A slide-by-slide walkthrough of all 9 topics in Force and Translational Dynamics. Each slide covers the big idea, key definitions, and a worked example for one topic.
To analyze motion, we pick a system — one or more objects we care about. The system can be treated as a single point located at its center of mass. Internal forces between parts of the system always cancel in pairs (Newton's third law), so only EXTERNAL forces can change the system's center-of-mass motion.
Two balls (3 kg and 1 kg) at positions x = 0 and x = 4 m. The center of mass is at:
x_cm = (3·0 + 1·4) / (3 + 1) = 1 m
Note: the CoM is closer to the heavier mass. If the balls push each other through the spring, the spring force is internal — the CoM stays at x = 1 m unless an external force acts.
A force is an interaction between two objects. It's a vector — has both magnitude and direction. A free-body diagram (FBD) shows every force acting ON a single chosen object, drawn as labeled arrows from a single dot. Every Unit 2 problem starts with a correct FBD.
Forces ON the box:
Tip: Tilt your coordinate axes so one axis is along the incline. This makes Newton's-law equations cleaner.
If object A pushes object B, then B pushes A with equal force in the opposite direction. The two forces of an action-reaction pair act on DIFFERENT objects, so they never cancel each other. Tension in an ideal (massless) string is the same throughout; an ideal pulley just redirects tension without changing its magnitude.
You might think: "the book sits still because gravity pulls it down and the normal force pushes it up — they're a third-law pair."
That's wrong. Both gravity and normal force are forces ON the book. Their third-law partners act on OTHER objects: gravity's partner is the book pulling Earth UP; normal's partner is the book pushing the table DOWN. The book sits still because the two forces on IT (gravity and normal) happen to be balanced — that's Newton's first law, not third law.
If the net force on a system is zero, its velocity stays constant — either at rest, or moving at constant velocity in a straight line. This is translational equilibrium. Forces can be balanced in one direction but unbalanced in another; the system accelerates only in the unbalanced direction.
Once a skydiver reaches terminal velocity, the air drag force equals their weight: F_g (down) = F_drag (up). Net force = 0.
By Newton's first law, the skydiver continues at constant velocity. They're NOT at rest — they're falling — but they're not accelerating either. This is equilibrium with non-zero velocity.
F_net = ma. The acceleration of a system equals the net external force divided by its mass, and points in the same direction as the net force. This single equation is the foundation of Unit 2 — and most of classical physics.
A 5 kg box is pushed with a horizontal force of 20 N on a frictionless floor.
Vertical: F_N = F_g = mg = 50 N (no vertical acceleration).
Horizontal: F_net = 20 N. So a = F_net / m = 20 / 5 = 4 m/s² in the direction of the push.
Every mass attracts every other mass: F_g = Gm₁m₂/r². Near Earth's surface, the gravitational force on a mass m is simply F_g = mg with g ≈ 10 m/s² downward. Weight is this gravitational force. Apparent weight is what a scale reads — the normal force on you, which can differ from true weight when accelerating.
A 60 kg person in an elevator accelerating upward at 2 m/s².
Newton's 2nd law (taking up as positive): F_N − mg = ma → F_N = m(g + a) = 60(10 + 2) = 720 N
True weight = mg = 600 N. The scale reads 720 N — you feel heavier than usual. If the elevator were in free fall (a = −10 m/s²), the scale would read zero — apparent weightlessness.
Kinetic friction acts when surfaces ARE sliding — magnitude F_k = μ_k · F_N, always opposing motion. Static friction acts when surfaces aren't sliding — it adjusts to whatever value is needed to prevent slipping, up to a maximum of F_s,max = μ_s · F_N. The coefficients depend on the materials, not on contact area or speed.
A 20 kg crate sits on a horizontal floor with μ_s = 0.5 and μ_k = 0.3.
Normal force F_N = mg = 200 N. Maximum static friction = 0.5 · 200 = 100 N.
If you push with 80 N: static friction matches at 80 N → no motion. If you push with 110 N: static friction maxes at 100 N, you exceed it, crate starts sliding. Once it's moving, kinetic friction takes over at 0.3 · 200 = 60 N.
An ideal spring exerts a force proportional to how far it's stretched or compressed from its relaxed length: F_s = kx. The force always points back toward the equilibrium position — it's "restoring." The spring constant k measures stiffness (N/m).
A 2 kg mass hangs from a vertical spring with k = 200 N/m. Find how far it stretches at equilibrium.
At equilibrium, the spring force balances gravity: kx = mg → x = mg/k = (2)(10)/200 = 0.10 m = 10 cm.
Pull it down further and let go: it oscillates around this equilibrium point (more in Unit 7: Oscillations).
An object moving in a circle is constantly accelerating toward the center: a_c = v²/r. By Newton's second law, this requires a net inward force: F_c = mv²/r. "Centripetal force" isn't a new kind of force — it's a ROLE played by whatever real force points inward (tension for a ball on a string, gravity for an orbit, friction for a car turning, normal force on a banked curve).
A 1000 kg car drives around a curve of radius 50 m at 20 m/s. What's the minimum coefficient of static friction needed?
Friction provides the centripetal force: μ_s · mg = mv²/r → μ_s = v²/(gr) = 400/(10·50) = 0.80.
If the road's friction coefficient is less than 0.80, the car will slide outward. That's why curves are often banked — the normal force can also help provide centripetal force.
Spend 1–2 minutes per slide. Read the Big Idea first, then the definitions, then the worked example. Nine slides total — one per topic.
Use the topic pills to jump to any topic, or use the arrow keys to step through them in order.
Unit 2 is the heaviest on the exam — this set works great for the night before the test to refresh every topic in about 20 minutes.