Force

Forces are why objects change their motion — or stay put. For the MCAT, be fluent in drawing free-body diagrams, resolving vectors into components, and recognizing how forces govern systems from a contracting muscle to a cell settling in a centrifuge.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Newton's First Law and Inertia

The Core Idea

Must know

An object continues in its current state of motion (constant velocity, including rest) unless a net external force acts on it. The property of matter that resists changes in motion is inertia — not a force, but a property of mass. More mass means more inertia, so a larger net force is needed for the same change in motion.

Equilibrium

Must know

When the net force is zero ($\sum \vec{F} = 0$), the object is in mechanical equilibrium:

• Static equilibrium: at rest ($v = 0$, $a = 0$).
• Dynamic equilibrium: constant velocity ($a = 0$, $v \neq 0$).

This is the basis for analyzing muscles, joints, and levers — a flexed elbow holding a weight stationary is a static-equilibrium problem.

Free-body diagrams (FBDs): Every force analysis begins here. Draw the object as a dot, then labeled arrows for every force acting on it. Never include forces the object exerts on something else. The MCAT rewards instinctively reaching for an FBD.

Quick check

Q: A person holds a 5 kg book stationary in midair. What is the net force on the book?

A: Zero — static equilibrium. The upward hand force cancels the downward weight. Any object with zero acceleration has zero net force, stationary or moving at constant velocity.

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Newton's Second Law

The Core Idea

Must know

A net force changes velocity — the object accelerates:

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

Both $\vec{F}$ and $\vec{a}$ are vectors (direction matters), and the "F" is the net force — the vector sum of all forces on the object.

The Common Forces

Must know

• Weight $W = mg$ — gravity's pull, always straight down; $g \approx 9.8\ \text{m/s}^2$ (use $10$ to simplify unless told otherwise).
• Normal force $N$ — the surface pushes back perpendicular to itself ("normal" = perpendicular). $N = mg$ on a flat surface for a stationary object, but changes on inclines or with vertical applied forces.
• Tension $T$ — the pull transmitted through a string/rope, acting along it toward the attachment point. For a massless rope, tension is the same throughout.

The Inclined Plane — A Classic MCAT Setup

Must know

Resolve weight into components relative to the slope (angle $\theta$):

• Parallel (down the ramp): $mg\sin\theta$
• Perpendicular (into the surface): $mg\cos\theta$

The normal force balances the perpendicular component: $N = mg\cos\theta$. The net force along the slope (before friction) is $mg\sin\theta$.

Worked Example: Block on a Frictionless Incline

Must know

Problem: A 10 kg block on a frictionless 30° incline. Acceleration down the slope? ($g = 10\ \text{m/s}^2$)

The only force along the slope is gravity's parallel component:

$F_{\parallel} = mg\sin\theta = (10)(10)\sin 30° = 50\ \text{N}$

Applying $\sum F = ma$ along the slope: $50 = 10a \implies a = 5\ \text{m/s}^2$.

Sanity check: $\theta = 0°$ gives $a = 0$; $\theta = 90°$ gives $a = g$ (free fall). $5\ \text{m/s}^2$ at 30° sits sensibly between.

Quick check

Q: Replace the block with a 20 kg block on the same frictionless 30° incline. New acceleration?

A: Still $5\ \text{m/s}^2$. Both $F_{\parallel}$ and $m$ double, so they cancel in $a = F/m$. On a frictionless incline, acceleration is independent of mass.

Apparent Weight (Accelerating Frames)

Know the logic

A scale reads the normal force it exerts, not true weight $mg$. For a person in an elevator with vertical acceleration $a$ (up positive):

$N - mg = ma \implies N = m(g + a)$

Accelerating up → $N > mg$ (feel heavier); down → $N < mg$ (lighter); free fall ($a = -g$) → $N = 0$ (apparent weightlessness).

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Newton's Third Law

The Core Idea

Must know

For every force A exerts on B, B exerts an equal and opposite force on A — action-reaction pairs. They:

• Are equal in magnitude, opposite in direction.
• Act on different objects (so they never cancel each other).
• Are always the same type of force (gravitational-gravitational, normal-normal, etc.).

Biological Examples

Know the logic

Your foot pushes back on the ground, the ground pushes you forward (walking); a fish pushes water back, water pushes the fish forward.

Pairs vs. Equilibrium

Must know

Don't confuse Third Law pairs with equilibrium. In equilibrium, forces on the same object sum to zero. Third Law pairs act on different objects and can never cancel each other.

Example — book on a table: Earth pulls book down ↔ book pulls Earth up is a Third Law pair. But weight (Earth on book) and normal force (table on book) are not a pair — they are two forces on the same object that cancel in equilibrium.

Quick check

Q: A 70 kg person stands on a scale reading 700 N. What is the Newton's Third Law partner of the normal force the scale exerts on the person?

A: The force the person exerts on the scale — 700 N downward. The partner of "scale pushes person up" is "person pushes scale down." This is NOT the weight (Earth on person), a separate force of a different type.

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Friction: Static and Kinetic

The Core Idea

Must know

Friction opposes relative motion (or attempted motion) between contacting surfaces. Two kinds:

• Static friction ($f_s$): acts when surfaces are not sliding; self-adjusts to match the applied force, up to a maximum.
• Kinetic friction ($f_k$): acts when surfaces are sliding; roughly constant for given surfaces and normal force.

The Friction Equations

Must know

$f_s \leq \mu_s N \qquad f_k = \mu_k N$

$\mu_s$ and $\mu_k$ are dimensionless coefficients depending only on the surface pair. A universal relationship:

$\mu_s > \mu_k$

It is always harder to start sliding than to keep it sliding. At the MCAT level, friction is independent of apparent contact area and (for kinetic) of sliding speed — it depends only on $\mu$ and $N$.

Friction on an Inclined Plane

Must know

Static friction acts up the slope to oppose sliding. The maximum angle before sliding is the angle of repose $\theta_r$, where $\tan\theta_r = \mu_s$. Once sliding, the net force down the slope is:

$F_{\text{net}} = mg\sin\theta - \mu_k mg\cos\theta = mg(\sin\theta - \mu_k\cos\theta)$

Worked Example: Friction on an Incline

Must know

Problem: A 5 kg box slides down a 37° incline, $\mu_k = 0.50$. Acceleration? ($g = 10$, $\sin 37° = 0.60$, $\cos 37° = 0.80$)

$N = mg\cos\theta = (5)(10)(0.80) = 40\ \text{N}$
$f_k = \mu_k N = (0.50)(40) = 20\ \text{N}$
$F_{\text{net}} = mg\sin\theta - f_k = 30 - 20 = 10\ \text{N} \implies a = \tfrac{10}{5} = 2\ \text{m/s}^2$

Sanity check: Frictionless would give $a = g\sin 37° = 6\ \text{m/s}^2$; friction reduces it to $2$ ✓.

Quick check

Q: You push a 25 kg crate horizontally with 80 N but it doesn't move; $\mu_s = 0.40$. Is the static friction 80 N, 100 N, or does the crate slide? ($g = 10$)

A: Max static friction $= \mu_s N = (0.40)(25)(10) = 100\ \text{N}$. Your 80 N is below the max, so the crate stays still and static friction equals exactly 80 N. Static friction is not always at its maximum.

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Center of Mass

The Core Idea

Must know

The center of mass (COM) is the single point whose motion represents a system's translational motion — the mass-weighted "average location of mass," where the system behaves as if all its mass were concentrated.

The Formula

Must know

$x_{\text{COM}} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots}$

In 2D, apply separately for $x$ and $y$. Symmetry shortcut: for any uniform, symmetric object the COM is at the geometric center.

COM vs. center of gravity: treat them as identical for the MCAT; they differ only in non-uniform gravitational fields (out of scope).

Stability and Biological Relevance

Know the logic

An object is stable when its COM lies above its base of support; it tips when the COM moves outside that base. So a wider stance and a lower COM both increase stability — why quadrupeds are more stable than bipeds, and why elderly individuals with high COMs and narrow bases face greater fall risk. The standing human body's COM is roughly in the lower abdomen/pelvis.

Worked Example: Two-Mass System

Must know

Problem: A 60 kg person at $x = 0$ and a 30 kg child at $x = 3$ m on a light seesaw. Where is the COM?

$x_{\text{COM}} = \frac{(60)(0) + (30)(3)}{60 + 30} = \frac{90}{90} = 1\ \text{m}$

The COM is 1 m from the adult — closer to the heavier mass, and where the fulcrum balances.

Sanity check: 1 m lies between 0 and 3 m, closer to the heavier mass ✓.

Quick check

Q: Three equal masses at $x = 0$, $x = 2$ m, $x = 4$ m. Where is the COM?

A: $x_{\text{COM}} = \frac{m(0) + m(2) + m(4)}{3m} = 2\ \text{m}$ — at the middle mass. For equally spaced, equal masses the COM is at the geometric center.

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Common Confusions & Tricks

1. Static friction ≠ always maximum. It equals $\mu_s N$ only at the threshold of sliding; below that it equals the applied force (opposite direction). Ask: "Is the object actually sliding?"

2. Third Law pairs vs. balanced forces. Forces that cancel in equilibrium (e.g., weight and normal force on a book) are NOT a Third Law pair. A pair always acts on two different objects and is the same type of force.

3. Normal force ≠ always mg. On an incline $N = mg\cos\theta$; pushing down gives $N > mg$; a rope pulling up gives $N < mg$. Normal force is whatever prevents surfaces from passing through each other.

4. Weight ≠ mass. Weight $W = mg$ is a force (N); mass is in kg. A 70 kg person weighs ~700 N on Earth, ~112 N on the Moon ($g \approx 1.6$), but still has mass 70 kg.

5. $\mu_s > \mu_k$, always. The force to start moving always exceeds the force to keep moving.

6. The COM doesn't have to be inside the object. A ring's COM is in the hole; a bent arm's may be in the air. It is simply the mass-weighted average position.

7. Constant velocity → equilibrium. Any object moving at constant speed in a straight line has zero net force. Look for changing velocity to identify net force.

8. $\sin$ vs. $\cos$ on inclines. Force parallel to the slope uses $\sin\theta$; perpendicular uses $\cos\theta$. Check at $\theta = 0°$: no pull along the slope ($\sin 0° = 0$), full weight into the surface ($\cos 0° = 1$).

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Key Equations

Equation	Variables & When to Use
$\sum \vec{F} = m\vec{a}$	Net force = mass × acceleration; the master equation of translational dynamics
$W = mg$	Weight (gravitational force); $m$ = mass (kg), $g = 9.8 \approx 10\ \text{m/s}^2$
$F_{\parallel} = mg\sin\theta$	Component of weight along an inclined plane at angle $\theta$
$N = mg\cos\theta$	Normal force on an inclined plane at angle $\theta$ (no additional vertical forces)
$f_s \leq \mu_s N$	Static friction; $\mu_s$ = static coefficient; use when object is not sliding
$f_k = \mu_k N$	Kinetic friction; $\mu_k$ = kinetic coefficient; use when object is sliding
$\tan\theta_r = \mu_s$	Angle of repose; maximum incline angle before object slides
$x_{\text{COM}} = \dfrac{\sum m_i x_i}{\sum m_i}$	$x$-coordinate of center of mass; apply separately for each coordinate axis