2.5 - Newton’s Second Law

From Balanced to Unbalanced Forces

Newton’s First Law dealt with situations where forces were balanced (net force zero), giving equilibrium and constant velocity. Newton’s Second Law picks up where forces become unbalanced (net force not zero).

First Law	Second Law
$\sum \vec{F} = 0$	$\sum \vec{F} \neq 0$
$a = 0$	$a \neq 0$
Constant velocity	Changing velocity (acceleration)

Once the net force stops being zero, the object or system starts to accelerate.

The Statement

Newton’s Second Law: the acceleration of a system’s center of mass is directly proportional to the net force acting on it and inversely proportional to its mass. The acceleration points in the same direction as the net force.

\sum \vec{F} = m\vec{a} \qquad \text{or equivalently} \qquad \vec{F}_{\text{net}} = m\vec{a}

where

\sum \vec{F}

is the net force (vector sum of everything acting on it),

m

is the mass, and

\vec{a}

is the resulting acceleration.

Rearranged to solve for acceleration:

\vec{a} = \frac{\sum \vec{F}}{m}

This version makes it clear that acceleration is directly proportional to net force (double the force, double the acceleration) and inversely proportional to mass (double the mass, half the acceleration).

On the AP equation sheet, it’s written for a whole system:

\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}

Here

\vec{a}_{\text{sys}}

is the acceleration of the system’s center of mass and

m_{\text{sys}}

is the total mass of everything in the system. Only external forces go in the sum, since internal forces cancel in pairs (2.3).

Direct and Inverse Proportionality

For a constant mass, acceleration grows as net force grows:

F \propto a

. For example, applying 10 N to a 2 kg object gives 5 m/s² of acceleration, and applying 20 N to the same object gives 10 m/s².

For a constant net force, acceleration shrinks as mass grows:

a \propto \frac{1}{m}

. For example, 20 N applied to a 2 kg object gives 10 m/s², but 20 N applied to a 4 kg object only gives 5 m/s².

Physically, small mass is easy to accelerate (low inertia), and large mass is hard to accelerate (high inertia), tying right back to inertia from the First Law.

Newton’s Original Framing

Newton himself never wrote

F = ma

. In the Principia (1687), he stated the law in terms of “motion,” what we now call momentum: the change in an object’s motion is proportional to the force applied, and it happens in the direction of that force. Today we usually write it with acceleration on the left, to show that force is the cause:

\vec{a} = \frac{\vec{F}_{\text{net}}}{m}

This framing emphasizes that force causes acceleration, not the other way around, that acceleration depends on both the net force and the mass, and that the acceleration always matches the direction of the net force.

Connection to Center of Mass

Newton’s Second Law technically applies to a system’s center of mass.

Recall from earlier:

\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{\sum m_i}

The center of mass velocity only changes when a nonzero external net force acts on the system:

\vec{F}_{\text{net, external}} \neq 0 \implies \vec{v}_{cm} \text{ changes}

Internal forces within a system can’t change the center of mass motion, only external forces can. Think of two ice skaters pushing off each other: the forces between them are internal, equal and opposite (Newton’s Third Law), and they don’t change the center of mass velocity at all. Only something external, like friction with the ice, would affect that.

Common Problem Types

Incline plane problems: objects sliding or being pulled on ramps at various angles, requiring weight to be resolved into components parallel and perpendicular to the surface, with normal force perpendicular to the incline and friction (if present) parallel to it, opposing motion.

Connected object problems: multiple objects connected by ropes or in contact with each other. This includes two blocks pushed together (contact forces between them), Atwood machines (two masses hanging from opposite sides of a pulley), and modified Atwood machines (one mass on a horizontal or inclined surface, one hanging vertically). You can analyze these as one combined system (simpler) or as separate systems (more detailed), and connected parts share the same acceleration if the rope doesn’t stretch.

Multiple force problems: objects with several forces acting at once, like applied force, friction, gravity, and normal force together, solved by adding everything as vectors to find net force, then applying

F_{\text{net}} = ma

Analyzing Incline Planes

Incline problems need careful coordinate choices and force resolution.

The best approach is to rotate your axes so one lines up with the direction of motion. The standard choice is an x-axis parallel to the incline surface (positive pointing down the slope) and a y-axis perpendicular to the surface (positive pointing away from it).

This helps because motion only happens along the x-axis (nothing moves perpendicular to the surface), which simplifies things to

\sum F_y = 0

(no acceleration perpendicular to the surface) and you only need to solve

\sum F_x = ma

for motion up or down the slope.

Force components on an incline: weight always points straight down, never perpendicular to the surface, so it needs resolving:

F_{gx} = mg\sin\theta \quad \text{(parallel to incline, down the slope)}

F_{gy} = mg\cos\theta \quad \text{(perpendicular to incline, into the surface)}

Normal force is always perpendicular to the contact surface:

N = mg\cos\theta \quad \text{(when no other perpendicular forces are present)}

Here’s a key geometric fact: the incline angle

\theta

equals the angle between the weight vector and its perpendicular component, which is why the parallel piece uses

mg\sin\theta

and the perpendicular piece uses

mg\cos\theta

. That’s actually the opposite of typical horizontal and vertical problems, where horizontal is

F\cos\theta

and vertical is

F\sin\theta

. On an incline, sine and cosine essentially swap roles compared to standard problems.

Worked Example: Frictionless Incline

A wooden crate with mass

m

sits on a frictionless incline at angle

\theta

above horizontal. Derive the acceleration down the incline, then describe and justify what happens as

\theta

increases.

Draw: the forces on the crate are weight (

F_g = mg

, straight down) and normal force (

N

, perpendicular to the incline, away from the surface). There’s no friction here since the problem says frictionless.

Sum: choosing the x-axis parallel to the incline (down as positive) and y-axis perpendicular to it.

In the x-direction, the only relevant force component is

F_{gx} = mg\sin\theta

(down the slope):

\sum F_x = ma_x \implies mg\sin\theta = ma

In the y-direction, there’s no acceleration perpendicular to the surface (the crate doesn’t sink into or lift off the incline):

\sum F_y = 0 \implies N - mg\cos\theta = 0 \implies N = mg\cos\theta

We don’t actually need the normal force value here, since there’s no friction to calculate.

Solve: starting from

mg\sin\theta = ma

, divide both sides by

m

g\sin\theta = a \implies \boxed{a = g\sin\theta}

Notice this doesn’t depend on mass at all. A heavy crate and a light crate slide down a frictionless incline at the same rate, as long as they’re at the same angle.

What happens as $\theta$ increases from 0° to 90°?

Since

a = g\sin\theta

, and sine grows from 0 to 1 across that range:

Angle ( $\theta$ )	$\sin\theta$	Acceleration ( $a = g\sin\theta$ )
0°	0	0
30°	0.5	0.5g = 5.0 m/s²
45°	0.707	0.707g ≈ 7.1 m/s²
60°	0.866	0.866g ≈ 8.7 m/s²
90°	1	1g = 10 m/s²

So acceleration steadily increases as the angle increases.

Justification: at

\theta = 0°

(a horizontal surface), the component of gravity parallel to the surface is zero, so there’s no acceleration along it. At

\theta = 90°

(fully vertical), the entire weight acts parallel to the “surface,” which is just free fall with

a = g

. In between, the parallel component

mg\sin\theta

keeps growing continuously, causing greater and greater acceleration.

Mathematically, sine increases monotonically from 0 to 1 as the angle goes from 0° to 90°, and since

a = g\sin\theta

with

g

constant, as

\theta

increases,

\sin\theta

increases, and so

a

increases too. Physically, steeper inclines put more of gravity’s pull along the slope, producing bigger acceleration, and at the extreme of a vertical drop, all of gravity contributes, giving free fall.

Problem-Solving Strategy

Draw: free body diagram with all forces, choosing a coordinate system aligned with motion when possible

Sum: apply $\sum \vec{F} = m\vec{a}$ in each direction. For inclines, this usually means both $\sum F_x = ma$ and $\sum F_y = 0$ , with angled forces resolved into components

Solve: algebra to find the unknowns, substitute numbers, and check units and reasonableness

Same demarcation as before: everything above the sum equations is physics (identifying forces, drawing diagrams, setting up equations with proper components), and everything below is math (algebra, substitution, calculating results).

Special Cases

Mass independence on frictionless inclines:

a = g\sin\theta

doesn’t depend on mass at all, which mirrors Galileo’s famous observation that everything falls at the same rate ignoring air resistance. That’s because

F_g = mg

is proportional to mass, but so is the resistance to accelerating, and the two effects cancel exactly,

a = \frac{F_g}{m} = \frac{mg}{m} = g

Extreme angles: at

\theta = 0°

a = g\sin(0°) = 0

, meaning the object just stays still or moves at constant velocity per the First Law. At

\theta = 90°

a = g\sin(90°) = g

, meaning free fall.

Normal force on an incline:

N = mg\cos\theta

decreases as the angle increases (less surface support needed), hits zero at 90° (no contact during free fall), and equals

mg

at 0° (full weight supported by a horizontal surface). Normal force is always adjusting just enough to prevent motion perpendicular to the surface.

Worked Example: Connected Objects (Modified Atwood Machine)

A 3.00 kg block sits on a frictionless table. A light string tied to it runs over an ideal pulley at the table’s edge to a 2.00 kg block hanging below. The blocks are released from rest. Use

g = 10 \text{ m/s}^2

. Find the acceleration of the blocks and the tension in the string.

Draw: for the table block (

m_1

): weight

m_1 g

down, normal force

N

up, and tension

T

toward the pulley. For the hanging block (

m_2

): weight

m_2 g

down and tension

T

up. Since the string and pulley are ideal, both blocks feel the same tension

T

, and both move with the same size of acceleration

a

Sum (whole system first): treat both blocks and the string as one system, and measure the direction of motion along the string (toward the pulley for

m_1

, then down for

m_2

). Tension is now internal, so it drops out.

N

and

m_1 g

cancel each other. The only unbalanced external force is the hanging block’s weight:

a = \frac{\sum F}{m_{\text{sys}}} = \frac{m_2 g}{m_1 + m_2} = \frac{(2.00)(10)}{3.00 + 2.00} = \frac{20.0}{5.00} = 4.00 \text{ m/s}^2

Sum (one block) to find the tension: tension is internal to the whole system, so to find it, pick a system where it’s external. The table block by itself has just one horizontal force, the tension:

T = m_1 a = (3.00)(4.00) = 12.0 \text{ N}

Check with the hanging block:

m_2 g - T = m_2 a

gives

20.0 - 12.0 = 8.00

N on the left and

(2.00)(4.00) = 8.00

N on the right. ✓

Notice the tension (12.0 N) is less than the hanging block’s weight (20.0 N). It has to be: if the tension matched the weight, the hanging block would have zero net force and couldn’t speed up. The string only pulls hard enough to give the 3.00 kg block its acceleration.

Key Takeaways

Newton’s Second Law: $\sum \vec{F} = m\vec{a}$ , net force causes acceleration

Acceleration form: $\vec{a} = \frac{\sum \vec{F}}{m}$ , emphasizing that force causes acceleration

Acceleration is directly proportional to net force at constant mass

Acceleration is inversely proportional to mass at constant force

Unbalanced forces cause acceleration, balanced forces cause equilibrium

Center of mass velocity only changes with a nonzero net external force

Common problem types: incline planes, connected objects, Atwood machines, modified Atwood machines

Incline strategy: rotate axes parallel and perpendicular to the surface

Component resolution on inclines: $F_{gx} = mg\sin\theta$ (parallel), $F_{gy} = mg\cos\theta$ (perpendicular)

Sine and cosine flip roles on inclines compared to standard horizontal/vertical setups

Frictionless incline acceleration: $a = g\sin\theta$ , independent of mass

As incline angle rises from 0° to 90°, acceleration rises from 0 to g

Problem-solving: Draw (FBD), Sum (Newton’s Second Law), Solve (algebra)

Most dynamics problems from here forward involve Newton’s Second Law

Equation sheet form: $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ , the acceleration of the system’s center of mass, with only external forces in the sum

Connected objects: treat them as one system to find $a$ (the tension is internal and drops out), then use one object to find the tension

Back Next