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.