Fundamental Laws of Motion to Know for AP Physics C: Mechanics

1. What You Need to Know

You’re being tested on how well you can translate physical situations into Newton’s laws, correctly choose a system, draw a clean free-body diagram (FBD), and write the right equations for translation and rotation. Nearly every Mechanics problem is “just” Newton’s laws + bookkeeping.

The core idea

• Newton’s laws govern how forces relate to motion.
• They’re simplest in inertial frames (non-accelerating frames).
• Most AP Physics C: Mech setups boil down to one of these forms:
  • Translation: $\sum \vec{F}_{\text{ext}} = m\vec{a}$
  • Momentum form (great for systems): $\sum \vec{F}_{\text{ext}} = \frac{d\vec{p}}{dt}$ with $\vec{p}=m\vec{v}$
  • Rotation about a fixed axis: $\sum \tau_{\text{ext}} = I\alpha$

Newton’s Three Laws (what they really mean)

1. 1st Law (Inertia): If $\sum \vec{F}_{\text{ext}} = \vec{0}$, then $\vec{v}$ is constant (including rest). This defines inertial frames.
2. 2nd Law (Dynamics): Net external force produces acceleration: $\sum \vec{F}_{\text{ext}} = m\vec{a}$ (valid for constant mass) or more generally $\sum \vec{F}_{\text{ext}}=\frac{d\vec{p}}{dt}$.
3. 3rd Law (Action–Reaction): Forces come in pairs on different objects: $\vec{F}_{A\to B}=-\vec{F}_{B\to A}$.

Why “system choice” matters

• For a single object: include all forces on that object.
• For a system of objects: internal forces cancel in pairs (Newton’s 3rd), so you can often relate motion to external forces only:
  $\sum \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}$
  where $M$ is total mass.

Critical reminder: Normal force, tension, and friction are forces (not “equal to” something automatically). They are whatever they need to be to satisfy Newton’s 2nd law + constraints.

2. Step-by-Step Breakdown

This is the high-yield workflow for 90% of force/torque problems.

A. Translational Newton’s Laws (FBD method)

1. Choose the system

   • One object? A block? A pulley? A whole set of connected masses?
   • Ask: “What forces become internal and cancel if I group things?”

2. Draw a correct FBD (only forces ON the system)

   • Include: weight $\vec{W}=m\vec{g}$, normal $\vec{N}$, tension $\vec{T}$, friction $\vec{f}$, applied forces, spring forces.
   • Do not include “$m\vec{a}$” as a force.

3. Pick axes strategically

   • Align axes with motion/constraints.
   • Common trick: for inclines, use axes parallel/perpendicular to the surface.

4. Resolve forces into components

   • Example on incline angle $\theta$:
     • Parallel component of weight: $mg\sin\theta$ down the slope
     • Perpendicular component: $mg\cos\theta$ into the plane

5. Write Newton’s 2nd law in components

   • For each axis: $\sum F_x = ma_x$, $\sum F_y = ma_y$.
   • If there’s no acceleration perpendicular to a surface: set $a_\perp=0$.

6. Add constraint equations (ropes/pulleys/rolling)

   • Ideal rope: same tension throughout (if massless rope, frictionless pulley).
   • Connected masses often share an acceleration magnitude.
   • No slip rolling: $v_{\text{cm}}=\omega R$ and $a_{\text{cm}}=\alpha R$.

7. Solve algebraically; sanity-check signs and limits

   • Check limiting cases (frictionless, very large mass, etc.).

Mini worked example (setup only): Block on incline with friction

• Mass $m$ on incline angle $\theta$, kinetic friction coefficient $\mu_k$, sliding down.
• Choose axes: $x$ down slope, $y$ perpendicular.
• Forces:
  • Down slope: $mg\sin\theta$
  • Normal: $N$
  • Friction up slope: $f_k=\mu_k N$
• Equations:
  • Perpendicular: $\sum F_y = 0 \Rightarrow N - mg\cos\theta = 0 \Rightarrow N=mg\cos\theta$
  • Parallel: $\sum F_x = ma \Rightarrow mg\sin\theta - \mu_k mg\cos\theta = ma$
  • Result: $a=g(\sin\theta-\mu_k\cos\theta)$

B. Rotational Newton’s Laws (torque method)

1. Choose axis of rotation (fixed axis or choose a point)
2. Draw forces and lever arms
3. Compute torques with sign convention
   • Magnitude: $\tau = rF\sin\phi$ (where $\phi$ is angle between $\vec{r}$ and $\vec{F}$)
4. Write $\sum \tau_{\text{ext}} = I\alpha$
5. Link translation and rotation if needed
   • Rolling without slipping: $a=\alpha R$

Decision point: If the object both translates and rotates (pulley, rolling cylinder), you often need both $\sum F=ma$ and $\sum \tau = I\alpha$ plus a constraint.

3. Key Formulas, Rules & Facts

A. Newton’s laws and momentum (core statements)

B. Common forces (models you must know cold)

On AP Physics C, friction is the #1 place students silently lose points: always state whether it’s static vs kinetic and whether the object is actually slipping.

C. Equilibrium conditions (translation + rotation)

D. Torque and rotational dynamics essentials

E. Quick moment of inertia facts (only what you commonly need with $\sum\tau=I\alpha$)

4. Examples & Applications

Example 1: Atwood machine (two hanging masses)

Two masses $m_1$ and $m_2$ connected by a light rope over a frictionless pulley. Assume $m_2>m_1$.

Setup (FBD + Newton’s 2nd):

• For $m_2$ (down positive): $m_2g - T = m_2 a$
• For $m_1$ (up positive): $T - m_1g = m_1 a$

Key insight: Add equations to eliminate $T$:
$m_2g-m_1g=(m_1+m_2)a \Rightarrow a=\frac{(m_2-m_1)g}{m_1+m_2}$
Then back-substitute for $T$.

Exam variation: If the pulley has rotational inertia, you must add $\sum\tau=I\alpha$ and $a=\alpha R$; then tension differs on each side.

Example 2: Elevator apparent weight

Person of mass $m$ stands on scale in elevator accelerating upward with acceleration $a$.

FBD: Forces on person: $N$ up, $mg$ down.

Equation (up positive):
$N - mg = ma \Rightarrow N = m(g+a)$

Key insight: Scale reads $N$. If elevator accelerates downward, $N=m(g-a)$. In free fall, $N=0$.

Example 3: Block pushed against a wall (static friction trap)

A horizontal force $F$ pushes a block of mass $m$ against a vertical wall. Coefficient of static friction $\mu_s$.

FBD:

• Horizontal: applied $F$ into wall, normal $N$ out.
• Vertical: weight $mg$ down, static friction $f_s$ up (if not slipping).

Equations:

• Horizontal equilibrium: $F-N=0 \Rightarrow N=F$
• Vertical equilibrium: $f_s-mg=0 \Rightarrow f_s=mg$
• Static friction condition: $f_s\le \mu_s N \Rightarrow mg \le \mu_s F$

Key insight: Friction is not automatically $\mu_s N$; it becomes whatever is needed up to the maximum.

Example 4: Rolling down an incline (translation + rotation)

A rigid body (mass $m$, radius $R$, moment of inertia $I$ about center) rolls without slipping down incline angle $\theta$.

Equations:

• Translation along slope: $mg\sin\theta - f = ma$
• Torque about center: $fR = I\alpha$
• No slip: $a=\alpha R$

Eliminate $f$ and $\alpha$:
$f=\frac{I a}{R^2}$
$mg\sin\theta - \frac{I a}{R^2} = ma \Rightarrow a=\frac{g\sin\theta}{1+\frac{I}{mR^2}}$

Key insight: Bigger $\frac{I}{mR^2}$ (more “rotational inertia”) means smaller acceleration.

5. Common Mistakes & Traps

1. Mixing up Newton’s 3rd law pairs

   • Wrong: Putting action–reaction forces on the same FBD so they cancel.
   • Why wrong: 3rd law pairs act on different objects.
   • Fix: Label forces as $\vec{F}_{A\to B}$; only include forces acting on your chosen system.

2. Assuming $N=mg$ automatically

   • Wrong: Writing $N=mg$ even on inclines, elevators, or with additional vertical forces.
   • Why wrong: $N$ comes from $\sum F_\perp = ma_\perp$.
   • Fix: Always write the perpendicular/vertical Newton’s 2nd equation first.

3. Using $f=\mu N$ without checking static vs kinetic

   • Wrong: Replacing friction with $\mu_s N$ in static problems.
   • Why wrong: Static friction satisfies $f_s\le \mu_s N$; it’s only equal at the threshold of slipping.
   • Fix: Solve for required $f_s$, then check if it exceeds $\mu_s N$.

4. Sign errors from sloppy axis choices

   • Wrong: Taking “down the incline” as positive for one mass and negative for the other without consistent constraints.
   • Why wrong: You’ll get contradictory accelerations/tensions.
   • Fix: Define one positive direction per object but relate them with a clear constraint (e.g., same magnitude $a$).

5. Forgetting that tension can differ with massive pulleys

   • Wrong: Assuming same $T$ on both sides when the pulley has rotational inertia.
   • Why wrong: Net torque requires $\Delta T \neq 0$: $\sum\tau=(T_2-T_1)R=I\alpha$.
   • Fix: Use $T_1$ and $T_2$ separately when pulleys are not ideal.

6. Treating centripetal force as an extra force

   • Wrong: Adding a “centripetal force” term in the FBD.
   • Why wrong: “Centripetal” describes the net inward force requirement: $\sum F_r = m\frac{v^2}{r}$.
   • Fix: Draw real forces (tension, normal, gravity) and set their radial sum equal to $m\frac{v^2}{r}$.

7. Dropping torque signs or wrong lever arm

   • Wrong: Using $\tau=rF$ with the full force when only a component is perpendicular.
   • Why wrong: Only perpendicular component contributes: $\tau=rF\sin\phi$.
   • Fix: Identify the perpendicular component or use $\tau=F\ell$ with moment arm $\ell$.

8. Using rolling constraints when slipping occurs

   • Wrong: Applying $a=\alpha R$ even when kinetic friction is present and slipping happens.
   • Why wrong: That constraint is only for pure rolling.
   • Fix: If slipping, relate translation and rotation through dynamics only (and use $f_k=\mu_k N$).

6. Memory Aids & Quick Tricks

7. Quick Review Checklist

• You can state and apply Newton’s laws, especially $\sum \vec{F}_{\text{ext}}=m\vec{a}$ and $\vec{F}_{A\to B}=-\vec{F}_{B\to A}$.
• You always start with a system choice and a clean FBD (only real forces on the system).
• You write Newton’s 2nd law in components aligned with constraints.
• You treat friction correctly: $f_k=\mu_k N$ when sliding, $f_s\le\mu_s N$ when not.
• You don’t assume $N=mg$ unless the perpendicular/vertical equation says so.
• For connected masses, you add the constraint that accelerations match (and handle pulley inertia if present).
• For rotation, you can compute torque with $\tau=rF\sin\phi$ and apply $\sum\tau=I\alpha$.
• For rolling without slipping, you remember the trio: $\sum F=ma$, $\sum\tau=I\alpha$, $a=\alpha R$.

You’ve got this—if your FBD and sign conventions are clean, the algebra almost always falls into place.