Laws of Motion, Friction and Circular Dynamics
Almost every mechanics problem you will ever meet is solved by the same three steps: isolate one body, draw every force on it, and write ΣF = ma along a chosen direction. The physics is one line; the marks are in the discipline. Students who lose them do so by forgetting a force, by drawing a force that does not exist (centrifugal force in a ground frame is the classic), or by writing a single equation for two bodies that have different accelerations. This chapter builds the free-body habit first and then applies it to the two contexts the paper returns to most — friction and circular motion.
Friction — Static, Kinetic and the Angle of Repose
The single most misused formula in mechanics is f = μN. For kinetic friction it is correct: once sliding, f = μₖN regardless of the applied force. For static friction it is wrong as written, because static friction is self-adjusting: f ≤ μₛN, and it takes whatever value is needed to prevent motion, up to that maximum.
| Feature | Static friction | Kinetic friction |
|---|---|---|
| Magnitude | Variable: equals the applied force, up to μₛN | Fixed at μₖN once sliding |
| Relative size | μₛ > μₖ, which is why starting a load is harder than keeping it moving | — |
| Depends on contact area? | No — only on N and the surfaces | No |
| Direction | Opposes impending relative motion | Opposes actual relative motion |
⚠️ A 10 N push on a block with μₛN = 30 N gives 10 N of friction, not 30
This is worth stating as bluntly as possible, because it is the error that recurs most. If the maximum static friction is 30 N and you push with 10 N, the block does not move, so the net force must be zero, so friction is exactly 10 N. Push with 25 N and friction is 25 N. Push with 31 N and the block breaks free, after which friction drops to the kinetic value μₖN, which is lower — this is why a heavy box lurches forward the instant it starts to move. So the procedure is always: compute μₛN, compare it with the applied force, and only then decide whether you are in the static case (friction = applied force) or the kinetic case (friction = μₖN).🎯 The angle of repose, and why it measures μ directly
Tilt a plane until a block just begins to slide. At that instant the component pulling it down the slope, mg sin θ, exactly equals the maximum static friction, μₛ(mg cos θ). The masses cancel — so tan θ = μₛ, and the angle at which sliding begins depends only on the surfaces, not on the weight of the block. That is the angle of repose, and it is genuinely useful: it lets you measure a coefficient of friction with nothing but a protractor. It also explains why sand and gravel piles always settle to the same characteristic slope whatever their size. Two related results worth holding: on an incline steeper than repose, a released block accelerates at g(sin θ − μ cos θ), and if the incline is frictionless that reduces to g sin θ.Circular Dynamics — and the Force That Does Not Exist
A body moving in a circle at constant speed is still accelerating, because its velocity is changing direction. That acceleration points to the centre and has magnitude a = v²/r = ω²r. Newton's second law then demands a real inward force of magnitude mv²/r, and the entire subject is the question: which real force is supplying it?
| Situation | What supplies mv²/r | Result |
|---|---|---|
| Car on a flat road turning | Static friction between tyre and road | v_max = √(μrg) — independent of mass |
| Banked road, no friction | Horizontal component of the normal force | tan θ = v²/rg |
| Stone on a string, horizontal circle | Tension | T = mv²/r |
| Satellite in orbit | Gravity | GMm/r² = mv²/r |
| Vertical circle, at the top | Tension plus weight, both pointing down | v_min at top = √(gr) when T = 0 |
⚠️ There is no centrifugal force in a ground frame
A free-body diagram drawn in the ground frame must contain only real forces — gravity, normal, friction, tension. Adding an outward "centrifugal force" is wrong, and it is wrong in a way that produces a plausible-looking equation summing to zero, which is why it survives. What you actually feel thrown outward in a turning car is your own inertia: your body continues straight while the car turns into you, and the inward force on you comes from the seat and the door. Centrifugal force is a legitimate pseudo-force and it is the right tool — but only when you deliberately work in the rotating frame, where you add mω²r outward and then treat the body as being in equilibrium. Pick one frame and stay in it; mixing them, by writing both mv²/r inward and a centrifugal term, double-counts and is the error the examiner is hoping for.🧠 The vertical circle, and the two different minimum speeds
For a stone on a string in a vertical circle, the string can only pull. At the top both tension and weight point downward, so mg + T = mv²/r; the limiting case is T = 0, giving v_top = √(gr). Applying conservation of energy from top to bottom then gives v_bottom = √(5gr) as the minimum speed at the lowest point to complete the loop. But if the stone is on a rod instead of a string, the rod can push as well as pull, so it can support the stone at the top at arbitrarily small speed — the minimum becomes v_top = 0 and v_bottom = √(4gr) = 2√(gr). Questions distinguish string from rod precisely to see whether you noticed, and the same distinction applies to a bead on a smooth wire or a ball inside a tube.🎯 Why roads are banked, and why the safe speed has no mass in it
On a flat road the only inward force available is friction, so v_max = √(μrg) — and friction is exactly what fails on a wet or icy surface. Banking tilts the road so that the normal force itself acquires an inward horizontal component, N sin θ, which supplies mv²/r with no help from friction at all. Setting N sin θ = mv²/r and N cos θ = mg and dividing gives tan θ = v²/rg. Notice what is absent from both formulae: the mass. A loaded truck and a motorcycle overturn at the same speed on the same curve, because the required force and the available force both scale with m, so it cancels. That is why speed limits on curves are posted for the curve and not for the vehicle.Key takeaways
- Isolate one body, draw every force, choose axes along the motion, write ΣF = ma for each body separately.
- F = dp/dt is the general second law; F = ma assumes constant mass.
- Action and reaction act on DIFFERENT bodies, so they never cancel within one free-body diagram.
- N = mg only on a horizontal surface with no vertical acceleration; on an incline N = mg cos θ, in a lift m(g ± a).
- Static friction is self-adjusting: f ≤ μₛN and equals the applied force until the block breaks free.
- Angle of repose: tan θ = μₛ, independent of mass — so a protractor measures the coefficient.
- Circular motion needs a real inward force mv²/r; centrifugal force exists only in the rotating frame.
- Vertical circle on a string: v_top = √(gr), v_bottom = √(5gr). On a rod both minima drop.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A block on a horizontal floor has maximum static friction 30 N. A horizontal push of 10 N is applied and the block does not move. The frictional force is:
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Answer: B — 10 N
Static friction is self-adjusting. Since the block does not move, the net force is zero, so friction must be exactly 10 N — equal and opposite to the push. The 30 N figure is only the maximum available; using it whenever friction appears is the single most common error in this topic.A block just begins to slide on an incline at 30°. The coefficient of static friction is:
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Answer: B — 1/√3 ≈ 0.58
At the angle of repose mg sin θ = μₛ mg cos θ, so μₛ = tan θ = tan 30° = 1/√3 ≈ 0.58. The masses cancel, which is why the angle depends only on the surfaces — and why a protractor is enough to measure a coefficient of friction. Choosing 0.5 comes from using sin 30° instead of tan 30°.A man of mass m stands in a lift that is in free fall. The normal force on him from the floor is:
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Answer: C — Zero
In a lift accelerating downward at a, N = m(g − a); in free fall a = g, so N = 0. This is exactly what weightlessness is — not the absence of gravity, which still acts, but the absence of any contact force, so the man and the floor fall together and press on each other not at all.The maximum speed at which a car can take an unbanked curve of radius r is:
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Answer: A — √(μrg), independent of the mass of the car
Friction supplies the centripetal force, so μmg = mv²/r and the mass cancels, giving v = √(μrg). This is why speed limits on curves are posted for the curve rather than the vehicle: the required force and the available force both scale with mass, so a loaded truck and a motorcycle have the same limiting speed.For a stone tied to a string to just complete a vertical circle of radius r, its minimum speed at the topmost point is:
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Answer: B — √(gr)
At the top both weight and tension point downward: mg + T = mv²/r. The limiting case is T = 0, which gives v = √(gr). The value √(5gr) is the minimum speed at the lowest point, obtained from this by energy conservation, and zero would apply only if the stone were on a rod, which can push as well as pull.Drawing a free-body diagram in the ground frame for a car turning a corner, one should:
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Answer: B — Include only real forces, with friction providing the inward mv²/r
In an inertial ground frame only real forces belong on the diagram, and here static friction supplies the centripetal force. Centrifugal force is a legitimate pseudo-force but only in the rotating frame; including both, as option 3 does, double-counts and is exactly the error the question is set to catch.It is easier to pull a heavy crate at an angle above the horizontal than to push it at the same angle below, because pulling:
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Answer: B — Reduces the normal force and hence the friction
Pulling upward at angle θ gives N = mg − F sin θ, while pushing downward gives N = mg + F sin θ. Since friction is μN, the pull reduces the resisting force and the push increases it. Note that the coefficient itself is a property of the surfaces and does not change, and the weight certainly does not change — only the normal force does.A road is banked at angle θ so that no friction is needed at speed v on a curve of radius r. Then:
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Answer: B — tan θ = v²/rg
With no friction, N sin θ = mv²/r horizontally and N cos θ = mg vertically; dividing eliminates both N and m to give tan θ = v²/rg. The mass vanishing is the physically interesting part — the correct banking angle is the same for a truck and a motorcycle, which is why it can be built into the road at all.