Banked track: from the survey stakes to the cornering force

SPH4U Grade 12 Physics · Dynamics

A closed test oval is being commissioned. On the banked north turn a surveyor levels across one 3.50 m lane and finds the outer edge of the pavement sitting 1400 mm above the inner edge; the turn itself is laid out on a 175 m radius. Find the bank angle, the speed a car could hold through the turn with no help at all from sideways friction, the centripetal acceleration at that speed, and the inward force the pavement must supply to a 1520 kg car.

Step 1 of 4 · solve for Acute angle

The bank angle is pure geometry — rise over horizontal run — and it has to come first, because every dynamics question about this turn is really a question about this one angle. The rise comes off the level in millimetres and the lane width in metres, so watch the conversion: 1400 mm across 3.50 m is a ratio of 0.4, not 400.

Rearranged for θ
θ=arctan(oa)\theta = \arctan\left(\frac{o}{a}\right)
Your values, in your units
θ=arctan((1,400 mm)(3.5 m))\theta = \arctan\left(\frac{\left( 1{,}400\ \text{mm} \right)}{\left( 3.5\ \text{m} \right)}\right)
Converted to base units
θ=arctan((1.4 m)(3.5 m))\theta = \arctan\left(\frac{\left( 1.4\ \text{m} \right)}{\left( 3.5\ \text{m} \right)}\right)
Answer
θ=21.8\theta = 21.8

Carried onward at full precision, not this rounded figure.

Open the Right-Triangle Tangent Ratio (TOA) solver →

Step 2 of 4 · solve for Design speed

Now the physics. At the design speed the tilted normal force alone supplies mv²/r, so tan θ = v²/rg and the angle from step 1 hands over a speed. Note the mass never enters — the same tilt works for a motorbike.

Rearranged for v
v=rgtanθv = \sqrt{r g \tan\theta}
21.8 °carried from step 1
Your values, in your units
v=(175 m)gtan(0.380506 rad)v = \sqrt{\left( 175\ \text{m} \right) \, g \, \tan \left( 0.380506\ \text{rad} \right)}
Converted to base units
v=(175 m)gtan(21.8014 )v = \sqrt{\left( 175\ \text{m} \right) \, g \, \tan \left( 21.8014\ ^{\circ} \right)}
Answer
v=26.2v = 26.2

Carried onward at full precision, not this rounded figure.

Open the Banked Curve Angle solver →

Step 3 of 4 · solve for Centripetal acceleration

Feed the design speed back into a = v²/r. The result is g tan θ, which is a useful thing to recognise on sight.

Rearranged for ac
ac=v2ra_c = \frac{v^2}{r}
26.2 m/scarried from step 2
Your values, in your units
ac=(26.2005 m/s)2(175 m)a_c = \frac{\left( 26.2005\ \text{m/s} \right)^2}{\left( 175\ \text{m} \right)}
Answer
ac=3.92a_c = 3.92

Carried onward at full precision, not this rounded figure.

Open the Centripetal Acceleration (a = v²/r) solver →

Step 4 of 4 · solve for Force

Newton's second law converts that acceleration into the force the road has to deliver. This is the number a structural engineer needs; the driver only feels it.

Rearranged for F
F=maF = m a
3.92 m/s²carried from step 3
Your values, in your units
F=(1,520 kg)(3.92266 m/s2)F = \left( 1{,}520\ \text{kg} \right) \, \left( 3.92266\ \text{m/s}^{2} \right)
Answer
F=5.96F = 5.96

Carried onward at full precision, not this rounded figure.

Open the Newton's Second Law solver →

Why this order

The order here is the whole lesson. A bank angle is a geometric fact about a piece of concrete, measurable with a level and a tape before a single car drives on it, so it goes first. Only then does dynamics get a say: tan θ = v²/rg turns that fixed geometry into one particular speed, the design speed, at which the normal force does all the work and friction does none. Below it a car tends to slide down the bank; above it, friction has to make up the difference, and when the tyres run out the car goes over the top. A single bank angle is not a safety margin — it is a single speed, with tolerance on either side paid for in grip.

Two things trip students between step 1 and step 2. The first is units, and it bites twice. The level reading arrives in millimetres and the lane width in metres, so a ratio taken off the raw numbers is 400 instead of 0.4 and the bank comes out at 89.86°; then the answer to step 1 is an angle, which this site carries in radians even though it shows you degrees, and a calculator left in the wrong mode produces a design speed that is wrong by nothing recognisable. The second is the disappearing mass. Students expect a heavy car to need a steeper bank, but writing N sin θ = mv²/r beside N cos θ = mg and dividing kills the m outright — which is why step 2 needs no mass at all and step 4 suddenly does. Mass decides how much force, never how much tilt. Railway engineers hit the same result in the 1830s and still call it cant; Daytona's 31° and Talladega's 33° are this equation built at the limit of what a paving machine can hold.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.