An 1891 Panhard et Levassor, 600 kg, engine out of gear, rolls down a slope. What pushes it along, what holds it back, and how strong does its brake need to be? The arrows are the real forces from the simulation.
θ = 5°, brake off
Without friction, a = g sin θ. Work it out before you watch. Then compare with the measured a. Why is the car a little slower?
g sin 5° = 9.81 × 0.0872 = 0.855 m/s². The car measures about 0.78 m/s². The difference is the road friction F of about 45 N. It is not the brake: the road has to push back on each tyre to spin its wheel up. Heavy wheels take some of the energy, so the car speeds up more slowly.
μ = 0.5, P = 720 N per block, two blocks
The brakes can resist at most 2μP. On what slope does that equal mg sin θ? Predict the angle, then try just below and just above it. Press Brake and see if the car stops.
2μP = 720 N = mg sin θ, so sin θ = 720 / 5886 = 0.122 and θ ≈ 7.0°. At 6.5° the car stops and holds. At 7.5° it keeps creeping faster, however long you hold the brake. This model's brake is weak compared with its weight: the same 180 N·m per wheel it was validated with.
θ = 10°, brake on
What brake friction coefficient μ is just enough to hold the car on this hill? Work it out, then try a little below and a little above your answer.
2μP = mg sin 10°, so μ = 1022 / 1440 ≈ 0.71. At 0.65 it creeps; at 0.75 it holds. Why not use μ = 10? Past a point the blocks lock the wheels, and the tyres slide on the road. Then the limit is the tyres' grip, about μroad × N on the rear wheels (μroad = 0.8 here), not the brake.
θ = 5°, μ = 0.5
Cows wander onto the road now and then. Let the car reach 3 m/s, then brake. How far will it roll before it stops? Use a = (2μP − mg sin θ) / m and v² = 2as, then watch the distance to the cow. Then press Surprise cow: a cow steps out when you least expect it, and the page measures your own reaction time.
a = (720 − 513) / 600 = 0.35 m/s², so s = 3² / (2 × 0.35) ≈ 13 m. The simulated car needs about 15.5 m. The brake takes a quarter of a second to bite, and the spinning wheels carry energy that the brake must also absorb (the same effect as in challenge 1). At 5 m/s the stop is nearly three times as long. Speed counts twice in v². On a real road, add the thinking distance: the car keeps going at full speed during your reaction time, so thinking distance is v × t.