C. Forces · Forces

Solving Problems with Friction and Newton's Laws

C2.4 C2.2 C3.1About 75 minutes

Learning goals

I can solve one-dimensional problems involving applied forces, normal force, gravity and friction using Newton's laws.
I can decide whether an object at rest will start to move.
I can find the acceleration and stopping distance of an object sliding to a stop.
I can combine dynamics with kinematics, for example to compare braking on dry, wet and icy roads.

Words to know

deceleration
An everyday word for an acceleration that is opposite to the velocity, so the object slows down. In physics we usually just give a negative acceleration.

stopping distance
The distance an object travels while slowing to a stop. For a vehicle, the total stopping distance also includes the distance travelled during the driver's reaction time.

locked wheels
Wheels that stop turning during braking, so the tires skid and kinetic friction acts.

The first snowfall

Every November, on the morning of the first real snowfall in Ontario, there are more fender-benders than usual. Drivers are still braking as if the roads were dry. In this lesson you will calculate exactly how much longer it takes to stop on a slippery road, and you will practise the full problem-solving method of dynamics: free-body diagram, friction, Newton's second law, then kinematics.

The friction problem-solving routine

Draw the FBD. Choose a + direction.
Vertical: if nothing accelerates vertically, the vertical forces balance. Find FN (usually FN = mg on a level surface, but not if someone pushes down or pulls up).
Friction: sliding? Use Ff,k = μk FN. Not sliding? Static friction is whatever is needed, up to μs FN.
Horizontal: Fnet = ma.
Kinematics (if needed): v₂ = v₁ + aΔt, Δd = v₁Δt + ½aΔt², v₂² = v₁² + 2aΔd.

Worked example

Example 1: Pushing a crate

A worker pushes a 30 kg crate across a warehouse floor with a horizontal force of 150 N. μk = 0.35. Find the crate's acceleration.

FBD: Fg down, FN up, Fa = 150 N right, Ff left. Right is +.
Vertical: FN = mg = 30 × 9.8 = 294 N.
Friction (sliding): Ff = μk FN = 0.35 × 294 = 102.9 N, to the left.
Horizontal: Fnet = 150 − 102.9 = 47.1 N.
a = Fnet ÷ m = 47.1 ÷ 30 ≈ 1.6 m/s² right.

a ≈ 1.6 m/s² in the direction of the push.

Example 1: the vertical forces balance, so FN = mg. Kinetic friction is μk FN = 0.35 × 294 ≈ 103 N. The net force is 150 − 103 ≈ 47 N to the right.

Sliding to a stop

When the only horizontal force on a sliding object is kinetic friction (on a level surface):

Fnet = −μk FN = −μk mg, so a = −μk g

The mass cancels. A heavy truck and a light car with the same tires on the same road slide to a stop with the same deceleration (they need the same distance from the same speed, if they skid). The truck has more friction on it, but also more inertia, and the two effects cancel exactly.

Combine with v₂² = v₁² + 2aΔd and v₂ = 0:

Δd = v₁² ÷ (2 μk g)

So the skidding distance is proportional to the square of the speed: double the speed and the distance is four times as long. And it is inversely proportional to μk: an icy road with one eighth of the friction gives eight times the distance.

Worked example

Example 2: A curling stone

A curler releases a 19 kg stone at 2.5 m/s. The coefficient of kinetic friction between stone and ice is 0.010. How far does it slide, and for how long? (Ignore sweeping and curl.)

a = −μk g = −0.010 × 9.8 = −0.098 m/s².
Distance: 0 = (2.5)² + 2(−0.098)Δd, so Δd = 6.25 ÷ 0.196 ≈ 32 m.
Time: 0 = 2.5 + (−0.098)Δt, so Δt = 2.5 ÷ 0.098 ≈ 26 s.
The mass (19 kg) was not needed: it cancels.

About 32 m, taking about 26 s.

Worked example

Example 3: Braking on dry, wet and icy roads

A car is travelling at 90 km/h (25 m/s) when the driver brakes hard and the wheels lock. Find the skidding distance on (a) dry concrete, μk = 0.8, (b) wet concrete, μk = 0.5, (c) an icy road, μk ≈ 0.1.

Δd = v₁² ÷ (2 μk g), with v₁² = 625 m²/s².
(a) Δd = 625 ÷ (2 × 0.8 × 9.8) = 625 ÷ 15.68 ≈ 40 m.
(b) Δd = 625 ÷ (2 × 0.5 × 9.8) = 625 ÷ 9.8 ≈ 64 m.
(c) Δd = 625 ÷ (2 × 0.1 × 9.8) = 625 ÷ 1.96 ≈ 320 m.
Real stopping distances are longer still: before braking, the car travels about 25 m during each second of the driver's reaction time.

About 40 m, 64 m and 320 m.

The same car at the same speed needs about 8 times as far to skid to a stop on ice as on dry pavement, because the friction (and so the deceleration) is 8 times smaller.

Worked example

Example 4: Finding μk from motion

A puck leaves a stick at 4.0 m/s and slides 16 m across a rink before stopping. Find μk between puck and ice.

v₂² = v₁² + 2aΔd: 0 = 16 + 2a(16), so a = −0.50 m/s².
a = −μk g, so μk = 0.50 ÷ 9.8 ≈ 0.051.

μk ≈ 0.051

Just for fun

Ice is slippery because of a very thin, watery layer on its surface. You might expect scientists to have figured out exactly why that layer forms long ago, but they are still studying and debating it today. Skaters, luckily, don't need to wait for the answer.

Hands-on activity

Measure μk with a sliding book

You need: a hardcover book, a long smooth table or a hard floor, a tape measure, and a phone camera that can record slow motion (optional).

Give the book a push so it slides and stops. Measure the sliding distance Δd from where your hand left it.
To find the starting speed, film the slide with a ruler in view, or use this trick: time the slide with a stopwatch (Δt). For constant deceleration, v₁ = 2Δd ÷ Δt.
Calculate a = −v₁ ÷ Δt, then μk = −a ÷ g.
Repeat 5 times and average. Then put a second book on top and repeat. Does μk change? (It shouldn't, much.)
Try a different surface (a tablecloth). Compare μk.
List your biggest sources of error (timing, measuring where the slide begins).

Your turn

A 20 kg box is pulled with a horizontal 90 N force. μk = 0.30. Find its acceleration.

A 50 kg box sits on a floor with μs = 0.40. Will a horizontal 150 N push move it?

A sled slides on snow with μk = 0.10. Find its acceleration once nothing is pulling it.

How far will the sled above slide if it starts at 5.0 m/s?

Why does mass not affect the stopping distance of an object sliding to a stop?

A car's skidding distance from 50 km/h is 15 m. Estimate it from 100 km/h on the same road.

More practice: warm-up

On a level floor with no other vertical forces, what is FN for a 10 kg box?

What is the kinetic friction on that box if μk = 0.20?

Write the acceleration of an object sliding to a stop on a level surface.

If μk is halved, what happens to the stopping distance (same speed)?

Which step comes first in a dynamics problem?

More practice: on your own

A 15 kg toboggan is pulled horizontally from rest with 40 N across snow (μk = 0.12). Find its speed after 4.0 s.

A hockey puck slides 25 m and stops. μk = 0.050. What was its initial speed?

A 1200 kg car on a wet road (μk = 0.50) skids to a stop in 3.2 s. How fast was it going?

A 40 kg crate is pushed at constant velocity with 120 N. Find μk. If the push increases to 160 N, find the acceleration.

A 2.0 kg book is pressed against a wall by a horizontal push of 40 N. μs between book and wall is 0.60. Will it slide down?

Challenge

A driver at 72 km/h (20 m/s) sees a moose on the road 50 m ahead. Her reaction time is 0.80 s, then she brakes with μk = 0.70. Does she stop in time?

A 5.0 kg box is pulled to the right with 30 N while also being pushed down from above with 20 N. μk = 0.40. Find the acceleration.

A box slides onto a rough patch at 3.0 m/s and stops after 1.5 m. A second, identical box with twice the mass slides onto the same patch at 3.0 m/s. How far does it slide? Explain.

Check your understanding

Quiz

A 10 kg box is pushed with 50 N; μk = 0.30. Its acceleration is about:

An object slides to a stop on a level surface. Its acceleration depends on:

If a car's speed doubles, its skidding distance:

A crate on a floor has a maximum static friction of 200 N. A 120 N push is applied. The friction force is:

A puck slows at 0.49 m/s² on a level rink. μk is:

Pressing down on top of a sliding box will:

For parents and teachers

Big idea: Using friction coefficients with Newton's laws to solve one-dimensional problems (C2.4, C2.2) and to understand how friction changes velocity (C3.1).

How to help: The braking calculations connect directly to safe driving. If your teen is learning to drive, compare the numbers with your own experience in winter: leave more distance, slow down, and use winter tires.

Common difficulties:

Forgetting to check whether an object actually moves before using kinetic friction.
Using the applied force instead of the net force in F = ma.
Mixing km/h and m/s (divide km/h by 3.6).
Losing the negative sign on the acceleration when an object slows down.