Newton's Laws aren't just abstract equations—they're the analytical toolkit you'll use to decode nearly every mechanics problem on the AP Physics C exam. Whether you're analyzing a skier descending a slope, a satellite orbiting Earth, or a mass oscillating on a spring, you're applying the same fundamental principle: net force determines acceleration. The exam tests your ability to identify all forces acting on a system, resolve them into components, and apply Fnet=ma to predict motion.
The applications in this guide demonstrate constraint forces, contact interactions, restoring forces, and velocity-dependent forces—concepts that appear repeatedly in both multiple-choice and free-response questions. You're being tested on your ability to construct free-body diagrams, choose appropriate coordinate systems, and recognize when forces like tension, normal force, or friction provide the net force needed for a particular type of motion. Don't just memorize formulas—know which physical principle each scenario illustrates and how to set up the equations from scratch.
Contact Forces and Surface Interactions
These forces arise whenever objects touch surfaces or each other. The key principle: contact forces are reaction forces that adjust based on the system's constraints and applied forces.
Friction Forces (Static and Kinetic)
Static frictionfs≤μsN—prevents relative motion between surfaces until the applied force exceeds the maximum static friction threshold
Kinetic frictionfk=μkN—acts on sliding objects and is typically smaller than maximum static friction (μk<μs)
Coefficient of frictionμ depends on surface materials—this dimensionless quantity appears in nearly every inclined plane and connected-mass problem
Normal Force
Perpendicular contact forceN—always acts perpendicular to the surface, not necessarily equal to mg when other forces or angles are present
Adjusts dynamically based on applied forces, inclination angles, and acceleration—calculate it from equilibrium in the perpendicular direction
Directly determines friction since f=μN—errors in normal force propagate through your entire solution
Inclined Planes
Component resolution splits gravity into mgsinθ (parallel) and mgcosθ (perpendicular)—choose coordinates aligned with the surface
Normal forceN=mgcosθ for objects on simple inclines—decreases as the angle increases
Net parallel force determines acceleration—combine gravity component, friction, and any applied forces along the slope
Compare: Static vs. kinetic friction—both depend on normal force, but static friction is an inequality (fs≤μsN) while kinetic friction is an equality (fk=μkN). FRQs often ask you to find the angle at which sliding begins, requiring you to set mgsinθ=μsmgcosθ.
Constraint Forces in Connected Systems
When objects are connected by ropes, cables, or pulleys, constraint forces ensure the system moves together. The key insight: constraints reduce degrees of freedom and create relationships between accelerations.
Tension in Ropes and Cables
Transmits force through the rope—for an ideal (massless, inextensible) rope, tension is uniform throughout
Direction always pulls along the rope away from the object—never pushes
Changes with angle in systems like conical pendulums—resolve tension into components when the rope isn't vertical
Pulleys and Atwood Machines
Ideal pulleys redirect tension without changing its magnitude—the rope has the same tension on both sides
Atwood machine accelerationa=m1+m2(m1−m2)g—derived by applying Newton's second law to each mass separately
System approach treats connected masses as one object with total mass m1+m2—useful for finding acceleration before solving for tension
Connected Mass Systems
Constraint equations link accelerations—if masses are connected by an inextensible rope, they share the same magnitude of acceleration
Apply Newton's second law separately to each mass—this generates enough equations to solve for all unknowns
Tension is internal to the system—it cancels when you add the equations, leaving only external forces
Compare: Simple pulley vs. Atwood machine—a simple pulley with one hanging mass has a=g, while an Atwood machine has a<g because the counterweight reduces net force. If an FRQ asks about limiting cases, note that a→0 as m1→m2.
Circular Motion Applications
Circular motion requires a centripetal (center-seeking) force to continuously change the velocity direction. The principle: identify which real force(s) provide the required centripetal accelerationac=rv2.
Centripetal Force in Circular Motion
Net radial forceFc=rmv2=mω2r—this is not a new force but the sum of real forces pointing toward the center
Provided by tension, gravity, friction, or normal force—your job is to identify which force(s) act radially inward
Vertical loops require minimum speed v=gr at the top—at this speed, gravity alone provides the centripetal force
Banked Curves
Normal force has horizontal componentNsinθ pointing toward the center—this provides centripetal force without friction
Design speedv=rgtanθ—at this speed, no friction is needed to maintain the circular path
Friction assists or opposes depending on whether the car travels faster or slower than design speed—static friction adjusts to provide additional centripetal force or prevent sliding outward
Compare: Flat curve vs. banked curve—on a flat curve, friction alone provides centripetal force (μsmg=rmv2), while a banked curve uses the normal force's horizontal component. Banked curves allow higher speeds because N>mg contributes more centripetal force than friction alone.
Velocity-Dependent and Restoring Forces
Some forces depend on velocity or displacement rather than being constant. These lead to non-constant acceleration and require different analytical approaches.
Drag Forces and Terminal Velocity
Drag forceFD=bv or FD=cv2—opposes motion and increases with speed, where the form depends on the flow regime
Terminal velocity occurs when FD=mg—net force becomes zero, so acceleration stops and velocity remains constant
Exponential approach to terminal velocity—acceleration decreases as speed increases, creating a characteristic curve on v(t) graphs
Springs and Hooke's Law
Hooke's LawFs=−kΔx—the negative sign indicates a restoring force directed toward equilibrium
Spring constantk (units: N/m) measures stiffness—larger k means more force required for the same displacement
Elastic potential energyU=21k(Δx)2—this energy transfers to kinetic energy during oscillation, connecting to SHM with ω=mk
Compare: Drag force vs. spring force—both create non-constant acceleration, but drag depends on velocity while spring force depends on position. Drag leads to terminal velocity (equilibrium), while springs lead to oscillation (SHM). FRQs may ask you to sketch a(t) or v(t) graphs for each.
Quick Reference Table
Concept
Best Examples
Contact forces perpendicular to surface
Normal force, friction on inclines
Constraint forces in connected systems
Tension, Atwood machines, pulley systems
Centripetal force providers
Tension (conical pendulum), friction (flat curves), normal force (banked curves), gravity (orbits)
Velocity-dependent forces
Drag force, terminal velocity
Position-dependent restoring forces
Spring force (Hooke's Law), pendulum restoring force
Component resolution
Inclined planes, banked curves, conical pendulums
System vs. individual analysis
Connected masses, Atwood machines
Non-constant acceleration
Drag approach to terminal velocity, spring oscillations
Self-Check Questions
On an inclined plane, how does increasing the angle affect (a) the normal force and (b) the acceleration of a sliding block? At what angle does sliding begin if μs=0.5?
In an Atwood machine with masses m1 and m2, what happens to the acceleration and tension as m1→m2? What physical situation does this limiting case represent?
Compare how centripetal force is provided in three scenarios: a car on a flat curve, a car on a banked curve at design speed, and a ball on a string in a vertical loop at the top. Which force(s) point toward the center in each case?
A falling object experiences drag force FD=bv. Sketch qualitative graphs of a(t) and v(t) from release until terminal velocity is reached. How would these graphs change if the object were more massive?
For a mass-spring system displaced from equilibrium and released, explain why the acceleration is maximum at the endpoints and zero at equilibrium—even though the velocity behavior is exactly opposite. How does this connect to the equation a=−ω2x?