A school corridor has just been mopped. A student hurries across and the back foot slips away backward. On a dry floor the same step works every time. On the frozen lake at Gulmarg in winter, visitors shuffle, flap their arms, and hardly move. The legs push just as hard on ice as on a dry road. Something that the dry floor gives, the ice does not.
Walking: the ground pushes you forward
When you take a step, your back foot pushes on the ground. The push is backward and a little downward. If the ground were perfectly slippery, the foot would slide backward, exactly as it does on a wet floor. On a dry floor the foot does not slip. So the foot tends to slide backward, and static friction on the foot points against that tendency, forward.
By Newton's third law, the foot pushes the ground backward and the ground pushes the foot forward, with equal forces. The forward friction on the foot is the only horizontal force from outside the body. So this friction is the force that accelerates you forward. Your muscles cannot move you by themselves; they can only push on the ground and let the ground push back.
The foot does not slide over the ground, so this friction is static. Its largest value is \(\mu_s N\). A sprinter who tries to accelerate faster than the ground can push asks for more than \(\mu_s N\), and the foot slips. That is why athletes wear spiked shoes, and why you take short steps on a wet floor.
Friction on a walking foot points forward, in the direction of motion of the walker. This does not break the rule of Lesson 1. Friction opposes the relative motion of the surfaces, and the sole tends to slide backward over the ground. Always ask which way the surface in contact would slide, not which way the body is going.
Cycles and cars
A cycle works the same way. The pedals turn the rear wheel. At the bottom of the wheel, the tyre pushes backward on the road, trying to spin in place. The road grips the tyre and pushes it forward with static friction. The rear wheel is the driving wheel, and friction on it points forward. In most small cars in India the engine turns the front wheels, so friction on the front wheels points forward.
A cyclist pedals to speed up on a level road. What is the direction of friction from the road on the rear tyre?
Why is it nearly impossible to start walking on smooth ice?
Braking
When the driver brakes, the brake pads slow the turning of the wheels. The tyres now tend to slide forward over the road. Road friction on each braked tyre therefore points backward, against that tendency. This backward friction is the force that stops the car. The brakes only slow the wheels; the road stops the car.
A tyre that rolls without slipping touches the road with a patch that is momentarily at rest. So the best braking uses static friction. With all four wheels braked, the largest friction is \(\mu_s N = \mu_s mg\), and the largest deceleration is \(\mu_s g\). If the driver brakes too hard, the wheels lock and the tyres slide over the road. This is skidding. A skidding tyre feels kinetic friction, which is smaller, so the car takes longer to stop. A skidding front wheel also cannot steer.
A car moves at \(72\,\text{km/h}\) on a level highway. Take \(g = 10\,\text{m/s}^2\). (a) Find the shortest stopping distance on a dry road with \(\mu_s = 0.8\). (b) Find it on a wet road with \(\mu_s = 0.4\). (c) On the dry road the wheels lock and the car skids with \(\mu_k = 0.6\). Find the stopping distance.
(a) \(v = 72 \times 1000 / 3600 = 20\,\text{m/s}\). The deceleration is \(\mu_s g = 8\,\text{m/s}^2\), so \(s = 400 / 16 = 25\,\text{m}\).
(b) The deceleration is \(4\,\text{m/s}^2\), so \(s = 400 / 8 = 50\,\text{m}\). Half the friction doubles the distance.
(c) The deceleration is \(\mu_k g = 6\,\text{m/s}^2\), so \(s = 400 / 12 \approx 33\,\text{m}\), eight metres more than without the skid.
A scooter moves at \(54\,\text{km/h}\) on a dry road with \(\mu_s = 0.75\). The rider brakes as hard as possible without skidding. Find the stopping distance in metres. Take \(g = 10\,\text{m/s}^2\).
Many cars have an anti-lock braking system (ABS), which releases and reapplies a brake many times a second whenever a wheel starts to lock. Why does this help the car stop sooner?
A wheel that keeps rolling touches the road with a patch at rest, so the road can supply static friction up to \(\mu_s N\). A locked wheel slides, and the road then supplies only kinetic friction, \(\mu_k N\), which is smaller. ABS keeps the wheels just short of locking, so the braking friction stays near the larger static limit and the deceleration stays near \(\mu_s g\). It also keeps the front wheels rolling, so the driver can still steer.
In which of these situations does friction on the named body point in its direction of motion?
The everyday numbers
The table gives rough values for common pairs of dry surfaces. The values depend strongly on how clean and how smooth the surfaces are, so they are guides, not constants of nature. Two patterns hold in every row: \(\mu_k\) is smaller than \(\mu_s\), and a film of water or oil lowers both.
A car moves at \(54\,\text{km/h}\) on a wet road, where \(\mu_s\) between the tyres and the road is about \(0.5\). Estimate the shortest stopping distance, in metres, ignoring the driver's reaction time.
On a dry road a smooth, wide racing tyre grips best, because more rubber touches the road. On a wet road, a smooth tyre floats on a film of water, and \(\mu\) falls almost to zero. The grooves of an ordinary tyre squeeze the water out of the contact patch, so the rubber can touch the road and friction returns. A tyre worn smooth loses this, which is why traffic police check the depth of the tread.
- Drawing friction backward on a walking foot or a driving wheel.Ask which way the contact surface would slip without friction. The foot and the driving tyre slip backward, so friction on them points forward.
- Using \(\mu_k\) for a car braking without skidding.A rolling tyre does not slide over the road. The friction is static, with a limit of \(\mu_s N\). Use \(\mu_k\) only for a skid.
- Forgetting to convert km/h before using \(v^2 = 2as\).Divide by \(3.6\): \(72\,\text{km/h} = 20\,\text{m/s}\), \(54\,\text{km/h} = 15\,\text{m/s}\).