Forces & Newton's Laws

Forces & Newton's Laws

Part 1 of 4 — Newton's Laws & Free-Body Diagrams

Kinematics described motion; dynamics explains it. Every dynamics question on the MCAT reduces to one ritual: draw the free-body diagram, sum the forces, apply $F_{net} = ma$.

The Three Laws

1. First law (inertia): with zero net force, velocity is constant — at rest stays at rest, moving keeps moving in a straight line. Motion needs no force to continue; it needs force to change.
2. Second law: $F_{net} = ma$. Acceleration is proportional to net force and inversely proportional to mass, and points in the direction of the net force (not necessarily the motion).
3. Third law: if A pushes on B, B pushes back on A with equal magnitude and opposite direction. The two forces act on different objects, so they never cancel each other in one object's free-body diagram.

Mass vs. Weight (Classic Discrimination)

A $60\;\text{kg}$ astronaut is $60\;\text{kg}$ everywhere; her weight is $600\;\text{N}$ on Earth ($g = 10\;\text{m/s}^2$) but only $96\;\text{N}$ on the Moon ($g = 1.6\;\text{m/s}^2$).

Free-Body Diagrams

Draw ONE object, then every force acting on it (never forces it exerts on other things):

• Weight $mg$, always straight down
• Normal force $N$, perpendicular to the contact surface
• Tension $T$, along the rope, pulling away from the object
• Friction $f$, along the surface, opposing sliding (Part 2)

Equilibrium ($a = 0$) means forces balance in each direction separately: $\Sigma F_x = 0$ and $\Sigma F_y = 0$. Constant velocity is equilibrium too — the first law again.

Worked Example — From Free-Body Diagram to Acceleration

A $10\;\text{kg}$ crate on a horizontal floor is pushed with a $60\;\text{N}$ horizontal force. Friction opposes the motion with $20\;\text{N}$. ($g = 10\;\text{m/s}^2$)

Step 1 — Free-body diagram. Four forces: weight $mg = 100\;\text{N}$ down, normal $N$ up, push $60\;\text{N}$ forward, friction $20\;\text{N}$ backward.

Step 2 — Vertical direction. No vertical acceleration:

$N - mg = 0 \Rightarrow N = 100\;\text{N}$

Step 3 — Horizontal direction.

$F_{net} = 60 - 20 = 40\;\text{N} \Rightarrow a = \frac{40}{10} = 4\;\text{m/s}^2$

Worked Example — Hanging in Equilibrium

A $2\;\text{kg}$ picture hangs at rest from a single vertical wire.

$T - mg = 0 \Rightarrow T = (2)(10) = 20\;\text{N}$

Now cut the wire: the only remaining force is gravity, so $a = g = 10\;\text{m/s}^2$ downward — regardless of mass. Heavier objects feel more gravitational force but also have proportionally more inertia; the two effects cancel, which is why all objects free-fall at the same $g$.

MCAT habit: always ask "what object am I analyzing, and what forces act ON it?" Most wrong answers come from mixing forces on different objects or from inventing a leftover "force of motion" that does not exist.

Newton's Laws & Free-Body Diagrams 🎯

Key Takeaways — Part 1

• First law: constant velocity (including rest) whenever the net force is zero — motion needs no maintaining force
• Second law: $F_{net} = ma$; acceleration points along the NET force
• Third law: equal-and-opposite forces act on DIFFERENT objects and never cancel in one free-body diagram
• Mass (kg, invariant) vs. weight ($mg$, in newtons, location-dependent) — a favorite MCAT discrimination
• Free-body diagrams list only forces acting ON the chosen object; solve $x$ and $y$ separately

Forces & Newton's Laws

Part 2 of 4 — Friction & Inclined Planes

Two Kinds of Friction

Key facts the MCAT tests:

• $\mu_s > \mu_k$ for a given pair of surfaces — it takes more force to start sliding than to keep sliding.
• Static friction is an inequality: push a heavy crate gently and friction matches your push exactly. Only at the verge of slipping does $f_s = \mu_s N$.
• Kinetic friction is independent of sliding speed and of contact area — it depends only on $\mu_k$ and the normal force.
• Friction depends on $N$, not directly on weight. Press down on a block and $N$ (hence friction) grows; lift up on it and both shrink.

Inclined Planes

Tilt the axes so $x$ runs along the incline. Weight $mg$ splits into:

• Along the incline (pulling downhill): $mg\sin\theta$
• Perpendicular (into the surface): $mg\cos\theta$

With no other vertical-direction forces, $N = mg\cos\theta$ — less than $mg$. Two limiting checks: at $\theta = 0$, the downhill pull is $0$ and $N = mg$; at $\theta = 90^\circ$, the pull is the full $mg$ and $N = 0$.

Angle-of-slip trick: a block on the verge of sliding at angle $\theta$ gives $\mu_s = \tan\theta$; a block sliding down at constant velocity gives $\mu_k = \tan\theta$. Both follow from setting the downhill pull equal to friction.

Worked Example — Sliding Down a 37-Degree Ramp

A $10\;\text{kg}$ crate slides down a $37^\circ$ loading ramp with $\mu_k = 0.25$. Find its acceleration. ($g = 10\;\text{m/s}^2$, $\sin 37^\circ \approx 0.6$, $\cos 37^\circ \approx 0.8$)

Step 1 — Resolve the weight. $mg = 100\;\text{N}$:

$mg\sin\theta = 100(0.6) = 60\;\text{N} \quad \text{(downhill)}$

$mg\cos\theta = 100(0.8) = 80\;\text{N} \quad \text{(into the ramp)}$

Step 2 — Normal force and friction.

$N = 80\;\text{N} \Rightarrow f_k = \mu_k N = (0.25)(80) = 20\;\text{N} \quad \text{(uphill, opposing the slide)}$

Step 3 — Newton's second law along the incline.

$F_{net} = 60 - 20 = 40\;\text{N} \Rightarrow a = \frac{40}{10} = 4\;\text{m/s}^2$

Note the mass cancels if you keep it symbolic: $a = g(\sin\theta - \mu_k\cos\theta)$ — a heavier crate slides with the same acceleration.

Worked Example — Will It Move at All?

A $10\;\text{kg}$ crate sits on a horizontal floor with $\mu_s = 0.4$. You push horizontally with $30\;\text{N}$.

Maximum static friction: $f_{s,max} = \mu_s N = (0.4)(100) = 40\;\text{N}$.

Your $30\;\text{N}$ push is below the $40\;\text{N}$ ceiling, so the crate does not budge — and the friction force is $30\;\text{N}$, exactly matching your push. Writing $40\;\text{N}$ here is the single most common friction error on the exam: the maximum is only reached at the verge of slipping.

Friction & Inclined Planes 🎯

Key Takeaways — Part 2

• Static friction adjusts up to a ceiling ($f_s \leq \mu_s N$); kinetic friction is fixed at $f_k = \mu_k N$; $\mu_s > \mu_k$
• Friction tracks the NORMAL force — pressing down or pulling up on an object changes its friction
• Kinetic friction ignores sliding speed and contact area
• Inclines: downhill pull $mg\sin\theta$, normal force $mg\cos\theta$ — check the $0^\circ$ and $90^\circ$ limits
• Constant-velocity slide down an incline: $\mu_k = \tan\theta$; verge of slipping: $\mu_s = \tan\theta$
• Acceleration on an incline is mass-independent: $a = g(\sin\theta - \mu_k\cos\theta)$

Forces & Newton's Laws

Part 3 of 4 — Tension, Apparent Weight & Circular Motion

Tension

An ideal (massless) rope transmits the same tension everywhere along its length, and pulleys merely redirect it. Tension pulls away from the object along the rope — ropes can pull, never push.

Apparent Weight (Elevators)

A scale reads the normal force it exerts, not your actual weight. In an elevator, apply $F_{net} = ma$ vertically to the person:

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

with $a$ positive for upward acceleration:

Direction of velocity is irrelevant; only acceleration matters. Moving down while slowing is an upward acceleration — the scale reads heavy.

Uniform Circular Motion

An object moving in a circle at constant speed still accelerates, because its velocity direction changes. The acceleration points toward the center:

$a_c = \frac{v^2}{r} \qquad F_c = \frac{mv^2}{r}$

Centripetal force is not a new force — it is the net inward force, always supplied by real forces:

• Car on a flat curve: static friction (the tires grip, they do not slide)
• Ball on a string: tension
• Moon around Earth: gravity
• Bottom of a vertical loop: $N - mg$ points up toward the center, so $N = mg + mv^2/r$ — you feel heavy

There is no outward "centrifugal force" acting on the object; the outward fling you feel is your inertia trying to continue in a straight line.

Worked Example — The Elevator Scale

A $50\;\text{kg}$ patient stands on a scale in an elevator ($g = 10\;\text{m/s}^2$).

Case 1 — accelerating upward at $2\;\text{m/s}^2$:

$N = m(g + a) = 50(10 + 2) = 600\;\text{N}$

Case 2 — accelerating downward at $2\;\text{m/s}^2$:

$N = m(g - a) = 50(10 - 2) = 400\;\text{N}$

Case 3 — constant velocity $3\;\text{m/s}$ upward:

$a = 0 \Rightarrow N = mg = 500\;\text{N}$

The scale cannot tell constant-velocity motion from standing still — only acceleration registers.

Worked Example — Rounding a Curve

A $1000\;\text{kg}$ car rounds a flat curve of radius $50\;\text{m}$ at $10\;\text{m/s}$.

Step 1 — Required centripetal force.

$F_c = \frac{mv^2}{r} = \frac{(1000)(10)^2}{50} = \frac{100000}{50} = 2000\;\text{N}$

Step 2 — What supplies it? Static friction from the road on the tires, pointing toward the center of the curve. If the road cannot supply $2000\;\text{N}$ (ice), the car does not fly outward — it continues in a straight line, which the curve then bends away from.

Step 3 — Speed sensitivity. $F_c \propto v^2$: doubling the speed to $20\;\text{m/s}$ quadruples the needed force to $8000\;\text{N}$ — why speeding into curves fails so abruptly.

Tension, Apparent Weight & Circular Motion 🎯

Key Takeaways — Part 3

• Ideal ropes carry one tension throughout; pulleys redirect it; ropes only pull
• Scales read the normal force: $N = m(g + a)$ with upward $a$ positive — velocity direction is irrelevant
• Free fall reads zero: "weightlessness" is missing support force, not missing gravity
• Circular motion at constant speed still accelerates inward: $a_c = v^2/r$, $F_c = mv^2/r$
• Centripetal force is the net inward force, always supplied by a real force (friction, tension, gravity, normal)
• $F_c \propto v^2$: doubling speed quadruples the required inward force

Forces & Newton's Laws

Part 4 of 4 — MCAT Integration: Torque, Equilibrium & the Body as Levers

Torque: Force with Leverage

A force's ability to rotate an object about a pivot is its torque:

$\tau = rF\sin\theta$

where $r$ is the distance from pivot to the point of application and $\theta$ is the angle between $r$ and $F$. Maximum torque comes from pushing perpendicular to the lever arm ($\sin 90^\circ = 1$); a force aimed straight through the pivot ($r = 0$ or $\theta = 0$) produces zero torque, no matter how large.

Complete Equilibrium — Two Conditions

For balance problems, pick the pivot cleverly: any force acting at the pivot drops out of the torque equation. Counterclockwise torques balance clockwise torques:

$r_1F_1 = r_2F_2 \quad \text{(perpendicular forces)}$

The Body Is Built from Levers

MCAT passages love musculoskeletal setups. Most limb muscles insert close to the joint, while the load sits far away (a weight in the hand). Torque balance then demands:

$F_{muscle} \times r_{small} = F_{load} \times r_{large} \Rightarrow F_{muscle} \gg F_{load}$

Muscles routinely exert forces several times the load's weight. The trade-off: the load end moves through a large arc quickly — anatomy sacrifices force advantage for speed and range of motion.

Center of gravity: an object balances when the support is under its center of gravity; for torque problems, treat the object's entire weight as acting at that single point (the middle of a uniform beam).

Worked Example — The Forearm as a Lever (Passage Style)

Biomechanics researchers model the forearm as a rigid bar pivoting at the elbow. The biceps inserts $5\;\text{cm}$ from the elbow and pulls vertically upward. A subject holds a $20\;\text{N}$ ball in the hand, $40\;\text{cm}$ from the elbow, with the forearm horizontal and stationary. (Neglect the forearm's own weight.)

Step 1 — Torque balance about the elbow. Choosing the elbow as the pivot eliminates the unknown joint-reaction force:

$F_{biceps}(0.05) = (20)(0.40)$

Step 2 — Solve.

$F_{biceps} = \frac{8}{0.05} = 160\;\text{N}$

The muscle pulls with eight times the ball's weight — the ratio of the lever arms ($40/5 = 8$).

Step 3 — Joint reaction force. Translational equilibrium: upward forces $160\;\text{N}$ (biceps) must balance downward forces $20\;\text{N}$ (ball) plus the elbow joint force $R$:

$160 = 20 + R \Rightarrow R = 140\;\text{N}$

The humerus presses down on the forearm at the joint with $140\;\text{N}$ — joints bear forces far exceeding the loads we carry, which is exactly the punchline such passages build toward.

Passage strategy: (1) draw the bar, pivot, and every force with its distance; (2) put the pivot where the ugliest unknown acts; (3) use torque balance for the muscle force, then force balance for the joint force. Data-table versions simply repeat this at several angles — remember only the perpendicular component $F\sin\theta$ contributes.

Torque, Equilibrium & Levers 🎯

Key Takeaways — Part 4

• $\tau = rF\sin\theta$: only the perpendicular component of force rotates; forces through the pivot give zero torque
• Complete equilibrium needs BOTH $\Sigma F = 0$ and $\Sigma\tau = 0$
• Choose the pivot where an unknown force acts to eliminate it from the torque equation
• Body levers: muscles insert near joints, so muscle forces exceed loads by the lever-arm ratio; joints bear even more
• Treat an object's weight as acting at its center of gravity (midpoint of a uniform beam)
• Passage recipe: torque balance first for the muscle, then force balance for the joint reaction