls behind it, the magnitude is:
Drop a blunt body from rest and the upward drag starts at zero and climbs as it speeds up, until it just balances gravity. The acceleration then vanishes and the body coasts at constant terminal speed. Setting \(D = F_{g}\) gives:
Uniform Circular Motion
Recall from Chapter 4 that a body moving in a circle at constant speed \(v\) has a centripetal acceleration of constant magnitude pointed at the center. By Newton's second law, some force must produce it — a centripetal force, also directed inward:
A centripetal force changes the direction of the velocity without changing its magnitude. The role can be filled by friction (a car cornering), gravity (a satellite orbiting), tension (a whirled string), or a normal force (a banked road or a loop). Identify which real force is doing the centripetal job, then set it equal to \(mv^{2}/R\).
On a level curve the inward force is static friction on the tires. On the verge of sliding, \(\mu _{s}F_{N} = mv^{2}/R\), which caps the safe cornering speed.
Tilt the road and the normal force gains an inward component. Even with no friction, the right bank angle holds the car on the curve: \(\tan \theta = v^{2}/gR\).
At the top, gravity and the normal force both point down, together supplying \(mv^{2}/R\). Contact is just barely kept when \(F_{N} = 0\), giving a minimum speed \(v = \sqrt{gR}\).
Putting It to Work
Problem. A car brakes with locked wheels and skids to a stop, leaving marks \(290 m\) long (a real record set by a Jaguar in 1960). If \(\mu _{k} = 0.60\), how fast was it going when the wheels locked?
Solution. The only horizontal force is kinetic friction, so \(-f_{k} = ma\) with \(f_{k} = \mu _{k}F_{N} = \mu _{k}mg\). The mass cancels: \(a = -\mu _{k}g\). Then use \(v^{2} = v_{0}^{2} + 2a(x - x_{0})\) with final speed zero:
And since the marks ended only because the car left the road, the true speed was at least 210 km/h. Notice the mass dropped out entirely — a loaded and an empty car skid the same distance.
Problem. A force \(F = 12.0 N\) is applied at an upward angle \(\theta\) to a 3.0 kg block sliding on a floor with \(\mu _{k} = 0.40\). What angle gives the greatest acceleration?
Solution. An upward tilt lifts some weight off the floor, so \(F_{N} = mg - F \sin \theta\) and the acceleration is \(a = (F \cos \theta )/m - \mu _{k}(g - (F \sin \theta )/m)\). Setting \(da/d\theta = 0\) gives a clean result:
Two effects compete: tilting up relieves the normal force (less friction, good) but also shrinks the horizontal pull (bad). They balance at \(\theta \approx 22^{\circ}\), where the acceleration peaks.
Problem. A spherical raindrop of radius \(R = 1.5 \mathrm{mm}\) falls from a 1200 m cloud with drag coefficient \(C = 0.60\) (water density 1000 kg/m³, air density 1.2 kg/m³). Find its terminal speed, and compare with the no-drag speed.
Solution. With \(A = \pi R^{2}\) and \(F_{g} = (4/3)\pi R^{3}\rho _{w}g\), the terminal-speed formula simplifies to \(v_{t} = \sqrt{8R\rho _{w}g / 3C\rho _{a}}\):
Drag is the difference between gentle rain and being pelted at the speed of a handgun bullet. The cloud height drops out of the terminal-speed result — only the drop's size and the fluids matter.
Problem. A stunt cyclist rides a vertical loop of radius \(R = 2.7 m\). What is the least speed at the very top that keeps the bike in contact with the track?
Solution. At the top both gravity and the normal force point down, toward the center, together giving the centripetal force: \(F_{N} + mg = mv^{2}/R\). "Just barely in contact" means the track pushes with \(F_{N} = 0\):
The mass cancels — a heavier rider needs exactly the same minimum speed. Below 5.1 m/s at the top, the bike falls away from the loop.
Problem. A car rounds a banked circular track of radius \(R = 190 m\) at a steady \(v = 20 m/s\). If friction is negligible, what bank angle \(\theta\) keeps it on the road?
Solution. Tilting the road tips the normal force inward. Its horizontal part supplies the centripetal force, \(F_{N} \sin \theta = mv^{2}/R\), while its vertical part carries the weight, \(F_{N} \cos \theta = mg\). Divide the two — both \(F_{N}\) and \(m\) cancel:
At exactly this speed the bank alone holds the car; faster or slower, friction makes up the difference. This is why curves are banked — and why wet, icy curves at speed are so dangerous.
Chapter Summary
Static adjusts to match the applied force up to a maximum; kinetic acts during sliding and is usually smaller.
\(f_{s,\max} = \mu _{s}F_{N}\) and \(f_{k} = \mu _{k}F_{N}\); friction is parallel to the surface, opposing the slide.
\(D = \tfrac{1}{2}C\rho Av^{2}\) opposes relative motion and grows with the square of speed.
Reached when \(D = F_{g}\): \(v_{t} = \sqrt{2F_{g}/C\rho A}\). Bigger A ⇒ slower fall.
\(a = v^{2}/R\) inward; net inward force \(F = mv^{2}/R\) — supplied by some real force.
Flat turn: friction caps the speed. Banked: \(\tan \theta = v^{2}/gR\). Loop top: \(v_{\min} = \sqrt{gR}\).
Problems
Draw the free-body diagram first, and decide whether friction is static (matching) or kinetic (μkFN). For circular motion, name the real force playing the centripetal role before writing \(F = mv^{2}/R\). Take \(g = 9.8 m/s^{2}\).
- A 45 kg bureau rests on a floor with μs = 0.45. (a) What minimum horizontal force starts it moving? (b) If 17 kg of contents are removed first, what is the new minimum?
- A person pushes horizontally with 220 N on a 55 kg crate, with μk = 0.35. Find (a) the kinetic frictional force and (b) the crate's acceleration.
- A 2.5 kg block on a horizontal surface has μs = 0.40 and μk = 0.25. A horizontal 6.0 N force is applied. Determine the frictional force, and state whether the block moves.
- The coefficient of static friction between Teflon and scrambled eggs is about 0.04. What is the smallest tilt angle of the pan that lets the eggs start to slide?
- A slide-loving pig slides down a 35° slide in twice the time it would take on a frictionless 35° slide. What is μk between pig and slide?
- A 3.5 kg block is pushed along a floor by a 15 N force directed 40° below the horizontal, with μk = 0.25. Find (a) the frictional force and (b) the block's acceleration.
- What is the terminal speed of a 6.00 kg spherical ball of radius 3.00 cm and drag coefficient 1.60, falling through air of density 1.20 kg/m³?
- A cat dozes on a merry-go-round 5.4 m from the center. The ride spins up to one rotation every 6.0 s. What least coefficient of static friction keeps the cat from sliding off?
- What is the maximum speed at which a car can round a flat, unbanked curve of radius 30.5 m if μs between tires and road is 0.60?
- A stuntman drives over the top of a hill whose crest is a circular arc of radius 250 m. What is the greatest speed at which the car stays on the road at the top?