Newton's laws of motion

Until this week we only described motion (kinematics). Newton’s laws give the cause of motion: force. On the exam this topic appears in two ways. Either a concept question (“what is net force, which objects do action-reaction forces act on”) or a calculation question where you draw a free-body diagram and write $\sum \vec F = m\vec a$. Both earn many points with few formulas.

Golden rules:

1. In every question, first the free-body diagram. Do not write equations without a diagram.
2. Forces are vectors. Resolve them into components and write $\sum F = ma$ separately on each axis.
3. Constant velocity = zero net force. Moving does not need a force; changing the velocity does.
4. The action-reaction pair is on different objects. They never appear together in the same diagram.
5. Mass (kg) and weight (N) are different things. $W = mg$.

Force and net force

Force is an influence that can change the state of motion of an object. Its unit is the newton: $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$. If more than one force acts on an object, what matters is the net force, that is, the vector sum:

$\vec F_{net} = \sum \vec F = \vec F_1 + \vec F_2 + \dots$

Forces along the same line are added with signs. Perpendicular or angled forces are added with components; the magnitudes are not added directly. The sum of perpendicular forces of 6 N and 8 N is not 14 N but $\sqrt{36+64} = 10\ \text{N}$.

The first law and the inertial frame

The first law (law of inertia): If the net force is zero, an object at rest stays at rest, and a moving object keeps going in a straight line at constant velocity. The tendency of an object to resist a change in velocity is called inertia. Mass is the measure of inertia.

Frames in which the first law holds are called inertial frames. A train moving at constant velocity is inertial. A braking bus is not inertial: a passenger inside seems to be “thrown” forward even though no force pushes. We will study such frames with fictitious forces in week 9.

Equilibrium: $\sum \vec F = 0$, that is, $\vec a = 0$. An object at rest and an object moving at constant velocity are both in equilibrium.

PITFALL · the “if there is motion there is a force” misconception. For a disc sliding on ice at a constant 4 m/s, the net force is zero. Eliminate the option that says “there must be a force in the direction of motion”. Force determines the acceleration, not the velocity.

The second law

$\sum \vec F = m\vec a \qquad\Longleftrightarrow\qquad \sum F_x = m a_x,\quad \sum F_y = m a_y$

The acceleration is in the same direction as the net force and is inversely proportional to the mass. The same force gives a small mass a large acceleration.

Example 1. A 5.0 kg box on a frictionless floor is acted on by 20 N toward the east and 8.0 N toward the west.

East +:  ΣF = 20 - 8.0 = 12 N
a = ΣF / m = 12 / 5.0 = 2.4 m/s², toward the east

Weight and mass

Weight is the gravitational force the Earth exerts on the object: $W = mg$, directed downward. We take $g = 9.80\ \text{m/s}^2$. The weight of a 70 kg person is $70 \cdot 9.80 = 686\ \text{N}$. On the Moon ($g = 1.62\ \text{m/s}^2$) the mass is still 70 kg, and the weight becomes 113 N. Mass is a property of the object, weight depends on location.

The third law

If object A exerts a force $\vec F_{AB}$ on B, then B exerts a force of the same magnitude in the opposite direction on A:

$\vec F_{AB} = -\vec F_{BA}$

This pair is called an action-reaction pair. It has three properties: equal magnitude, opposite direction, different objects. When a small car collides with a big truck, the forces the two exert on each other are equal. The car is damaged more because its mass is small, that is, its acceleration is large.

PITFALL · the normal force is not the reaction to the weight. For a book on a table, $N$ and $mg$ are equal and opposite, but both act on the book, so they are not an action-reaction pair. The reaction to the weight is the book pulling the Earth upward. The reaction to the normal force is the book pushing the table downward.

Commonly used forces

The free-body diagram method

The same five steps in every Newton question:

1. Choose the object you are examining and draw it alone as a point.
2. Add all the forces acting on it as arrows. Do not add the forces the object exerts on other objects.
3. Choose axes. If possible, let one axis be along the acceleration (along the incline on an incline).
4. Write $\sum F = ma$ for each axis. On an axis with no acceleration, $\sum F = 0$.
5. If there is more than one object, draw a separate diagram for each. The same rope means the same $T$ and the same $|a|$.

Example 2: Elevator. A 70 kg person stands on a scale in an elevator that is speeding up at 1.2 m/s² while going up. What does the scale read?

Forces: N upward, mg downward. The acceleration is upward. Up +.
N - mg = m a
N = m(g + a) = 70 (9.80 + 1.2) = 770 N

The scale reads the normal force. At constant velocity it would be $mg = 686\ \text{N}$, and if the acceleration were 1.2 m/s² downward, it would read $m(g-a) = 602\ \text{N}$. Find the direction of the acceleration not from the velocity but from the change of the velocity: “going down and slowing” means the acceleration is upward.

Example 3: Blocks pushed side by side. Two blocks of 2.0 kg and 6.0 kg are on a frictionless floor, and a 16 N force pushes the 2.0 kg block. What is the contact force between the blocks?

System (internal forces cancel): a = 16 / (2.0 + 6.0) = 2.0 m/s²
The 6.0 kg block alone: the only horizontal force is P
P = 6.0 · 2.0 = 12 N
Check, the 2.0 kg block: 16 - P = 2.0 · 2.0  →  P = 12 N

Example 4: A hanging lamp. A 120 N lamp hangs from two symmetric ropes making 35° with the horizontal. What is the tension?

Horizontal: T cos35° - T cos35° = 0 (symmetry, automatic)
Vertical: 2 T sin35° - 120 = 0
T = 120 / (2 · 0.574) = 105 N

As the angle gets smaller (as the ropes approach horizontal), $\sin\theta$ gets smaller and the tension gets larger. That is why it is impossible to stretch a clothesline perfectly straight.

Example 5: Two forces at an angle. Two forces act on a 2.0 kg object on a frictionless horizontal plane: $F_1 = 10\ \text{N}$ in the $+x$ direction ($0^\circ$) and $F_2 = 8.0\ \text{N}$ at $120^\circ$ from the $+x$ axis. What are the magnitude and direction of the acceleration?

x: F1x + F2x = 10 + 8.0 cos120° = 10 - 4.0 = 6.0 N
y: F1y + F2y = 0 + 8.0 sin120° = 6.93 N
|F_net| = √(6.0² + 6.93²) = 9.17 N
a = 9.17 / 2.0 = 4.58 m/s²
direction: tan⁻¹(6.93 / 6.0) = 49.1°  (from +x, quadrant I because both are positive)

If you had added the magnitudes directly, you would get $18/2.0 = 9.0\ \text{m/s}^2$; this is correct only if the two forces are in the same direction. In angled-force questions you use exactly the component method from the vectors week. Note that $\cos 120^\circ$ is negative: the $x$ component of the second force works against the first.

Checklist while solving

When you finish a Newton question on the exam, ask these three questions before marking the answer:

• Is the unit right? Force N, acceleration m/s², mass kg. If you see “kg” where “N” should come out, you forgot to multiply by $g$.
• Is the magnitude sensible? In an elevator accelerating upward the scale should read more than $mg$, in one accelerating downward less. The force on the rear block should be smaller than the applied force.
• Does the limiting case hold? Put $a = 0$: it should return to the equilibrium formula. Send a mass to zero: the system should behave as you expect.

These checks eliminate most of the wrong options without any calculation. With +5 / −1 scoring, being able to eliminate even one option makes guessing profitable.

Frictionless incline

On an incline of angle $\theta$, choose the axes parallel ($x$) and perpendicular ($y$) to the incline. The components of the weight:

$mg\sin\theta \ (\text{down along the incline}),\qquad mg\cos\theta\ (\text{perpendicular to the surface})$

There is no acceleration on the perpendicular axis: $N = mg\cos\theta$. Along the incline: $a = g\sin\theta$. If the object is held by a rope parallel to the incline, $T = mg\sin\theta$.

TIP · limit check. Put $\theta = 0$: horizontal ground, you should get $a = 0$ and $N = mg$. Put $\theta = 90^\circ$: free fall, $a = g$. If your formula does not pass these two checks, you have mixed up sin and cos.

Common mistakes