B. Dynamics · Dynamics: Forces and Motion in a PlaneInclined Planes: Gravity Components, Friction and the Angle of Slipping
B2.3 B2.5 B3.2 A1.12About 75 minutes
Learning goals- I can resolve gravity into components parallel and perpendicular to a slope: mg sin θ and mg cos θ.
- I can find the acceleration of an object sliding down or up an incline, with or without friction.
- I can find the force needed to push an object up a ramp at a constant velocity.
- I can show that an object just begins to slip when tan θ = μs, and use this to measure μs and μk.
- I can explain how friction helps and hinders on slopes (parked cars, icy hills, ramps).
Words to know
inclined plane
A flat surface tilted at an angle θ to the horizontal: a ramp.
parallel component of gravity
mg sin θ, the part of an object's weight that acts down the slope.
perpendicular component of gravity
mg cos θ, the part of an object's weight that presses into the slope. On a simple incline the normal force balances it.
angle of repose (critical angle)
The steepest angle at which an object can rest on a slope without sliding. tan θ = μs at that angle.
The toboggan hill
The best toboggan hill is steep enough to be fast but not so steep that you cannot climb back up it. Why does a steeper hill give a bigger acceleration? Why does a toboggan on a gentle slope sometimes not move at all until you give it a push? And why do cars parked on a steep, icy street sometimes slide?
All of these come down to one idea: on a slope, gravity splits into a part that pulls the object along the surface and a part that presses it into the surface. Friction depends on the second part and fights the first.
Tilt your axes
On a ramp, the acceleration (if any) is along the slope. So choose the x-axis parallel to the slope and the y-axis perpendicular to it. Then only gravity needs to be resolved:
- Down the slope: mg sin θ
- Into the slope: mg cos θ
The angle between the weight vector and the perpendicular to the slope equals the ramp angle θ (the two angles have perpendicular sides). Check the extremes: at θ = 0 (flat ground), the parallel part is 0 and the perpendicular part is mg. At θ = 90° (a wall), the parallel part is all of mg.
If nothing else pushes into or pulls away from the ramp, ΣFy = 0 gives FN = mg cos θ, which is less than mg. That is why friction is smaller on a steep slope.
Which way does friction point?
Friction always opposes sliding (or the tendency to slide). Sliding down: friction points up the slope. Sliding up (after a push): friction points down the slope, adding to mg sin θ. At rest: static friction points whichever way prevents slipping, usually up the slope.
Worked exampleExample 1: Down the toboggan hill
A 30 kg toboggan (with rider) slides down a 20° snowy hill with μk = 0.10. (a) Find its acceleration. (b) Starting from rest, how fast is it moving after 40 m?
- Axes: +x down the slope. ΣFy = 0: FN = mg cos 20° = 30 × 9.8 × 0.940 = 276 N.
- Ff = μk FN = 0.10 × 276 = 27.6 N, up the slope.
- ΣFx = ma: mg sin 20° − Ff = ma → 30 × 9.8 × 0.342 − 27.6 = 30a → 100.6 − 27.6 = 30a → a = 2.43 m/s².
- In general, a = g(sin θ − μk cos θ); the mass cancels.
- (b) v₂² = v₁² + 2aΔd = 0 + 2(2.43)(40) = 194.5, so v₂ = 13.9 m/s.
(a) 2.4 m/s² down the slope (b) about 14 m/s (50 km/h). That is why toboggan hills need a long, flat run-out at the bottom.
Worked exampleExample 2: Measuring μs with a ramp
A textbook rests on a board. One end of the board is slowly raised, and the book begins to slide when the board makes 28° with the floor. Find μs between the book and the board.
- Just before it slides, static friction is at its maximum and the book is still in equilibrium.
- Along the slope: μs FN = mg sin θ. Perpendicular: FN = mg cos θ.
- Divide: μs mg cos θ = mg sin θ, so μs = tan θ. The mass cancels.
- μs = tan 28° = 0.53.
μs ≈ 0.53. In the same way, if you lower the board until the book slides at a constant speed after a nudge, μk = tan θ at that smaller angle.
Worked exampleExample 3: Pushing a box up a ramp
A 50 kg box is pushed up a 15° ramp at a constant velocity by a force parallel to the ramp. μk = 0.25. Find the pushing force.
- Constant velocity, so ΣF = 0 along the slope. The box moves up, so friction points down the slope.
- Down-slope forces: mg sin 15° = 490 × 0.259 = 126.8 N, and Ff = μk mg cos 15° = 0.25 × 490 × 0.966 = 118.3 N.
- F = 126.8 + 118.3 = 245 N.
About 245 N: half the box's weight, even though the ramp is fairly gentle. Without friction, 127 N would do.
Friction on slopes: help and harm
Static friction is what keeps a parked car, a hiker's boots and a pile of gravel in place on a slope. When winter ice lowers μs below tan θ, objects that were safely at rest start to slide, which is why municipalities spread sand and salt on steep streets first. On the other hand, friction makes it harder to push loads up ramps and wastes energy as heat; rollers, wheels and conveyor belts are used so that only rolling resistance has to be overcome.
Just for funGalileo did not have a stopwatch, so he used inclined planes to "dilute" gravity: a ball rolling down a gentle ramp speeds up slowly enough to time. He described measuring the time by weighing the water that flowed out of a container while the ball rolled. One of history's most important experiments was timed with a balance, not a clock.
Hands-on activityMeasure the angle of slipping
You need: a hardcover book or a cutting board as a ramp, a ruler, a coin, an eraser, a small plastic container, and a pencil and paper.
- Put the coin near the top of the closed book lying flat on the table.
- Slowly lift one end of the book. Stop the moment the coin starts to slide. Hold the book still and measure the height h of the raised end and the length L of the book along the table (the horizontal distance from the bottom end to directly under the raised end).
- tan θ = h ÷ L, so μs = h ÷ L. Repeat 3 times and average.
- Repeat for the eraser and the container. Which has the largest μs? Does putting a coin inside the container (more mass) change the angle? Explain using μs = tan θ.
- For μk: lower the book a little below the slipping angle, give the object a tiny tap, and find the angle at which it keeps sliding at a steady speed. Is μk smaller than μs?
Your turn
1A 5.0 kg block is on a frictionless 30° incline. Find the components of its weight parallel and perpendicular to the slope, and its acceleration.
Parallel: 5.0 × 9.8 × sin 30° = 24.5 N. Perpendicular: 5.0 × 9.8 × cos 30° = 42.4 N. a = g sin 30° = 4.9 m/s² down the slope.
2Find the acceleration of a block sliding down a 30° incline with μk = 0.20.
a = 9.8(sin 30° − 0.20 cos 30°) = 9.8(0.500 − 0.173) = 3.2 m/s².
3A block just starts to slide when a ramp reaches 35°. Find μs.
4At what angle will a box with μs = 0.40 begin to slide?
5A cart rolls (with negligible friction) from rest down a 10° ramp that is 50 m long. Find its speed at the bottom.
a = 9.8 sin 10° = 1.70 m/s². v = √(2 × 1.70 × 50) = 13 m/s.
6Why is the normal force on a ramp less than the object's weight?
The normal force only has to balance the perpendicular component of gravity, mg cos θ, which is less than mg for any θ > 0.
More practice: warm-up
1On a ramp, which component of gravity pulls an object down the slope?
2What is the normal force on a block on a simple incline (no other perpendicular forces)?
3What is the acceleration of an object on a frictionless 90° "ramp"?
9.8 m/s²: it is in free fall.
4A box slides up a ramp after a push. Which way does friction point?
5Does the acceleration down a frictionless ramp depend on the object's mass?
More practice: on your own
1A skier slides from rest down a 25° slope with μk = 0.15. Find her acceleration and her speed after 20 m.
a = 9.8(sin 25° − 0.15 cos 25°) = 9.8(0.423 − 0.136) = 2.8 m/s². v = √(2 × 2.81 × 20) = 11 m/s.
2A book slides at constant speed down a board tilted at 12° after a small nudge. Find μk.
At constant velocity, μk = tan 12° = 0.21.
3Find the force parallel to a 10° ramp needed to push an 80 kg cart up it at constant velocity if the effective μk is 0.10.
F = mg(sin 10° + 0.10 cos 10°) = 784(0.174 + 0.098) = 213 N.
4A puck is given a speed of 5.0 m/s up a 20° ramp with μk = 0.20. How far up the ramp does it go, and how long does that take?
Going up, friction and gravity both point down: a = −9.8(sin 20° + 0.20 cos 20°) = −9.8(0.342 + 0.188) = −5.19 m/s². d = 5.0² ÷ (2 × 5.19) = 2.4 m. t = 5.0 ÷ 5.19 = 0.96 s.
5A road has a slope of 5% (it rises 5 m for every 100 m horizontally). Find its angle. If a parked car's tires have μs = 0.10 on glare ice, will it stay put?
θ = tan⁻¹(0.05) = 2.9°. It stays put if μs ≥ tan θ = 0.05. Since 0.10 > 0.05, yes, it stays (though with little margin).
Challenge
1A 20 kg block is held at rest on a frictionless 30° incline by a horizontal push. Find the push and the normal force.
Use horizontal and vertical axes. Vertical: FN cos 30° = mg → FN = 196 ÷ 0.866 = 226 N. Horizontal: F = FN sin 30° = 226 × 0.500 = 113 N (equivalently F = mg tan 30°). Here FN is greater than mg, because the push presses the block into the ramp.
2A block on an adjustable ramp starts to slide when the angle reaches 25°. When the ramp is lowered to 18°, the block slides at a constant speed after a nudge. Find μs and μk, and explain why the angles differ.
μs = tan 25° = 0.47; μk = tan 18° = 0.32. Static friction can hold until a larger angle because μs > μk: it takes more force to break the surfaces apart than to keep them sliding.
3A sled is given a speed of 6.0 m/s up a 15° snowy hill (μk = 0.10, μs = 0.15). How far up does it go? Does it slide back down, and if so, how fast is it going when it returns to its starting point?
Up: a = −9.8(sin 15° + 0.10 cos 15°) = −3.48 m/s²; d = 36 ÷ (2 × 3.48) = 5.2 m. At the top, tan 15° = 0.27 > μs = 0.15, so static friction cannot hold it: it slides back. Down: a = 9.8(sin 15° − 0.10 cos 15°) = 1.59 m/s²; v = √(2 × 1.59 × 5.17) = 4.1 m/s, slower than its launch speed because friction acted both ways.
Quiz
1. On a frictionless ramp at angle θ, a block's acceleration is:
2. A block rests on a 30° ramp. The normal force on it is:
3. A coin just starts to slide down a tilted book at 20°. Its coefficient of static friction is about:
4. A box is sliding up a ramp after a push. The friction on it points:
5. Two blocks of different masses slide down the same ramp with the same μk. Their accelerations are:
6. A block is pushed up a ramp at constant velocity. The push must balance:
Big idea: On a slope, tilt the axes. Gravity splits into mg sin θ (along the slope) and mg cos θ (into the slope), and friction depends on the normal force mg cos θ (B2.3, B2.5). The angle at which something starts to slide gives μs = tan θ, which links to the advantages and disadvantages of static friction (B3.2).
How to help: Check the extremes with your teen: "What happens at 0°? At 90°?" If their formula gives a nonsense answer at an extreme, sine and cosine are swapped.
Common difficulties:
- Swapping sin and cos for the components.
- Writing FN = mg on a slope.
- Pointing friction down the slope when an object slides down.
- Forgetting that friction changes direction when an object slides up and then back down.