Friction problems, air resistance and the fundamental forces
Real problems rarely have a horizontal push on a flat floor. Ropes pull at angles, boxes slide down ramps, and objects fall through air. This lesson puts the second law and friction together, then looks at air resistance and at the four forces behind everything else.
By the end of this lesson you can
- solve friction problems where a force acts at an angle;
- solve problems with objects sliding on ramps;
- explain air resistance and terminal speed;
- name the four fundamental forces.
A method for friction problems
The key fact is that the normal force is not always \(mg\). A rope pulling up at an angle lifts part of the weight; a push angled downward presses the object harder into the floor. Always find \(F_n\) from the forces perpendicular to the surface before you calculate friction.
Step by step
- Draw a free-body diagram. Choose \(x\) along the surface (the direction of motion) and \(y\) perpendicular to it.
- Split every slanted force into \(x\) and \(y\) components.
- Perpendicular to the surface the object does not accelerate, so \(\sum F_y = 0\). Solve this for \(F_n\).
- Calculate friction: \(F_k = \mu_k F_n\) if sliding, or compare with \(F_{s,\text{max}} = \mu_s F_n\) if it might not move.
- Along the surface, use \(\sum F_x = ma_x\).
Worked example 1: a pull at an angle
A student drags a 25.0 kg crate across a level floor with a rope that pulls at 30.0° above the horizontal with a force of 120 N. The coefficient of kinetic friction is 0.40. Find the crate's acceleration.
- Forces. Weight \(mg = (25.0)(9.81) = 245\ \text{N}\) down; normal force up; kinetic friction backward; the rope's pull at 30.0°.
- Components of the pull. \(F_x = 120\cos 30.0^\circ = 103.9\ \text{N}\) and \(F_y = 120\sin 30.0^\circ = 60.0\ \text{N}\).
- Normal force from \(\sum F_y = 0\). \(F_n + 60.0 - 245 = 0\), so \(F_n = 185\ \text{N}\). The rope lifts some of the weight, so the floor pushes less than \(mg\).
- Friction. \(F_k = \mu_k F_n = 0.40 \times 185 = 74.1\ \text{N}\)
- Second law along the floor. \(\sum F_x = 103.9 - 74.1 = 29.8\ \text{N}\), so $$a = \frac{29.8\ \text{N}}{25.0\ \text{kg}} = 1.19\ \text{m/s}^2$$
\(a = 1.2\ \text{m/s}^2\) in the direction of the pull
Worked example 2: sliding down a ramp
An 8.0 kg box slides down a ramp tilted at 35.0°. The coefficient of kinetic friction is 0.30. Find its acceleration.
- Tilt the axes. \(+x\) down the slope, \(+y\) out of the ramp. Gravity is the only slanted force: \(F_{g,x} = mg\sin\theta\) and \(F_{g,y} = -mg\cos\theta\).
- Normal force. \(F_n = mg\cos\theta = (8.0)(9.81)\cos 35.0^\circ = 64.3\ \text{N}\)
- Friction points up the slope: \(F_k = 0.30 \times 64.3 = 19.3\ \text{N}\).
- Second law along the slope. \(mg\sin\theta = (8.0)(9.81)\sin 35.0^\circ = 45.0\ \text{N}\). Net: \(45.0 - 19.3 = 25.7\ \text{N}\), so \(a = 25.7 / 8.0 = 3.2\ \text{m/s}^2\).
- Look at the algebra. Written out, \(a = g(\sin\theta - \mu_k\cos\theta)\). The mass cancels, so a heavier box of the same material slides down with the same acceleration.
\(a = 3.2\ \text{m/s}^2\) down the ramp
Worked example 3: finding \(\mu_k\) from the motion
A sled slides down a snowy hill tilted at 20.0° with an acceleration of 1.50 m/s². What is the coefficient of kinetic friction?
- Start from the result above. \(a = g(\sin\theta - \mu_k\cos\theta)\). The sled's mass is not needed.
- Solve for \(\mu_k\). $$\mu_k = \frac{g\sin\theta - a}{g\cos\theta}$$
- Substitute. \(g\sin 20.0^\circ = 3.36\ \text{m/s}^2\) and \(g\cos 20.0^\circ = 9.22\ \text{m/s}^2\), so $$\mu_k = \frac{3.36 - 1.50}{9.22} = 0.20$$
\(\mu_k = 0.20\)
Air resistance
Think first
Drop a flat sheet of paper and a crumpled ball of the same paper from the same height. Which lands first? They have the same mass, so why the difference?
Compare your answer
The crumpled ball lands first. Both feel the same gravity, but the flat sheet has to push a lot of air out of the way, so the air pushes back on it much harder.
Air resistance, \(\vec F_R\), is a kind of friction between an object and the air it moves through. It always points opposite to the velocity, and it grows as the object goes faster. For a slow object it is roughly proportional to the speed; for fast objects like cars it grows with the square of the speed.
Follow a skydiver after she jumps:
- At first she is slow, air resistance is small, and the net force is almost her full weight: she speeds up quickly.
- As she speeds up, air resistance grows, the net force shrinks, and her acceleration gets smaller.
- Eventually air resistance equals her weight. The net force is zero and she falls at a constant speed, called the terminal speed.
Opening a parachute suddenly makes air resistance much bigger than her weight. The net force points up, she slows down, and she settles at a new, much lower terminal speed.
Falling with air resistance
InteractiveScroll sideways to see the whole demo, or turn your phone sideways.
Try this
- Drop the object and watch the two arrows and the speed graph. When does the graph level off?
- Increase the mass and drop again. Does a heavier object reach a higher or lower terminal speed?
- Open the parachute and compare the terminal speeds.
How friction drives a car
What pushes a car forward? Not the engine directly. The engine turns the wheels, and each tire pushes backward on the road. By the third law, the road pushes forward on the tire. Because the bottom of a rolling tire is momentarily at rest against the road, this forward push is static friction.
That is also why skidding is dangerous. Once tires slide, friction switches from static to kinetic, and kinetic friction is smaller. Anti-lock brakes pump the brakes rapidly to keep the tires rolling, so the car can use the larger static friction to stop.
The four fundamental forces
Friction, tension, the normal force and air resistance look like very different forces. At the level of atoms they are all the same thing: electric and magnetic forces between charged particles. In fact every force in nature comes from just four fundamental forces:
| Force | Acts between | Where you meet it |
|---|---|---|
| Strong force | Particles in the nucleus | Holds atomic nuclei together. The strongest, but only acts over about the width of a nucleus. |
| Electromagnetic force | Charged particles | Holds atoms and molecules together; behind friction, the normal force, tension and magnetism. |
| Weak force | Particles in the nucleus | Causes some kinds of radioactive decay. |
| Gravitational force | Any two masses | Keeps you on the ground and planets in orbit. By far the weakest, but it reaches across the universe. |
Check yourself
Question 1
A box that weighs 300 N slides across a level floor at constant velocity while you push it with 250 N directed 30.0° below the horizontal. Find the coefficient of kinetic friction.
Show solution
The push presses down, so \(F_n = 300 + 250\sin 30.0^\circ = 300 + 125 = 425\ \text{N}\).
Constant velocity means friction equals the horizontal part of the push: \(F_k = 250\cos 30.0^\circ = 217\ \text{N}\).
\(\mu_k = 217 / 425 = 0.51\)
Question 2
A child slides down a playground slide tilted at 40.0°. The coefficient of kinetic friction is 0.25. What is the child's acceleration?
Show solution
\(a = g(\sin\theta - \mu_k\cos\theta) = 9.81(\sin 40.0^\circ - 0.25\cos 40.0^\circ) = 9.81(0.643 - 0.192) = 4.4\ \text{m/s}^2\)
Question 3
A 40.0 kg trunk is pulled across a level floor by a rope at 20.0° above the horizontal with a force of 150 N. The coefficient of kinetic friction is 0.35. Find its acceleration.
Show solution
\(F_x = 150\cos 20.0^\circ = 140.9\ \text{N}\); \(F_y = 150\sin 20.0^\circ = 51.3\ \text{N}\).
\(F_n = (40.0)(9.81) - 51.3 = 392.4 - 51.3 = 341.1\ \text{N}\); \(F_k = 0.35 \times 341.1 = 119.4\ \text{N}\).
\(a = (140.9 - 119.4)/40.0 = 0.54\ \text{m/s}^2\)
Question 4
A skydiver is falling at terminal speed. Which statement is true?
- Air resistance is larger than her weight.
- There is no air resistance at terminal speed.
- The net force points down, so she is still speeding up.
- Air resistance equals her weight, so the net force is zero.
Show solution
Terminal speed is constant speed, so the acceleration and the net force are zero: air resistance balances weight exactly.
Question 5
Which fundamental force is responsible for the friction between a shoe and the floor?
- Gravitational
- Strong
- Electromagnetic
- Weak
Show solution
Friction comes from electric forces between the atoms and molecules of the two surfaces.
Summary
- Find the normal force from \(\sum F_y = 0\) before calculating friction. An upward pull reduces it; a downward push increases it.
- On a ramp, \(F_n = mg\cos\theta\), and a sliding object accelerates at \(a = g(\sin\theta - \mu_k\cos\theta)\).
- Air resistance grows with speed. When it equals the weight, a falling object reaches terminal speed.
- All forces come from four fundamental forces: strong, electromagnetic, weak and gravitational.
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