Physics I formula sheet

A = size of the vector. Flat (2-D) problem: keep A_z = 0. Add or subtract vectors right in the boxes, part by part: for C = A − B type 10.4−6.25 in the Aₓ box (same for y). 2A + B: type 2×10.4+6.25. Vectors given as size + angle: type the parts, e.g. Aₓ box: 175cos(60)+80cos(90), Aᵧ box: 175sin(60)+80sin(90). Boat and river, plane and wind, rain seen from a car: add the velocity parts the same way.

θ = direction, counterclockwise from the +x axis (east). If Aₓ < 0: add 180°. If Aₓ > 0 and Aᵧ < 0: add 360°. Aₓ = 0: θ = 90° (Aᵧ > 0) or 270° (Aᵧ < 0). Angle from vertical or like 59° W of N: vertical (N) part in Aₓ, other part in Aᵧ, both positive. Right triangle: θ is the corner next to side Aₓ. Empty Aᵧ gives the far side (river width). Sagging rope: Aᵧ = sag, Aₓ = half the span. Radians: answer × π/180.

θ is measured from the +x axis. Angle given from the y axis: type 90−angle. Angle in radians: type it ×180/π. Ramp: A = m×9.8, θ = ramp angle gives Aₓ = normal force N. Cross a river straight: A = boat speed, Aₓ = river speed gives θ = aim angle upstream from the bank. 3-D, angle from +z given: type it as θ to get A_z. For Aₓ and Aᵧ type A×sin(z angle) in the A box. Empty θ gives 0° to 180°; if Aᵧ < 0 use 360° − θ.

θ is measured from the +x axis. Ramp: A = m×9.8, θ = ramp angle gives the weight part down the ramp. Sagging rope: first find θ = tan⁻¹(sag/half span). Then A = tension in one half, Aᵧ = m×9.8/2. Two cables at 90° to each other holding a weight: A = m×9.8, θ = one cable's angle above level, Aᵧ = that cable's tension. Swim straight across a river: A = swim speed, Aᵧ = river speed, θ = aim angle from straight across.

φ = angle between A and B (put their tails together). From parts: type AₓBₓ+AᵧBᵧ+A_zB_z in the A·B box, e.g. 2×(−3)+5×0+0×4. Sizes from parts: type A as √(2²+5²+0²). A·B = 0: the vectors are at 90°. A·B < 0: φ is over 90°.

φ = angle between A and B (tails together). Empty φ gives two angles that both work. Direction: right-hand rule. Fingers along A, curl toward B, thumb = A×B. B×A points the opposite way. x×y = z, y×z = x, z×x = y. Parts by hand: x = AᵧB_z − A_zBᵧ, y = A_zBₓ − AₓB_z, z = AₓBᵧ − AᵧBₓ. Parallel vectors: A×B = 0.

Type kg and m³ (answer in kg/m³). For g/cm³ divide the answer by 1000. 1 g = 0.001 kg. 1 L = 0.001 m³. 1 cm³ = 1E−6 m³. Volume: cube the factor. 1 ft³ = 0.3048^3 m³. Sphere: V = 4/3×π×(d/2)^3. Box: V = length×width×height. Unit changes: 1 in = 2.54 cm, 1 ft = 0.3048 m, 1 mi = 1609 m, 1 h = 3600 s, 1 yr = 365 days. Prefixes: k = 1E3, M = 1E6, c = 0.01, m = 0.001, μ = 1E−6, n = 1E−9.

Use when time is in the problem (steady a). Up/right is +. Free fall: a = −9.8. Stops, or top of a throw: v = 0. Lands at launch height: v = −v₀ gives the flight time. Projectile: do x and y apart. y: a = −9.8, v₀ = speed×sin(θ), like 22×sin(40). x: a = 0. Straight-line a-t graph: a = (a₁+a₂)/2, t = t₂−t₁. v(t) = 3 + 4t: v₀ = 3, a = 4. x(t) = A + Bt + Ct²: v₀ = B, a = 2C. Sliding to a stop on friction: a = −μ×9.8.

Average velocity, or steady speed (a = 0). Can be −. Δx = end − start position. t = end − start time. Average SPEED: Δx = total distance. Steady a (speed-up or braking): vₐ = (v₀+v)/2. Legs: Δx = sum of v×t (west −), t = total. Projectile sideways: vₐ = speed×cos(θ). River crossing: Δx = width, vₐ = speed across. Circle speed: Δx = 2πr, t = 1/f (f = rpm/60). mph with hours is OK. km/h to m/s: type 90/3.6.

Use when Δx and t are in the problem. Empty t: earliest first. Up/right is +. Free fall: a = −9.8. Up/down: Δx is Δy. Dropped, or thrown level: y part v₀ = 0, Δy = −height. Sideways: a = 0, v₀ = speed (type it +). At an angle: y part v₀ = speed×sin(θ), like 22×sin(40). Lands at launch height (Δy = 0): flight time is the "or" answer. Catch-up: Δx = gap, v₀ = v₁−v₂, a = a₁−a₂ (1 = chaser). km/h to m/s: type 150/3.6.

Use when there is no time in the problem. Up/right is +. Free fall: a = −9.8. Top of a throw: v = 0. Leave Δx empty to get the max height. Projectile max height: y part v₀ = speed×sin(θ), like 22×sin(40). v or v₀ comes out + or −: + first. Moving down or left: use the − one. From rest: v₀ = 0. Stops: v = 0. Friction: a = −μ×9.8.

Only when it lands at the height it was launched from. R = sideways distance. θ = angle above level. Empty θ: 2 angles work (they add to 90°). Longest range: θ = 45°. Max height: v² formula, v₀ = speed×sin(θ), v = 0, a = −9.8. Time in air: Δx formula, Δx = 0, a = −9.8, v₀ = speed×sin(θ); take the "or" answer. Not level: do x and y apart (x: a = 0, y: a = −9.8). Speed + angle from x and y parts: Measurement & vectors.

Use when x = A + Bt + Ct² + Dt³. B, C, D = the numbers in front of t, t², t³. Missing term: 0. Keep minus signs. Position at time t: type x(t) on the home screen with your t, like 5−12(2)+2(2)³. Given v(t)? Put its t, t², t³ numbers in B, C, D (skip its plain number): answer is a (m/s²). Leave t empty to find when v hits a value (v = 0: it stops or turns). 2-D: do x and y one at a time, then speed = √(vₓ² + vᵧ²).

Use when x = A + Bt + Ct² + Dt³. C, D = the numbers in front of t², t³. Missing term: 0. Keep minus signs. No t³ term: a = 2C at every time. Given v(t)? Use v = B + 2Ct + 3Dt² with B, C, D = v(t)'s t, t², t³ numbers: answer is a. 2-D: do x and y one at a time, then size = √(aₓ² + aᵧ²).

Use when a = A + Bt + Ct². A = plain number, B and C = numbers in front of t and t². Missing term: 0. v₀ = velocity at t = 0 (from rest: 0). Know v at another time? Leave v₀ empty first. Given v(t), want x? A, B, C = v(t)'s plain, t, t² numbers; x₀ (or 0) in v₀: answer is x (m). Δx from t₁ to t₂: do t₂, then t₁, subtract. Find t (a(t) has no t²): in v = B + 2Ct + 3Dt² type B = v₀, C = A/2, D = (a(t)'s t number)/6.

Fnet = all forces along the motion, with signs (type 310+420−610 or 26cos(22)). At rest or steady speed: a = 0. Scale, rope or floor holding m: type 9.8+a in the a box (a − if down). Tied blocks: Fnet = weights pulling, with signs (Atwood 5*9.8−3*9.8), m = total (5+3). Then use Fnet = ma on one block with every force on it, e.g. Atwood T − m₁g = m₁a so T = m₁(g + a). 3rd law: same force on both.

Level floor, nothing else pushing up or down: normal force N = mg. Hanging at rest: rope tension = mg (one rope holding two: m box 92+58). Pulley, 2 rope pieces hold the load: each pulls half (m box 42/2). Ramp: N = mgcosθ, type 7cos(38) in the m box. Pushed down with F: N = mg + F. Pulled up: N = mg − F. Moon: g = 1.62. Mass is the same everywhere.

μₛ (static): most it can hold = push that just starts it moving. μₖ (kinetic): while it slides. Not moving, not about to slip: friction = the push. N on level floor = mg: type 25*9.8 in the N box. On a ramp: type 7*9.8cos(38) in the N box. Rope pulling up at θ: N = mg − Fsinθ, type 20*9.8−90sin(25).

a = acceleration DOWN the ramp. μ = μₖ while it slides. Frictionless: μ = 0. Level floor: θ = 0, so a = −μg (slowing down). Angle where it just starts to slide: a = 0, μ = μₛ. Sliding UP the ramp: type μ as −μₖ (a is still + down the ramp). Ramp normal force: W = mg with 7cos(38) in the m box.

a points to the center. Divide by 9.8 for g's. Force toward the center (string, tube): Fnet = ma with this a (a box: 16²/15). Speed from lap time T: type 2π*r/T in the v box; PLUG IN shows v (3 rev/s: T = 1/3; rpm: T = 60/rpm). Know v and T: type v*T/(2π) in the r box. Flat curve, top speed before skidding: a = μₛ*9.8. Banked curve, no friction: a = 9.8tan(θ). Top of a hill, car leaving the road: a = 9.8.

Terminal speed: air drag = weight, so it stops speeding up. ρ = 1.2 for air (change it if the problem gives one). C = drag coefficient (no unit). A = area facing the air (m²). 4 times the mass: 2 times the speed.

θ = angle between the force and the way it moves. Along the motion: θ = 0. Against it: θ = 180. Sideways (normal force): W = 0. Friction: F = μₖ×m×g, θ = 180. On a ramp: F = μₖ×m×g×cos(ramp angle). Gravity: F = m×g, d = height. Up θ = 180, down θ = 0. Stopping distance: W = −½mv₀² (e.g. −0.5×2×5²), F = friction, θ = 180. F vs x graph: W = area under it. F = c×x^n from x₁ to x₂: W = c×(x₂^(n+1) − x₁^(n+1))/(n+1).

Kinetic energy = energy of motion. v is the speed (always +, the direction does not matter). 2 times the speed = 4 times the K. Momentum p given instead of v: type v = p/m (e.g. 12/3).

Net work = change in kinetic energy (work-energy theorem). Starts from rest: v₀ = 0. Comes to a stop: v = 0. Wₙₑₜ = work of ALL forces added up (friction work is −). Only friction on flat ground: Wₙₑₜ = −μₖ×m×g×d (e.g. −0.3×2×9.8×5). Stopping distance: use W = Fd cosθ with W = −½mv₀².

h = height above the zero level you pick (often the ground). Below it, h is −. Change in U: type the height change as h (e.g. 12−3). Work done BY gravity = −ΔU. Work to lift something at steady speed = mgh.

Energy at start + Wnc = energy at end. No friction: leave Wnc = 0. No mass given: leave m = 1. Ramp of length d: h = d×sin(θ) (e.g. 4×sin(30)). Friction: Wnc = −μₖ×m×g×d. On a ramp: −μₖ×m×g×cos(θ)×d. Spring pushes it off: Wnc = ½kx² (e.g. 0.5×400×0.1²). It squeezes a spring: −½kx². Pendulum: h₀ = L−L×cos(θ). Zero height = lowest point. Energy lost = −Wnc. To find m or Wnc, press DEL on its box first.

Pendulum or swing: how high the bob is above its lowest point. θ = angle from straight down. Then use the energy formula with this h. "To what angle does it swing?": find h with the energy formula, then θ here.

Hooke's law: force to hold a spring stretched or squeezed by x. Find k here first, then use U = ½kx². Mass hanging at rest: F = m×g (e.g. 2×9.8). The spring pulls back the other way (F = −kx); type sizes here. cm to m: 5 cm = 0.05.

Energy stored in a stretched or squeezed spring. x = change from its natural length, in m (5 cm = 0.05). Work from x₁ to x₂: type x = √(x₂²−x₁²), e.g. √(0.1²−0.05²). Spring launches a mass: energy formula with Wnc = ½kx². How far it gets squeezed (flat floor): get Wnc from the energy formula (v = 0), type it here as U without the −. Dropped from h onto it: x = (m×g + √((m×g)² + 2×k×m×g×h))/k.

Power = how fast work is done (1 W = 1 J per second). W can be any energy change: lifting mgh (e.g. 70×9.8×5), speeding up ½mv². Energy used in a time: leave W empty. 1 hp = 746 W. 1 kW = 1000 W.

Power while moving at speed v. Lifting at steady speed: F = m×g (e.g. 500×9.8). Steady speed against friction or drag: F = that force. Up a slope at steady speed: F = m×g×sin(θ) + friction (e.g. 1200×9.8×sin(5)+500). Force at an angle: type F×cos(θ) as F.

Momentum = mass × velocity. Sign = direction: pick one way as +, the other way is −. Total momentum: add up every object's mv (with signs). 2-D crash at right angles, stuck together: type p as √(pₓ²+pᵧ²), e.g. √((1200*20)²+(1500*15)²), and m as m₁+m₂. Then v = speed after. Direction after: θ = tan⁻¹(pᵧ/pₓ); add 180° if pₓ < 0.

Any collision or explosion: total momentum before = after. Pick a + direction; moving the other way is negative. At rest = 0. Explosion or recoil from rest (gun, skaters push apart): v₁ = 0 and v₂ = 0. Moving together at first: type that same v in v₁ and v₂. 2-D: use it once with x parts (e.g. 5*cos(30)), once with y parts (5*sin(30)). Speed = √(vₓ²+vᵧ²), angle = tan⁻¹(vᵧ/vₓ) (+180° if vₓ<0).

Objects stick together and move as one (perfectly inelastic). Target at rest: v₂ = 0. Ballistic pendulum (block swings up h): type v' = √(2*9.8*h), solve for v₁. Know the swing angle instead? h = L(1 − cos θ). Target at rest: KE lost = ½m₁v₁² × m₂/(m₁+m₂). Other KE: K = ½mv² (Work & energy) before and after. 2-D: see the tip on p = mv.

Elastic = they bounce apart and no KE is lost (1-D). Target at rest: v₂ = 0. For v₂': swap the objects. Type object 2's mass and velocity in the 1 boxes, object 1's in the 2 boxes. The answer is v₂'. Equal masses: they just trade velocities. Check: v₁ − v₂ = v₂' − v₁'.

Average force that changes velocity in time Δt (bat, airbag, catching). Bounces back: v gets the opposite sign of v₀. Stops: v = 0. Angled hit: use only the part into the wall, e.g. v₀ = 20*cos(30), v = −20*cos(30). Dropped from h, up is +: v₀ = −√(2*9.8*h). Bounces up to h₂: v = √(2*9.8*h₂). F is the net force. Floor force on someone landing = F + mg. Hose or thrust: m = kg per second, Δt = 1.000 (not 1).

Impulse = change in momentum (J = Δp). Signs matter! Ball hits a wall at 20 m/s and bounces back at 15 m/s: v₀ = 20, v = −15. Stops: v = 0. Starts at rest: v₀ = 0. Angled hit: use only the part into the wall, e.g. v₀ = 20*cos(30), v = −20*cos(30). The other object gets −J (same size, opposite way).

Impulse = average force × time it pushes. Force vs time graph: J = area under it. Triangle spike: type F as half the peak (e.g. 1500/2). J points the same way as the force. Know the mass and speeds instead? Use J = m(v − v₀) (one up in the list).

2 objects: leave m₃ and x₃ at 0. 3 objects: type over them. Put x = 0 anywhere you like (often on one object). 2-D: use it once for x, once for y. Velocity of center of mass: type velocities in the x boxes; answer = vcm (units m/s, not m). Uniform rod, ball or plate: its mass sits at its middle. Person walks L on a boat: m₁ = person, x₁ = L, m₂ = boat, x₂ = 0. Answer xcm = how far the boat slides back.

Speed a rocket gains by burning fuel. In space: leave Δt = 0. Straight-up launch near Earth: Δt = burn time. mᵢ = full mass at the start. m = mass after the burn (mᵢ − fuel used). u = exhaust speed (how fast the gas leaves the rocket). Thrust force = u × kg of fuel burned per second. Acceleration = thrust / mass − 9.8 (no − 9.8 in space).

From rest: ω₀ = 0. Comes to a stop: ω = 0. Slowing down: α is negative. Given rpm? Type rpm*2π/60 to get rad/s (300 rpm: 300*2π/60). Given rev/s (Hz)? Type f*2π. One turn every T s? Type 2π/T. No time given? Use ω² = ω₀² + 2αΔθ. Rope or rim accel. a given? Type a/r for α.

Use when you know the time and α. From rest: ω₀ = 0. Slowing down: α is negative. Steady spin (no speed-up): α = 0, so Δθ = ω₀t. Given revolutions? Type rev*2π (12 rev: 12*2π). Given degrees? Type deg*π/180. Want revolutions? Divide the Δθ answer by 2π. Given rpm? Type rpm*2π/60. Rope unwound or distance s on the rim: Δθ = s/r. Two answers for t? Use the smaller positive one.

Use when no time is given. From rest: ω₀ = 0. Stops: ω = 0 (α comes out negative). Revolutions: type rev*2π for Δθ (12 rev: 12*2π). Degrees: type deg*π/180. Given rpm? Type rpm*2π/60. Two answers for ω (+ and −)? Same speed, two spin directions: use + unless it turned back.

Use when α is not given and not asked. From rest: ω₀ = 0. Stops: ω = 0. Want revolutions? Divide the Δθ answer by 2π (6.283). Given revolutions: type rev*2π. Given rpm: type rpm*2π/60. (ω₀ + ω)/2 is the average ω.

Speed of a point on the rim, or of a belt, rope or rolling wheel. ω in rad/s: rpm → type rpm*2π/60 (900 rpm: 900*2π/60). One turn every T s → type 2π/T. Same formula, other uses: s = rθ: put θ (rad) in the ω box, answer is s aₜ = rα: put α in the ω box, answer is aₜ ac = rω²: put ω² in the ω box (or ac = vₜ²/r) Total accel.: a = √(aₜ² + ac²)

r = distance from the pivot (axis) to where F pushes. θ = angle between r and F. Push at a right angle (like a door handle): θ = 90. Push straight toward the pivot: θ = 0, no torque. Type τ as a positive size here. Add the sign after: counterclockwise +, clockwise −. Several forces: add the torques (net torque Στ).

Net torque: add the torques, counterclockwise +, clockwise −. Force on the rim at a right angle: τ = FR, type F*R (20 N on 0.3 m: 20*0.3). Type I right in, e.g. solid disk ½MR²: 1/2*4*0.3² Rope on a wheel: α = a/R. Rod on an end pivot, held level: weight acts at the center, type Στ = M*9.8*L/2 and I = 1/3*M*L². Then use ω = ω₀ + αt etc. for speed and turns. Block hanging on the wheel? Use a = mgR²/(I + mR²).

c by shape (type a fraction so sig figs stay right): Point mass, hoop, ring, thin pipe: 1/1 Solid disk, cylinder, pulley: 1/2 Solid sphere (ball): 2/5 Hollow sphere: 2/3 Rod about its center: 1/12 (L in the R box) Rod about one end: 1/3 (L in the R box) Axis a distance d from the center: add Md². Many parts: add them, I = I₁ + I₂ + … Point masses: I = m₁r₁² + m₂r₂² + … Type a sum right in any I box: 2*0.5²+3*0.2²

A block hangs from a rope on a pulley or wheel and falls. Disk pulley: type I as 1/2*M*R² (2 kg, 0.1 m: 1/2*2*0.1²). Then: tension T = m(g − a), α = a/R. Fall h from rest: v = √(2ah), t = √(2h/a) Also pulls a block on a smooth table? Add its m₂R² to I. Two hanging blocks (m₁ heavier): type m₁−m₂ for m and I+2*m₂*R² for I. Tensions: T₁ = m₁(g − a), T₂ = m₂(g + a)

Spin energy. ω in rad/s. No friction: energy at start = energy at end. Only spinning, and its center drops h: K = Mgh, type M*g*h (2 kg, 0.5 m: 2*9.8*0.5). Rod swinging down from level on an end pivot: h = L/2. Work by a torque: W = τΔθ, and W = K − K₀ (from rest: K = W). Rolling (moving + spinning)? Use the Rolling topic.

Power of a motor or engine. ω in rad/s. Given rpm? Type rpm*2π/60 for ω (3000 rpm: 3000*2π/60). Work by a steady torque: W = τΔθ (Δθ in rad). Power is also P = W/t (work / time). 1 hp = 746 W.

Rolling without slipping: center speed v = ωR. The point touching the ground is not moving; the top moves at 2v. Same idea for a = αR and distance rolled = Rθ: use aₜ = rα and s = rθ in Rotation. rpm given: type ω as rpm*2π/60, e.g. 300*2π/60. Revolutions: θ = 2πN (Rotation).

Speed after rolling down h, no slipping (v₀ = 0 from rest). c = shape, typed as a fraction: hoop/pipe 1/1, solid disk/cylinder 1/2, solid sphere 2/5, hollow sphere 2/3. Other I: c = I/(mR²). Rolling up: h negative (−2 = up 2 m). Ramp length d: h = d*sin(θ); d unknown: a = g sinθ/(1 + c), then Motion topic. Slides, no friction: c = 0. Block on pulley: c = (I/R² + table block m)/hanging m (no table block: I/(mR²)).

Acceleration rolling down a ramp, no slipping. θ = ramp angle. c = shape, typed as a fraction: hoop/pipe 1/1, solid disk/cylinder 1/2, solid sphere 2/5, hollow sphere 2/3. Other I: type I/(mR²) for c. Yo-yo falling: θ = 90, c = I/(mr²), r = axle radius. Block on a pulley: θ = 90, c = (I/R² + table block m)/hanging m. No table block: c = I/(mR²). Time or distance: use this a in Motion & projectiles (rolling up: −a).

Total KE of a rolling object (moving + spinning). c = shape, typed as a fraction: hoop/pipe 1/1, solid disk/cylinder 1/2, solid sphere 2/5, hollow sphere 2/3. Other I: type I/(mR²) for c, e.g. 0.8/(5*0.3²). Moving part only (½mv²): c = 0. Spinning part only: K = ½Iω² in Rotation. Fraction that is spinning = c/(1 + c), e.g. solid sphere 0.4/1.4 = 0.29.

Smallest static friction coefficient that lets it roll down the ramp without slipping. Less than this: it slides. c = shape, typed as a fraction: hoop/pipe 1/1, solid disk/cylinder 1/2, solid sphere 2/5, hollow sphere 2/3. Steepest ramp it can roll on: leave θ empty. Friction force on it: f = μN (Forces) with the μₛ found here and N = m*9.8*cos(θ).

Angular momentum of a spinning object. Type I right in the box: disk 0.5*m*R², hoop or point m*R², sphere 0.4*m*R², rod about its middle m*length²/12, e.g. 0.5*2*0.3². rpm given: type ω as rpm*2π/60, e.g. 300*2π/60. Period T given: type ω as 2π/T (Earth spin: 2π/86400). Counterclockwise +, clockwise −. Rolling: ω = v/R. Final ω after parts join: L = total starting L (person: m*v*r), I = final total I.

No outside torque, so L stays the same: skater pulls arms in, person jumps on or walks across a turning disk, two disks stick together. Add the parts: I₂ = 0.5*100*2²+40*2² (disk + person at the edge; person = mr²). Turning opposite ways: one ω is negative. rpm in gives rpm out, no change needed. Parts at different ω at the start, or a person running on: use L = Iω. KE lost: K = ½Iω² (Rotation).

Angular momentum of a small object about a point. φ = angle between r (point to object) and v. Moving in a circle, or r = closest distance to its straight path: φ = 90. Moving straight at the point: L = 0. Person running onto a merry-go-round edge: L = mvr (φ = 90). r and v as x,y parts: L = m(x·vᵧ − y·vₓ), + = counterclockwise.

Net torque = change in angular momentum / time. ΔL = L after − L before. One object: type I*(ω after − ω before), e.g. 4*(0−30). Slowing down: ΔL and Στ come out negative. Know α instead of time? Use Στ = Iα in Rotation.

Gyroscope or spinning top: how fast its axis swings around in a circle (rad/s). r = pivot to center of mass, M = mass. I and ω = its spin about its own axis, e.g. disk I = 0.5*M*R². rpm given: type ω as rpm*2π/60. rev/s given: rev/s*2π, e.g. 40*2π. Time for one swing around: T = 2π/ωₚ. T given: type ωₚ as 2π/T.

Torque one way = torque other way. Pivot at hinge or unknown force. F₁ turns it one way; F₂, F₃ the other. Weight: m*9.8 at beam middle. Unused: F₃ = r₃ = 0. More: add F×r in one box (30*1.5+20*2), r = 1/1. Slant: r = dist×sin(force-to-beam angle). Ladder 53° to floor: 2.5sin(37). − answer: force points the other way. About to tip: other support = 0. Balance point: F₁ = total mass, r₁ empty, F₂ = m₁x₁+m₂x₂, r₂ = 1/1.

Up forces = down forces. Sideways: right = left (right in F₁, F₂). 2+ forces on a side: add them in one box (200+350). Unused box: 0. Weight: m*9.8. Hinge (cable T at θ), leave F₁ empty: Up: F₂ = Tsin(θ), F₃, F₄ = weights. Sideways: F₃ = Tcos(θ), F₂ = F₄ = 0. − answer: it points the other way. Hinge size and angle: vectors topic. Ladder: friction = wall force; then μ from f = μN (Forces topic).

Stretch or squeeze along the length (wire, rod, bone, column). Round: A = π(d/2)², e.g. π(0.004/2)². Hanging mass: F = m*9.8. mm: ×0.001. cm²: ×1E−4. GPa: ×1E9. Steel Y = 2.0E11, aluminum 7.0E10, copper 1.1E11, bone 1.6E10 (squeezed 9E9). Squeezed: ΔL = how much shorter (no minus). Shear (top slides sideways): S in Y, slide Δx in ΔL, height in L₀, A = face pushed. Steel S = 7.5E10, aluminum 2.5E10.

Stress = force per area (Pa = N/m²). Round wire, rod or bone: A = π(d/2)². Hanging mass: F = m*9.8. Breaking: P = breaking stress (ultimate strength). Leave F empty: biggest force. Leave A empty: smallest area. Bulk stress is just the pressure.

Strain = fraction it stretched (no unit). ×100 for %. ΔL = how much longer or shorter. Same units for ΔL and L₀. Shear strain: put Δx in ΔL (height in L₀). Know stress and modulus instead: use stress = modulus × strain.

Works for all three: Y (stretch/squeeze), S (shear), B (squeezed on all sides). Know 2 of stress, modulus, strain: get the 3rd. Type stress as F/A and strain as ΔL/L₀ if you like, e.g. 500/(π(0.002)²). Biggest strain before breaking: P = breaking stress.

Squeezed on all sides (pressure, deep water). Δp = pressure increase. ΔV comes out − (it shrinks); type ΔV with − if given. Depth h: Δp = 1000*9.8*h (sea water 1025). 1 atm = 1.013E5 Pa. Fraction ΔV/V₀: type V₀ = 1/1 (×100 for %). Water B = 2.2E9, steel 1.6E11. Compressibility = 1/B.

Pull between any two masses. r = center to center. Weight at a height h: m₁ = planet mass (Earth 5.97E24), m₂ = object, r = 6.37E6+h. Touching balls: r = radius 1 + radius 2. Two pulls on one mass: find each F, then add (same way) or subtract (opposite ways). 2× the distance → 1/4 the force. Moon 7.35E22 kg, Sun 1.99E30 kg. Earth to Moon 3.84E8 m, Earth to Sun 1.50E11 m.

Free-fall acceleration at distance r from a planet's center. On the surface: r = planet radius (Earth 6.37E6). At a height h: r = 6.37E6+h. Find the height: solve r, then height = r − 6.37E6. Other planet: clear M, type its mass. Find a planet's mass: clear M (CLEAR), type g and r, solve M. Where 2 pulls cancel (d apart): x = d/(1+√(m₂/m₁)) from m₁. Earth-Moon: 3.46E8 m. Weight = m×g.

Speed of a satellite or moon in a circle orbit. r = planet center to satellite: 6.37E6+height. The satellite's mass does not matter. Around the Sun: M = 1.99E30. Around the Moon: M = 7.35E22. Find a planet's or star's mass from a moon's speed and r: clear M (CLEAR), solve M. Orbit KE = ½mv².

Time for one full orbit (Kepler's 3rd law). T in seconds: hours×3600, days×86400, years×3.156E7. Stays over one spot: T = 24×3600. Height = r − 6.37E6. Oval orbit: r = (closest + farthest)/2. Farthest = 2r − closest. Sun: M = 1.99E30. Find a star's or planet's mass from its moon: clear M (CLEAR), solve M. Two planets, same star, no mass given: clear M, solve M from planet 1, use it for planet 2.

Slowest launch speed that never falls back (no air, no engine). From the surface: R = planet radius (Earth's is filled in). From an orbit or height: R = 6.37E6+height. The object's mass does not matter. Other body: Moon 7.35E22 kg, 1.74E6 m. Sun 1.99E30 kg, 6.96E8 m. Black hole size: type its M, vₑ = 3E8 (light), clear R (CLEAR), solve R.

Gravity potential energy. Always negative; 0 when very far away. r = from the planet center: 6.37E6+height. Work to lift from r₁ to r₂: U₂ − U₁ (solve twice, subtract). Use this, not mgh, when the height is big (like 100 km+). Total energy of an orbit: use E = −GMm / 2r. Launch or falling speeds: use the ½v₁² − GM/r₁ formula.

Total energy (KE + U) of a circle orbit. Always negative. Orbit KE = the same number made positive. Orbit U = 2×E. Energy to move up to a higher orbit: E₂ − E₁ (solve twice, subtract). Energy to put it in orbit from the ground: E − U, with U from U = −GMm / r using r = 6.37E6.

Speed at one distance → speed at another (no air, no engine). Mass cancels. Every r is from the planet center: 6.37E6+height. Shot up, how high: v₂ = 0, solve r₂, then height = r₂ − 6.37E6. Falls from rest: v₁ = 0. Very far away: type 10^30 for that r. Escape speed: use vₑ = √(2GM / R). Comet's closest and farthest points: use v₁r₁ = v₂r₂.

Kepler's 2nd law: closer = faster. Only at an oval orbit's closest point and farthest point. r = distance from the Sun's (or planet's) center. No masses needed. Units cancel: both r in AU, or both speeds in km/s, works too.

Water 1000 kg/m³, seawater 1025, ice 917, air 1.29, mercury 13600. g/cm³ × 1000 = kg/m³. 1 L = 0.001 m³. 1 cm³ = 1E−6 m³. Volume: box L*W*H, ball 4/3*π*r^3 (e.g. 4/3*π*0.1^3), cylinder π*r²*h. Know the weight W? Type W/9.8 for m. Specific gravity = ρ / 1000.

Hydraulic lift: the pressure is the same on both pistons. Type F₁/A₁ in p, e.g. 300/(π*0.01²), the other piston's area in A, solve F. Round area: π*r² (diameter d: π*d²/4). Window, lid or roof: type the pressure difference in p. Hatch at depth h (air inside): type ρgh in p, e.g. 1025*9.8*30. Dam or tank wall: type ρgh/2 in p (average), wall area in A. 1 atm = 1.013E5 Pa = 760 mmHg = 14.7 psi. Weight: F = m*9.8.

p₀ = pressure on top of the fluid (open to air: 1.013E5, filled in). Gauge pressure (tire, blood, "above air"): p₀ = 0. Two points in one fluid: p₀ = pressure at the higher point, p = at the lower, h = how much deeper. Seawater 1025, mercury 13600. mmHg × 133.3 = Pa. Barometer (vacuum on top): p₀ = 0, p = air pressure. U-tube, two liquids: type ρgh of one side in p (e.g. 1000*9.8*0.05), p₀ = 0, the other ρ, solve h.

ρ = density of the FLUID (water 1000, seawater 1025, air 1.29), not the object. V = volume under the fluid (the whole object if it is fully under). Floating: Fb = its weight. Type m*9.8 in Fb, solve V (the part under). Weighed in air (W) and in water (Wₐ): type W−Wₐ in Fb (e.g. 25−21), solve V. Fb does not change with depth.

Wₐ = string tension or scale reading while it is under the fluid. ρ = fluid density, V = object volume. Know its density? Type m/ρobj in V, e.g. 2/2700. Negative answer = it floats up; the string pulls down that hard. Floating on its own: Wₐ = 0, V = part under. Weighed in air (W) and in water (Wₐ): type W/9.8 for m, solve V, then ρ = m/V. Raft/balloon max load: Wₐ = 0, solve m, minus its own mass (+gas). Air 1.29.

Only for a FLOATING object. f = fraction under the fluid (0.8 = 80%). Part above = 1 − f. Floats with 2/3 under: type 2/3 for f. ρ = fluid density (water 1000, seawater 1025). Ice: ρobj = 917. Block: depth under = f × its height.

Pipe gets narrower or wider (A₁v₁ = A₂v₂): Q is the same everywhere. Type A₁v₁ in Q, e.g. π*0.02²*1.5, the new A, solve v. Round pipe: A = π*r² (diameter d: π*d²/4). Several holes (shower head): A = n*π*r². Volume per time given: type V/t in Q, e.g. 0.1/72. 1 L = 0.001 m³. 1 min = 60 s.

Time to fill a tank, or volume that flows in a time. Know the area and speed? Type A*v in Q, e.g. π*0.01²*3. 1 L = 0.001 m³. 1 cm³ = 1E−6 m³. 1 min = 60 s. L/min to m³/s: × 0.001 / 60.

1 and 2 = two spots in the same moving fluid. y = height (y = 0 at the lower spot). Open to air at both: type 0 for p₁ and p₂. Big tank surface or still fluid: v = 0. Hole in a tank: p₁ = p₂ = 0, v₁ = 0, y₁ = water height above hole, y₂ = 0, solve v₂ (= √(2gh)). Level pipe: y₁ = y₂ = 0. Know Q? Type Q/A in each v box, e.g. 0.002/(π*0.01²).

Mass bouncing on a spring (sideways or hanging). Know f? Type 1/f as T. Know ω? Type 2π/ω as T. Need ω? For a spring ω = √(k/m): type it right in any ω box, e.g. √(200/0.5). Only told how far a hanging mass stretched it? Get k from mg = kΔy first. Mass added to change T? Solve m for each T and subtract. Twisting (torsion) pendulum: type I as m and the torsion constant as k.

Ball on a string, small swings. The mass does not matter. Find g (other planet): press CLEAR on the g box (9.8) and leave it empty. Swinging rod, pivot at one end: type 2/3 of the rod length as L. Any swinging object: type I/(m*d) as L (d = pivot to center of mass). Hanging spring that stretched Δy: type Δy as L. Pendulum ω = √(g/L).

Period T = seconds for 1 full back-and-forth. Frequency f = back-and-forths per second (Hz). Counted N cycles in t seconds? Type t/N as T, e.g. 20/10. End to middle takes T/4; end to end takes T/2. Type 4t or 2t as T. Know ω? Type ω/(2π) as f to get T.

ω (rad/s) is what goes in x = A cos(ωt + φ) and the speed formulas. ω is not f. Know T? Type 1/T as f, e.g. 1/0.5. Spring: ω = √(k/m). Pendulum: ω = √(g/L). Type these right in the ω box, e.g. √(200/0.5), to get f.

Mass hanging still on a spring: Δy = how far it stretched from its natural length. Use it to find k, then the period. Pulled down more and let go? That extra pull is the amplitude A, not Δy. Spring force at a stretch x: F = kx (Work & energy). Max force = kA. No mass given? Period = 2π√(Δy/g): use T = 2π√(L/g) with Δy as L.

Position at time t. Angles here are RADIANS (π = 180°). φ: pulled to +A and let go → 0. At the middle moving + → −π/2. At −A → π. Given x = A sin(ωt)? Use φ = −π/2. Given x = 0.25 cos(0.4πt)? A = 0.25, ω = 0.4π, φ = 0. t answer = first time it is at x; the 2nd answer is the next time. Starts at x₀ moving at v₀? Get A from v = ω√(A² − x²), then t = 0, leave φ empty: pick the φ with the opposite sign to v₀.

Speed at a spot x (x measured from the middle). Max speed (at the middle): type 0 as x. That is vₘₐₓ = Aω. Amplitude from a speed at a spot: leave A empty. ω: type √(k/m) for a spring, 2π/T, or 2πf. Speed is 0 at the ends (x = A).

Acceleration at a spot x. It always points back to the middle (opposite sign to x). Max acceleration aₘₐₓ = Aω²: type A as x and drop the minus sign. Know aₘₐₓ and A? Type them as a and x, leave ω empty. At a time t: find x first with x = A cos(ωt + φ). Spring: type √(k/m) as ω.

Total energy. It stays the same the whole time. Spring energy at a spot x: type x as A. Then kinetic energy = E − that. Kinetic energy ½mv²: type m in the k box and v in the A box. Max speed from E: type m as k and leave A empty (answer is vₘₐₓ). No k? Type m*ω² as k.

Velocity at time t. + or − shows the direction. RADIANS, and the same φ as in x = A cos(ωt + φ). vₘₐₓ = Aω: type it as A*ω, e.g. 0.1*20. t answer = first time it has that velocity. ω: type 2π/T, 2πf or √(k/m).

Damped (friction slows it): the amplitude shrinks over time. Only told a fraction? Half the amplitude: type 1/1 as A₀ and 1/2 as A. Energy goes with A²: half the energy → type 1/1 as A₀ and √(1/2) as A. Critical damping (no bouncing): b = 2√(mk). Damped ω = √(k/m − (b/2m)²): type it in an ω box.