P2 Forces and Newton's laws

In this chapter

• State Newton's three laws, and explain inertia, net force, and why action and reaction never cancel.
• Use \(F_\text{net} = ma\) in both directions: force from acceleration, and acceleration (or mass) from force.
• Tell mass from weight (\(W = mg\)), and explain why everything falls at the same rate without air.
• Draw a free-body diagram for any object, and use it with the normal force, tension, springs (Hooke's law) and friction.
• Solve slopes, pulleys and projectiles by splitting forces and motion into components.
• Explain circular motion with an inward (centripetal) force, \(mv^2/r\), and recognise fictitious forces in accelerating frames.

What keeps things moving?

Checkpoint: What keeps things moving? (1 of 2)

Checkpoint: What keeps things moving? (2 of 2)

Forces and the second law

Forces and the second law

The same net force gives a small mass a big acceleration and a big mass a small one: that is what mass (inertia, in kg) means.

Forces and the second law

Checkpoint: Forces and the second law (1 of 3)

Checkpoint: Forces and the second law (2 of 3)

Checkpoint: Forces and the second law (3 of 3)

Mass and weight

Mass and weight

Heavy things don't fall faster: the bigger weight is exactly matched by the bigger inertia, so without air everything falls with the same acceleration \(g\).

Mass and weight

Mass and weight

Checkpoint: Mass and weight (1 of 2)

Checkpoint: Mass and weight (2 of 2)

Free-body diagrams

Free-body diagrams

Checkpoint: Free-body diagrams (1 of 2)

Checkpoint: Free-body diagrams (2 of 2)

Action and reaction

Action and reaction

Checkpoint: Action and reaction (1 of 2)

Checkpoint: Action and reaction (2 of 2)

Normal force and tension

Normal force and tension

Checkpoint: Normal force and tension (1 of 3)

Checkpoint: Normal force and tension (2 of 3)

Checkpoint: Normal force and tension (3 of 3)

Springs and friction

Springs and friction

• Springs (Hooke's law): \(k\) is the spring constant in N/m: a stiffer spring has a bigger \(k\).
• Friction: roughly independent of how big the contact area is. Static friction adjusts itself to whatever is needed, up to its limit.

Springs and friction

Checkpoint: Springs and friction (1 of 3)

Checkpoint: Springs and friction (2 of 3)

Checkpoint: Springs and friction (3 of 3)

Slopes and pulleys

Slopes and pulleys

Checkpoint: Slopes and pulleys (1 of 3)

Checkpoint: Slopes and pulleys (2 of 3)

Checkpoint: Slopes and pulleys (3 of 3)

Projectiles

Projectiles

Projectiles

Checkpoint: Projectiles (1 of 2)

Checkpoint: Projectiles (2 of 2)

Circular motion

Circular motion

Circular motion

Circular motion

Let go, and the object goes straight on along the tangent. Throw a projectile fast enough and it falls all the way round the Earth: an orbit (P4).

Circular motion

Checkpoint: Circular motion (1 of 3)

Checkpoint: Circular motion (2 of 3)

Checkpoint: Circular motion (3 of 3)

Accelerating frames

Accelerating frames

Accelerating frames

Newton's laws hold in inertial frames; in an accelerating frame you see fictitious forces with no source and no partner (\(-m\mathbf{a}\)).

• A bus: a strap in a bus accelerating at 2 m/s² hangs at \(\tan^{-1}(2/9.81) \approx 11.5^\circ\).
• Centrifugal force is the fictitious force of a rotating frame; from outside, the only real force is inwards.
• Coriolis: in a rotating frame, moving things seem pushed sideways, which makes storms spin (but is far too weak to decide how a sink drains).
• Einstein's "happiest thought": a person falling freely doesn't feel their weight.

Checkpoint: Accelerating frames (1 of 3)

Checkpoint: Accelerating frames (2 of 3)

Checkpoint: Accelerating frames (3 of 3)

Key terms

Inertia

The tendency of an object to keep its velocity.

Net (resultant) force

The vector sum of all the forces on an object.

Newton's first, second and third laws

No net force, no change of velocity / \(F_\text{net} = ma\) / forces come in equal and opposite pairs on different objects.

Newton (N)

The force that gives 1 kg an acceleration of 1 m/s².

Key terms

Mass / weight

Inertia, in kg / the pull of gravity, \(mg\), in N.

Free-body diagram

One object, and every force on it.

Normal force / tension

A surface's push at right angles / a rope's pull along it.

Hooke's law / spring constant

\(F = kx\) / \(k\), in N/m.

Key terms

Static / kinetic friction; coefficient of friction

Friction before / during sliding; \(\mu\), friction ÷ normal force.

Projectile

An object moving under gravity alone after launch.

Centripetal force

The inward net force needed for circular motion, \(mv^2/r\).

Inertial frame / fictitious force

A frame that isn't accelerating / an apparent force in an accelerating frame.

Aristotle (4th century BC)

Galileo's ramps (1632, 1638)

Descartes (1644)

Newton's Principia (London, 1687)

The newton (1948)

SLAMMD (ISS)

Saturn V (Apollo 11, 16 July 1969)

Hooke's anagram (1676, 1678)

Leonardo's friction notes (probably 1493)

Galileo's parabola (1638)

Huygens and centrifugal force (1659, 1673)

Newton's cannon (published 1728)

Coriolis (1835)

Einstein's "happiest thought" (1907)

Elisha Otis (New York, 1854)

Numbers used:

The problem

Problem. A 60 kg person stands on a bathroom scale in a lift. The lift speeds up from rest to 2 m/s in 2 s, cruises for 10 s, then slows to a stop in 2 s. What does the scale read in each stage, going up and going down? What would it read if the cable snapped and there were no safety brake?

Solution

1. The ride: accelerations +1.0, 0 and −1.0 m/s²; distance = area under the velocity–time graph = 2 + 20 + 2 = 24 m (about eight floors).
2. Free-body diagram: weight \(mg = 60 \times 9.81 = 588.6\) N down; normal force \(N\) from the scale up. Up positive: \(N - mg = ma\), so \(N = m(g + a)\). The scale shows \(N/9.81\).

Solution

3. Speeding up (a = +1.0): \(N = 60 \times 10.81 = 648.6\) N, reading ≈ 66.1 kg.
4. Cruising (a = 0): \(N = 588.6\) N, reading 60.0 kg.
5. Slowing at the top (a = −1.0): \(N = 60 \times 8.81 = 528.6\) N, reading ≈ 53.9 kg.
6. Going down: starting down is \(a = -1.0\) (53.9 kg), stopping at the bottom is \(a = +1.0\) (66.1 kg). The reading depends on the acceleration, not the direction of motion.

Solution

7. Free fall (a = −g): \(N = m(g - g) = 0\): the scale reads zero, and you feel weightless.
8. Checks: units kg × m/s² = N; \(a = 0\) gives the normal reading; \(1.0/9.81 \approx 10\%\), and 66.1 kg is about 10% more than 60 kg.

★ Warm-up

1. A 2.0 kg book rests on a table, and you press down on it with 5.0 N. Draw its free-body diagram, then find the normal force on the book, in newtons.

   Your answer NCheck

   Solution

   Weight \(2.0 \times 9.81 = 19.6\) N down, your push 5.0 N down, normal force \(N\) up. At rest, so \(N = 19.6 + 5.0 = 24.6\) N.

Problem 2 (★ Warm-up)

2. A 1500 kg car accelerates at 2.0 m/s² while air resistance and friction total 600 N. What forward force does the road exert on the car, in newtons?

   Your answer NCheck

   Solution

   Net force \(= ma = 1500 \times 2.0 = 3000\) N. Forward force \(- 600 = 3000\), so the forward force is 3600 N.

Problem 3 (★ Warm-up)

3. A 70 kg person stands on a scale in a lift moving up at a steady speed. What is the normal force on the person, in newtons? Then say what the scale shows.

   Your answer NCheck

   Solution

   Steady speed: \(a = 0\), so \(N = mg = 70 \times 9.81 = 686.7\) N, and the scale shows 70.0 kg.

Problem 4 (★ Warm-up)

4. A book rests on a table. Name the third-law partner of (a) the book's weight and (b) the normal force on the book.

   Your prediction or attempt, before you look

   Show solution

   Solution

   (a) The book's gravitational pull on the Earth, upwards. (b) The book's push down on the table. The weight and the normal force both act on the book, so they are not a pair.

★★ Standard

5. A 0.50 kg mass stretches a spring by 4.0 cm. Find the spring constant, in newtons per metre.

   Your answer N/mCheck

   Solution

   \(F = 0.50 \times 9.81 = 4.905\) N, \(x = 0.040\) m, so \(k = F/x = 4.905/0.040 \approx 123\) N/m.

Problem 6 (★★ Standard)

6. A 5.0 kg box slides down a 25° slope with \(\mu_k = 0.20\). Find the acceleration, in m/s². Then find the normal force and the friction.

   Your answer m/s^2Check

   Solution

   Along the slope, the weight's component is \(mg\sin25^\circ = 5.0 \times 9.81 \times 0.423 \approx 20.73\) N, and friction opposes it. The normal force is \(N = mg\cos25^\circ = 5.0 \times 9.81 \times 0.906 \approx 44.5\) N, so friction \(= 0.20 \times 44.5 \approx 8.89\) N. Then \(a = (20.73 - 8.89)/5.0 \approx 2.37\ \text{m/s}^2\), the same as \(g(\sin25^\circ - 0.20\cos25^\circ) = 9.81 \times (0.423 - 0.181) \approx 2.37\ \text{m/s}^2\) ✓.

Problem 7 (★★ Standard)

7. Masses of 5.0 kg and 3.0 kg hang either side of a light pulley on one string. Find the acceleration, in m/s². Then find the tension.

   Your answer m/s^2Check

   Solution

   Heavy side (down positive): \(5g - T = 5a\); light side (up positive): \(T - 3g = 3a\). Add: \(2g = 8a\), so \(a = g/4 \approx 2.45\ \text{m/s}^2\). \(T = 3(g + a) \approx 36.8\) N; check \(5(g - a) \approx 36.8\) N ✓.

Problem 8 (★★ Standard)

8. A 4.0 kg block on a horizontal table is joined by a string over a pulley at the edge to a 1.0 kg mass hanging below. Find the acceleration, in m/s², with no friction. Then find the tension, and repeat both with \(\mu_k = 0.15\) between block and table.

   Your answer m/s^2Check

   Solution

   With no friction, the hanging weight \(1.0g\) accelerates both: \(a = 9.81/5.0 \approx 1.96\ \text{m/s}^2\); \(T = 4.0 \times 1.96 \approx 7.85\) N. With \(\mu_k = 0.15\), friction \(= 0.15 \times 4.0 \times 9.81 \approx 5.89\) N, so \(a = (9.81 - 5.89)/5.0 \approx 0.785\ \text{m/s}^2\); \(T = 4.0a + 5.89 \approx 9.03\) N. Check on the hanging mass: \(1.0 \times 9.81 - 9.03 \approx 0.785\) ✓.

Problem 9 (★★ Standard)

9. A stone is thrown horizontally at 12 m/s from a 20 m cliff. How far from the foot of the cliff does it land, in metres? Then find how long it is in the air and how fast it is going when it lands. (Ignore air.)

   Your answer mCheck

   Solution

   Vertical, from rest: \(20 = \tfrac12 \times 9.81\,t^2\), \(t = \sqrt{40/9.81} \approx 2.02\) s. Horizontal: \(x = 12 \times 2.02 \approx 24.2\) m. Landing: \(v_y = 9.81 \times 2.02 \approx 19.8\) m/s, speed \(\sqrt{12^2 + 19.8^2} \approx 23.2\) m/s.

★★★ Stretch

10. A car goes round a flat bend of radius 40 m. The tyres' static coefficient on the road is 0.60. What is the fastest it can go without sliding, in metres per second?

    Hint

    Friction is the only inward force, and it can be at most \(\mu_s mg\).

    Your answer m/sCheck

    Solution

    \(\mu_s mg = mv^2/r\), so \(v = \sqrt{\mu_s g r} = \sqrt{0.60 \times 9.81 \times 40} \approx 15.3\) m/s (about 55 km/h). The mass cancels.

Problem 11 (★★★ Stretch)

11. A banked curve with friction. A bend of radius 50 m is banked at 15°, and \(\mu_s = 0.30\). (a) At what speed would a car need no friction at all, in m/s? (b) What is the fastest it can take the bend? (c) Can a parked car slide down the bank?

    Hint

    Draw the normal force at right angles to the road (tilted 15° from vertical) and friction along the road. At top speed, friction points down the slope and equals \(\mu_s N\). Resolve vertically (net zero) and horizontally (net \(mv^2/r\) inwards).

    Your answer m/sCheck

    Solution

    (a) No friction: \(N\cos\theta = mg\) and \(N\sin\theta = mv^2/r\), so \(v = \sqrt{rg\tan\theta} = \sqrt{50 \times 9.81 \times 0.268} \approx 11.5\) m/s. (b) Vertical: \(N\cos\theta - \mu N\sin\theta = mg\). Horizontal: \(N\sin\theta + \mu N\cos\theta = mv^2/r\). Divide: \(v^2 = rg\,\dfrac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = 50 \times 9.81 \times \dfrac{0.259 + 0.290}{0.966 - 0.078} \approx 303\), so \(v \approx 17.4\) m/s (about 63 km/h). (c) No: it stays put as long as \(\tan15^\circ \le \mu_s\), and \(\tan15^\circ \approx 0.268 < 0.30\).

Problem 12 (★★★ Stretch)

12. A 60 kg person sees a lift scale read 50 kg. What is the lift's acceleration, in m/s² with up positive? Then say what the lift might be doing.

    Your answer m/s^2Check

    Solution

    \(N = 50 \times 9.81\) N, so \(a = N/m - g = 9.81\,(50/60 - 1) \approx -1.64\ \text{m/s}^2\): accelerating downwards, either starting to go down or slowing down while going up.

Problem 13 (★★★ Stretch)

13. A ring-shaped space station turns once a minute. How big must its radius be for the rim to feel like 1 g, in metres? Then find how fast the rim moves.

    Your answer mCheck

    Solution

    Rim speed \(v = 2\pi r/T\) and \(v^2/r = g\) give \(r = gT^2/4\pi^2 = 9.81 \times 60^2/(4\pi^2) \approx 895\) m. Rim speed \(2\pi \times 895/60 \approx 93.7\) m/s.

Answer these out loud, from memory. Then try again in a few days.

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Summary