Topic 2.4: Newton's First Law — Equilibrium
Name: _______________________________ Date: _______________
Before We Start: Yesterday's Recap
Yesterday we learned Newton's third law is unconditional — always equal, no exceptions. Today's law has a condition attached to it. Based on the word "equilibrium," what condition do you think has to be true?
Today's New Concepts
AP CED Alignment: Unit 2, Topic 2.4
Equilibrium and Newton's First Law
- Equilibrium means the net force on an object is exactly zero
- An object in equilibrium moves at constant velocity — including the special case of constant velocity equal to zero (at rest)
- Equilibrium does NOT mean "no forces are acting" — it means the forces that ARE acting sum to zero
- Symbol: $\Sigma F = 0$
Worked Example: A ball thrown straight up is momentarily at rest at the very top of its path. It is NOT in equilibrium at that instant, because gravity is still pulling down on it with nothing balancing that force — the net force is nonzero even though the velocity is momentarily zero.
Solving Axis by Axis
- Full equilibrium requires BOTH $\Sigma F_x = 0$ AND $\Sigma F_y = 0$, checked independently
- Changing one force's component on only one axis can break equilibrium in that direction while the other axis stays balanced
- Standard process: (1) draw the FBD, (2) write $\Sigma F_x = 0$ and/or $\Sigma F_y = 0$ symbolically with every force term, (3) solve symbolically for the unknown, (4) THEN substitute numbers
Worked Example: A box hangs from a string with a scale below it reading $F_N$. $\Sigma F_y = 0$: $F_T + F_N - mg = 0$, solved symbolically first: $F_T = mg - F_N$ — only then would numbers get substituted if given.
Symbolic Derivations (No Numbers Given)
- Some equilibrium problems never provide numbers at all — the entire answer is a symbolic expression, solved the same way: write $\Sigma F = 0$, then solve algebraically for the requested quantity
Worked Example: A skydiver system reaches terminal speed $v_T$ where air resistance $Av^2$ balances gravity $mg$: $Av_T^2 - mg = 0 \implies v_T = \sqrt{\dfrac{mg}{A}}$ — solved entirely in symbols.
Keep This Sheet!
These sheets build into your semester study guide. Keep them in order in a binder or folder — you'll want to flip back through them before quizzes, unit tests, and when AP exam review starts in the spring.