AP Physics 1 Unit 2: Force and Translational Dynamics Cheat Sheet

Quick-reference study sheet · uses g = 9.8 m/s2

2.1 Systems and Center of Mass

System:The object or group of objects you choose to study. Everything else is the environment. Forces between objects inside the system are internal; forces from the environment are external.
Treat as one object:When the internal structure doesn't matter, a system can be modeled as a single object located at its center of mass.

Center of Mass (1-D)

xcm = ΣmixiΣmi

Mass-weighted average position; it sits closer to the heavier object.

Motion of the Center of Mass

acm = ΣFextmtotal

Internal forces cancel in pairs, so only external forces change the motion of the center of mass.

Example: A 2 kg mass sits at x = 0 and a 6 kg mass sits at x = 4 m. Find the center of mass.

  1. xcm = (2)(0) + (6)(4)2 + 6 = 248
  2. Answer: xcm = 3 m (closer to the 6 kg mass)

2.2 Forces and Free-Body Diagrams

A force is a push or pull from an interaction between two objects. It is a vector measured in newtons (1 N = 1 kg·m/s2). Contact forces (normal, friction, tension, spring) need touching; field forces (gravity) act at a distance.

ForceSymbol / FormulaDirection
Weight (gravity)Fg = mgStraight down, toward Earth's center
NormalFNPerpendicular to the surface, pushing away from it
TensionFTAlong the rope, pulling away from the object
Frictionf ≤ μFNParallel to the surface, opposing sliding (or the tendency to slide)
SpringFs = kΔxBack toward the spring's natural length

Drawing a Free-Body Diagram

  1. Represent the object (or system) as a dot or box.
  2. Draw each force as an arrow starting on the object.
  3. Label every arrow (Fg, FN, FT, f).
  4. Make arrow lengths roughly match force sizes.
  5. Do not draw net force, velocity, or "centripetal force" arrows.
θFNmgfs

Block at rest on a rough incline

Inclines: tilt your axes so x runs along the slope

Fg,∥ = mg sinθ (down the slope)
Fg,⊥ = mg cosθ (into the slope)
No friction, no push: FN = mg cosθ
Frictionless slide: a = g sinθ
⚠ Watch out:The normal force is not always mg. It equals mg only on a flat surface with no other vertical forces and no vertical acceleration.

2.3 Newton's Third Law

FA on B = −FB on A

Forces come in pairs. The two forces in a pair have equal magnitude, opposite direction, are the same type of force, and act on different objects.

Example: A truck hits a small car head-on. Which feels the larger force?

  1. The forces form a third-law pair, so they are equal in size.
  2. The car has less mass, so by a = F/m it has the larger acceleration.
  3. Answer: equal forces; the car gets the bigger acceleration.
⚠ Watch out:Weight and the normal force on a book on a table are not a third-law pair: they act on the same object. The pair to Earth's pull on the book is the book's pull on Earth.

2.4 Newton's First Law

Law of Inertia

If the net force on an object is zero, its velocity stays constant: at rest stays at rest, moving keeps moving in a straight line at constant speed. Inertia is resistance to a change in velocity, and mass measures it.

Translational Equilibrium

ΣFx = 0   and   ΣFy = 0

Applies to objects at rest and objects at constant velocity.

⚠ Watch out:Motion does not require a force. A hockey puck sliding at constant velocity on frictionless ice has zero net force on it.

2.5 Newton's Second Law

Second Law

asys = ΣFmsys

Acceleration points in the direction of the net force. Apply it separately on each axis: ΣFx = max, ΣFy = may.

Atwood Machine (ideal pulley, m2 > m1)

a = (m2 − m1)gm1 + m2

Treat both masses as one system: the net external force is the difference in weights. Tension is internal.

Apparent Weight in an Elevator (scale reading = FN)

AccelerationNormal forceFeels
a = 0 (rest or constant velocity)FN = mgNormal weight
a upward (magnitude a)FN = m(g + a)Heavier
a downward (magnitude a)FN = m(g − a)Lighter
Free fall (a = g downward)FN = 0Weightless

Example: A 24 N push moves a 5 kg block and a 3 kg block (touching) across a frictionless floor. Find the acceleration and the force the 5 kg block exerts on the 3 kg block.

  1. Whole system: a = 24 N8 kg = 3 m/s2
  2. The only horizontal force on the 3 kg block is the contact force: F = (3 kg)(3 m/s2)
  3. Answer: a = 3 m/s2, contact force = 9 N

Example: A 60 kg person rides an elevator accelerating upward at 2.0 m/s2. What does the scale read?

  1. Take up as positive: FN − mg = ma
  2. FN = m(g + a) = 60(9.8 + 2.0)
  3. Answer: FN ≈ 708 N (more than the person's 588 N weight)
⚠ Watch out:"ma" is not a force. Never draw it on a free-body diagram; it is the result of the forces you drew.

2.6 Gravitational Force

Newton's Law of Universal Gravitation

Fg = Gm1m2r2

G = 6.67 × 10−11 N·m2/kg2; r is center-to-center distance. Always attractive.

Gravitational Field Strength

g = Fgm = GMr2

Near Earth's surface, g ≈ 9.8 N/kg = 9.8 m/s2. Weight is Fg = mg.

Ratio Reasoning

Double one mass → force doubles
Double both masses → force × 4
Double the distance r → force ÷ 4
Triple the distance r → force ÷ 9
Apparent weight:The normal force a scale pushes with. It differs from true weight (mg) whenever the object accelerates vertically.
Inertial = gravitational mass:The mass that resists acceleration is the same mass gravity pulls on, which is why all objects free-fall with the same acceleration.

Example: A satellite orbits at 3 Earth radii from Earth's center. Find g at that distance.

  1. g ∝ 1r2, and r is 3 times larger than at the surface
  2. g = 9.832 = 9.89
  3. Answer: g ≈ 1.1 m/s2
⚠ Watch out:Measure r from the center of the planet, not from its surface. An orbit 300 km up is at r = RE + 300 km.

2.7 Kinetic and Static Friction

Static Friction (not sliding)

fs ≤ μsFN

Adjusts to whatever is needed to prevent sliding, up to a maximum of μsFN.

Kinetic Friction (sliding)

fk = μkFN

Constant while sliding; points opposite the direction of sliding. Usually μk < μs.

Friction on Inclines

Sliding down: a = g(sinθ − μkcosθ)
About to slip: tanθ = μs

Block Sliding Down a Rough Incline (θ = 40°, μk = 0.27)

θ = 40°mg sin θmg cos θW = mgN = mg cos θf = μkN

x down the slope, y ⊥ slope

Perpendicular (y)

ΣFy = 0

N = mg cosθ

Along the slope (x)

ΣFx = ma

mg sinθ − μkN = ma

g sinθ − μkg cosθ = a

Result

a = g(sinθ − μkcosθ)

a = 9.8(0.643 − 0.27 × 0.766)

a ≈ 4.27 m/s2 down the slope

Example: A 10 kg box is pulled across a level floor by a 50 N horizontal rope. μk = 0.30. Find the acceleration.

  1. FN = mg = (10)(9.8) = 98 N
  2. fk = μkFN = (0.30)(98) = 29.4 N
  3. a = 50 − 29.410 = 20.610
  4. Answer: a ≈ 2.1 m/s2
⚠ Watch out:Static friction is not automatically μsFN. A box at rest with no push on it has zero friction. Use the formula only for the maximum, right before slipping.

Applying Newton's Laws: Pulleys, Rope Tension, and Angled Forces

Pulleys

Ideal pulley rules (massless, frictionless pulley and rope)

  • Tension is the same everywhere along one rope.
  • Masses tied by the rope share the same size of acceleration (and speed).
  • A fixed pulley only changes the direction of the force.
  • Fast method: a = net force along the ropetotal mass, then isolate one mass to find T.
m1m2

Atwood Machine (m2 > m1)

a = (m2 − m1)gm1 + m2

T = 2m1m2gm1 + m2

m1m2

Block on Table + Hanging Mass

a = m2gm1 + m2   T = m1m2gm1 + m2

with friction: a = (m2 − μkm1)gm1 + m2

θm1m2

Incline + Hanging Mass (frictionless)

a = (m2 − m1sinθ)gm1 + m2

Positive a: m2 falls and m1 slides up. Negative a: m1 slides down instead.

TTm

Movable Pulley (at rest or constant speed)

2T = mg  →  T = mg2

Two rope segments hold the load, so you pull with half the weight, but you must pull twice as much rope.

Example: A 4.0 kg block on a frictionless table is tied over a pulley to a hanging 2.0 kg mass. Find a and T.

  1. System: only the hanging weight drives it. a = (2.0)(9.8)4.0 + 2.0 ≈ 3.3 m/s2
  2. Isolate the block on the table: the rope is the only horizontal force, so T = m1a = (4.0)(3.27)
  3. Answer: a ≈ 3.3 m/s2, T ≈ 13 N (less than the hanging weight of 19.6 N, so the hanging mass accelerates down)

Example: A 4.0 kg block sits on a frictionless 30° incline, tied over a pulley at the top to a hanging 3.0 kg mass. Find a and T.

  1. Driving force: m2g = 29.4 N. Resisting force: m1g sin30° = 19.6 N
  2. a = 29.4 − 19.64.0 + 3.0 = 9.87.0 = 1.4 m/s2 (hanging mass falls)
  3. Isolate the hanging mass: m2g − T = m2a, so T = (3.0)(9.8 − 1.4)
  4. Answer: a = 1.4 m/s2, T ≈ 25 N

Rope Tension at Angles (hanging object in equilibrium)

θ1θ2T1T2m

Angles measured from the horizontal

Break Each Tension into Components

x: T1cosθ1 = T2cosθ2

y: T1sinθ1 + T2sinθ2 = mg

Two equations, two unknowns: solve for T1 and T2.

Symmetric Ropes (θ1 = θ2 = θ)

T = mg2 sinθ

Flatter ropes (smaller θ) mean much larger tension.

One Rope at θ + One Horizontal Rope

Tangled = mgsinθ   Thorizontal = mgtanθ

The angled rope holds all of the weight; the horizontal rope only balances its sideways pull.

10 kg sign (mg = 98 N) on two symmetric ropes

Rope angle θ90°60°45°30°10°5°
Tension in each rope49 N57 N69 N98 N282 N562 N

Example: A 10 kg sign hangs from two ropes at 30° and 60° above the horizontal. Find each tension.

  1. x: T1cos30° = T2cos60°, so T2 = 0.8660.5T1 = 1.73T1
  2. y: T1(0.5) + (1.73T1)(0.866) = 98, so 2.0T1 = 98
  3. Answer: T1 = 49 N (30° rope), T2 ≈ 85 N (60° rope). The steeper rope carries more of the load.
⚠ Watch out:A perfectly horizontal rope can't hold up any weight. As θ gets close to 0°, the tension needed grows without limit, which is why a clothesline always sags.

Pulling or Pushing at an Angle

θFFNmgfk

Box pulled at θ above horizontal

Split the applied force F into components

Fx = F cosθ   Fy = F sinθ

Fx moves the box forward. Fy changes the normal force, which changes friction. Vertical forces balance (ay = 0) as long as the box stays on the floor.

SituationNormal forceAcceleration
Horizontal pull (θ = 0)FN = mga = F − μkmgm
Pull at θ above horizontalFN = mg − F sinθa = F cosθ − μk(mg − F sinθ)m
Push at θ below horizontalFN = mg + F sinθa = F cosθ − μk(mg + F sinθ)m
Any angle, frictionless(doesn't affect a)a = F cosθm

Example: A 20 kg sled is moved with a 100 N force at 30° (μk = 0.20). Compare pulling it at 30° above horizontal with pushing it at 30° below.

  1. Both: Fx = 100 cos30° = 86.6 N, Fy = 100 sin30° = 50 N, mg = 196 N
  2. Pull: FN = 196 − 50 = 146 N, fk = (0.20)(146) = 29.2 N, a = 86.6 − 29.220 ≈ 2.9 m/s2
  3. Push: FN = 196 + 50 = 246 N, fk = (0.20)(246) = 49.2 N, a = 86.6 − 49.220 ≈ 1.9 m/s2
  4. Answer: pulling gives 2.9 m/s2, pushing only 1.9 m/s2. Pulling up lifts some weight off the floor and reduces friction.
⚠ Watch out:When a force is applied at an angle, FN is not mg. Always write ΣFy = 0 first to find the normal force, then use it for friction.

2.8 Spring Forces

Hooke's Law

Fs = −kΔx

k = spring constant (N/m); Δx = stretch or compression from natural length. The minus sign means it is a restoring force.

Springs in Parallel

keq = k1 + k2

Side by side, sharing the load: stiffer than either spring.

Springs in Series

1keq = 1k1 + 1k2

End to end: softer than either spring.

Example: A 0.50 kg mass hangs at rest from a spring with k = 49 N/m. How far does the spring stretch?

  1. Equilibrium: kΔx = mg
  2. Δx = mgk = (0.50)(9.8)49
  3. Answer: Δx = 0.10 m
⚠ Watch out:Δx is measured from the spring's natural (unstretched) length, not from wherever the object started.

2.9 Circular Motion

Centripetal Acceleration

ac = v2r

Points toward the center of the circle.

Net Force Toward Center

ΣFc = mv2r

Supplied by real forces: tension, gravity, friction, normal.

Speed, Period, Frequency

v = 2πrT   T = 1f

T = time for one revolution (s); f = revolutions per second (Hz).

Changing Speed on a Curve

a = √ac2 + at2

Tangential acceleration at changes speed; ac changes direction.

rvac

v is tangent; ac points inward

SituationWhat provides ΣFcKey result
Car on a flat curvefs = mv2rvmax = √μsgr
Top of a vertical loopFN + mg = mv2rvmin = √gr (when FN = 0)
Bottom of a vertical loopFN − mg = mv2rFN > mg (feels heavier)
Circular orbitGMmr2 = mv2rv = √GMr, T2 = 4π2GMr3

Example: A 1200 kg car rounds a flat curve of radius 50 m at 15 m/s. Find the friction force needed and the minimum μs.

  1. fs = mv2r = (1200)(15)250 = 5400 N
  2. μs = v2gr = 225(9.8)(50)
  3. Answer: fs = 5400 N, μs ≈ 0.46
⚠ Watch out:"Centripetal force" is not a new force. Never draw it on a free-body diagram; it is the net of the real forces pointing toward the center. If the string breaks, the object flies off along the tangent, not outward.