Mechanical Comprehension: mixed review

$W=mg$: weight W in newtons, mass m in kg, g about 9.8 m/s² (often 10 on tests). Mass never changes with location; weight does. An 8 kg toolbox weighs about 78.4 N.

$F_{net}=ma$: net force (N) equals mass (kg) times acceleration (m/s²); acceleration points along the net force. Example: 3,000 N on a 1,500 kg car gives 2 m/s².

They are equal and opposite but act on different objects. Only forces acting on the same object go in its free-body diagram and can cancel.

$R=\sqrt{F_1^2+F_2^2}$. Example: 6 N east and 8 N north give 10 N. Opposite forces subtract; only same-direction forces add directly.

$\tau =Fd\sin \theta$: force F (N) times distance d (m) from pivot, giving N·m. Largest when the push is perpendicular (90°); zero when the force points through the pivot.

Net force is zero and net torque is zero. Equal, opposite but offset forces (a couple) give zero net force yet still rotate the object.

Static friction matches the push up to a maximum ${\mu }_{s}N$; it is not always at that maximum. Once sliding, kinetic friction is ${\mu }_{k}N$, usually smaller than the static maximum.

Down-slope part $W\sin \theta$; into-slope part $W\cos \theta$, which equals the normal force with no other forces. A block stays put if $\tan \theta \le {\mu }_s$.

Look at what is in the middle (memory aid F-L-E for 1-2-3). Fulcrum in the middle: first class. Load in the middle: second class. Effort in the middle: third class.

$F_e\times d_e=F_L\times d_L$: effort times its distance from the fulcrum equals load times its distance. Example: 300 N rock at 0.5 m needs 100 N at 1.5 m.

$\text{IMA}=\frac{d_{effort}}{d_{load}}$, both measured from the fulcrum; unitless. A 1.6 m effort arm and 0.40 m load arm give IMA 4, so effort = load / 4.

$\text{AMA}=\frac{F_{out}}{F_{in}}$; efficiency = AMA / IMA × 100%. AMA is always less than IMA because of friction. IMA 6 with AMA 4.8 gives 80% efficiency.

Second class (wheelbarrow, nutcracker) always has IMA above 1. Third class (tweezers, broom, forearm) always has IMA below 1 but gains speed and reach. First class can go either way.

No. Work in equals work out at best. If force is multiplied by 5, the load moves only 1/5 as far. Any choice claiming more work out than in, or over 100% efficiency, is wrong.

Multiply the stages: ${\text{IMA}}_{total}={\text{IMA}}_1\times {\text{IMA}}_2$. Stages of 3 and 4 give 12, not 7. Nail clippers are an example.

Convert mass to weight first: $W=mg$. A 20 kg box at 0.5 m makes about 200 N × 0.5 m = 100 N·m (with g = 10), not 10 N·m. Also convert cm to m before computing N·m.

$N_1{n}_1=N_2{n}_2$: teeth times rpm is equal on both gears. An 18-tooth pinion at 1,200 rpm drives a 54-tooth gear at 400 rpm. The bigger gear always turns slower.

Each external mesh reverses direction. Odd number of meshes: output turns opposite the input. Even number: same direction. An internal (ring) mesh does not reverse.

It sits between driver and driven gear, reversing the output direction so it matches the input and bridging a gap. Its tooth count cancels out, so it does not change the overall speed ratio.

Gears on a shared shaft turn at the same rpm and direction. Multiply the stage ratios (driven/driver for each mesh): $i=\frac{N_2}{N_1}\times \frac{N_4}{N_3}$. Two 3:1 stages give 9:1.

$T_{out}=\eta i{T}_{in}$: torque rises by the ratio i as speed falls, times efficiency η. A 5:1 reducer at 90% with 6 N·m in gives 27 N·m. Power never increases.

$i=\frac{N_{wheel}}{\text{starts}}$. A 3-start worm driving a 60-tooth wheel gives 20:1. Worm sets give big reductions, are often self-locking, and connect shafts crossing at right angles.

Spur and helical: parallel shafts. Bevel: intersecting shafts, usually 90°. Worm and wheel: crossed, non-intersecting shafts. Rack and pinion: converts rotation to straight-line motion.

Rack moves $s=\pi d$ per turn (pinion pitch diameter d). Module $m=\frac{d}{N}$ in mm per tooth; mating gears need the same module. Center distance $a=\frac{d_1+d_2}{2}$.

A fixed pulley only changes direction: IMA 1, effort equals load. A movable pulley rides with the load and is held by two segments: IMA 2, so effort is half the load.

Count only the rope segments that hold up the moving block (not the wheels). A dead end tied to the moving block counts; ideal effort $F=\frac{W}{n}$.

$d_{rope}=n\times d_{load}$ for n supporting segments. A 3-segment tackle lifting 0.4 m needs 1.2 m of rope. Less force always costs more rope travel.

$\text{AMA}=\frac{F_{load}}{F_{effort}}$ using measured effort; efficiency $\eta =\frac{\text{AMA}}{\text{IMA}}$. 900 N lifted with 400 N on 3 segments: AMA 2.25, η = 75%.

Add the weight of the moving block, hook and sling to the load before dividing by n. 760 N motor plus 40 N block on 4 segments needs 200 N, not 190 N.

$D_1{N}_1=D_2{N}_2$: driver diameter times rpm equals driven diameter times rpm. An 80 mm pulley at 1,800 rpm driving a 240 mm pulley gives 600 rpm. The larger pulley turns slower.

Open belt: both pulleys turn the same direction. Crossed (figure-eight) belt: the driven pulley turns the opposite direction. Pulley sizes never change direction.

$v=\frac{\pi DN}{60}$: pulley diameter D in meters, speed N in rpm, gives m/s. A 0.10 m pulley at 1,500 rpm drives the belt at about 7.85 m/s.

$\text{IMA}=\frac{L}{h}$: sloped length divided by vertical rise (same units). An 800 N cart on a 6 m ramp rising 1.5 m needs an ideal 200 N push.

No. Lifting to height h always needs at least $W\times h$ of work. A longer ramp lowers the force but spreads the same work over more distance; friction can even add work.

$\text{IMA}=\frac{\text{length}}{\text{thickness}}$ at the wide end. A 15 cm long, 3 cm thick wedge has IMA 5. Long thin wedges give more advantage; blunt ones need more force.

$\text{lead}=\text{pitch}\times \text{starts}$: axial travel per full turn. A triple-start screw with 1.5 mm pitch advances 4.5 mm per turn. Single-start: lead equals pitch.

$\text{IMA}=\frac{2\pi r}{\ell }$: handle circle 2πr divided by lead ℓ (same units). A 0.25 m handle and 5 mm lead give about 314. Longer handle or finer lead raises IMA.

Friction always opposes motion. Raising: push = $W\sin \theta +{\mu }_{k}W\cos \theta$. Lowering at steady speed: hold back only $W\sin \theta -{\mu }_{k}W\cos \theta$. Normal force on a slope is $W\cos \theta$.

Low-efficiency screws (roughly below 50%) are self-locking: thread friction keeps the load from turning the screw back. Vibration or wear can still let it slip, so use jack stands.

$\text{efficiency}=\frac{\text{AMA}}{\text{IMA}}\times 100\%$, with AMA = load / actual effort. A 5 m ramp rising 1 m (IMA 5) needing 225 N for 900 N (AMA 4) is 80% efficient.

$P=\frac{F}{A}$: force (N) divided by area (m²) gives pascals; 1 bar = 100 kPa, sea-level air is about 101 kPa (14.7 psi). Note $1{\text{ m}}^2=10,000{\text{ cm}}^2$.

Pressure is equal on both pistons, so $\frac{F_1}{A_1}=\frac{F_2}{A_2}$ and $F_2=F_1\frac{A_2}{A_1}$. A 60 N push on 5 cm² drives a 40 cm² piston with 480 N.

Area grows with diameter squared: $A=\frac{\pi d^2}{4}$. A 4 cm piston has 4 times the area of a 2 cm piston, not 2 times, so square the diameter ratio.

Volume is conserved, $A_1{d}_1=A_2{d}_2$, so the large piston moves less. Six times the force means one sixth the travel; work in equals work out ideally.

$v=\frac{Q}{A}$: flow rate (volume per time) divided by piston area. 200 cm³/s into a 40 cm² piston gives 5 cm/s. Pressure sets force; flow sets speed.

$P_1{V}_1=P_2{V}_2$ at constant temperature, using absolute pressures. Air at 100 kPa abs in 12 L squeezed to 3 L reaches 400 kPa abs (about 300 kPa gauge).

$P_{abs}=P_{gauge}+P_{atm}$, with the atmosphere about 100 kPa. A tire gauge reading 250 kPa means about 350 kPa absolute. Gas-law problems require absolute pressure.

Relief valve: opens to tank to cap maximum pressure. Check valve: allows flow one way only, holding loads. Directional valve: routes flow to extend or retract a cylinder. A pump makes flow, not pressure.

$\rho =\frac{m}{V}$ in kg/m³ (water 1,000 kg/m³ = 1 kg/L). Specific gravity = $\frac{{\rho }_{material}}{{\rho }_{water}}$, unitless; below 1 floats in fresh water, above 1 sinks.

$\Delta P=\rho gh$: liquid density (kg/m³) times g (m/s²) times depth (m), in pascals. With g = 10, each meter of fresh water adds about 10 kPa. Doubling depth doubles pressure.

No. Pressure at a depth depends only on liquid density, g and vertical depth, not on container shape, width or total volume. A thin pipe and a pool 3 m deep have equal bottom pressure.

$F_b={\rho }_{fluid}g{V}_{disp}$: equals the weight of fluid displaced. Use the fluid's density, not the object's. A fully submerged 0.05 m³ object in fresh water gets 500 N (g = 10).

$W_{app}=W-F_b$: true weight minus buoyant force is the scale reading. A 54 N aluminum casting with 20 N of buoyancy reads 34 N; its mass is unchanged.

$\frac{V_{sub}}{V}=\frac{{\rho }_{obj}}{{\rho }_{fluid}}$. A 300 kg/m³ foam float is 30% submerged in fresh water; ice (about 917 kg/m³) is about 92% under.

Compare its average density (total mass / outside volume, including sealed air) with the fluid's. Less dense: floats. Denser: sinks. Equal: neutrally buoyant. That is why a steel ship floats.

No. Pressure rises equally on top and bottom, so the difference, and the buoyant force, stays the same. Only squeezable objects (balloons, wetsuits) lose buoyancy as they compress.

$W=Fd\cos \theta$: force (N) times displacement (m) times cosine of the angle between them, in joules. 40 N at 60° over 5 m does 100 J. A force opposing motion does negative work.

When nothing moves (holding a barbell still) or the force is perpendicular to the motion (carrying a tray level, cos 90° = 0). Effort alone is not work.

$KE=\frac{1}{2}m{v}^2$: mass in kg, speed in m/s, giving joules. Speed is squared, so doubling speed gives four times the energy; tripling gives nine times.

Gravity: $PE=mgh$ (weight times height). Spring: $P{E}_s=\frac{1}{2}k{x}^2$, with spring constant k in N/m and stretch x in m. Doubling the stretch quadruples stored energy.

Net work on an object equals its change in kinetic energy: $W_{net}=K{E}_f-K{E}_i$. Positive net work speeds it up, negative slows it, zero leaves speed unchanged.

$v=\sqrt{2gh}$, from $mgh=\frac{1}{2}m{v}^2$; mass cancels. A wrench dropped 3.2 m with g = 10 lands at 8 m/s.

$P=\frac{W}{t}=Fv$: work (J) over time (s), or force (N) times speed (m/s), in watts (1 W = 1 J/s). Same work in half the time means twice the power.

1 hp is about 746 W. 1 kWh = 3,600,000 J and is an energy unit, not power. A 500 W freezer running 6 hours uses 3 kWh.

Distance is the total path length traveled (never negative). Displacement is the net change in position, $\Delta x=x_f-x_i$, with a sign or direction. A round trip has zero displacement but nonzero distance.

$v=v_0+at$, where $v_0$ is starting velocity (m/s), a is acceleration (m/s²) and t is time (s). Example: from rest at 2.5 m/s² for 8 s gives 20 m/s.

$\Delta x=v_{0}t+\frac{1}{2}a{t}^2$ (meters). The first term is travel with no acceleration; the second grows with time squared, so doubling t quadruples that part.

$v^2=v_0^2+2a\Delta x$. Setting v = 0 gives braking distance $\Delta x=\frac{v_0^2}{2|a|}$, so doubling speed makes braking distance 4 times longer.

Same sign for velocity and acceleration: speeding up. Opposite signs: slowing down. Negative acceleration alone does not mean braking; an object moving in the negative direction speeds up with it.

Position-time: slope = velocity. Velocity-time: slope = acceleration, area under the line = displacement (area below the axis counts negative; add area sizes for total distance).

About 9.8 m/s² (often 10) downward for every mass if air drag is ignored. At the top of a throw velocity is zero but acceleration is still g downward. Dropped from rest: $v=gt$, $d=\frac{1}{2}g{t}^2$.

Yes, because direction changes. Centripetal acceleration points to the center: $a_c=\frac{v^2}{r}$ (v in m/s, r in m). Halving the radius doubles it; doubling speed quadruples it.

Friction between surfaces that are not slipping. It adjusts to exactly balance the push trying to cause slipping, up to a maximum $f_{s,max}={\mu }_{s}N$. A 30 N push on a crate at rest meets 30 N of friction.

$f_k={\mu }_{k}N$: kinetic coefficient (no unit) times normal force (N). It is roughly constant regardless of sliding speed and points opposite the slipping, along the surface.

Usually ${\mu }_s>{\mu }_k$, so starting an object moving takes more force than keeping it sliding. Both are dimensionless ratios, $\mu =f/N$, describing a pair of surfaces.

From the force balance perpendicular to the surface: N = mg on level ground; a downward push adds to N, an upward pull subtracts; on a ramp $N=mg\cos \theta$. N equals weight only in the simple level case.

Down-slope part $mg\sin \theta$, into-ramp part $mg\cos \theta$. It stays put if $\tan \theta \le {\mu }_s$; the angle of repose is where $\tan \theta ={\mu }_s$, independent of weight.

Deceleration $a={\mu }_{k}g$ (mass cancels, so heavy and light objects slow alike). Stopping distance $d=\frac{v_0^2}{2\mu g}$; e.g. 20 m/s, μ = 0.8, g = 10 gives 25 m.

No. It opposes relative slipping between the surfaces. A box on an accelerating conveyor, a walker's foot and a driven tire all get forward static friction.

No. A brick slides with the same friction on its broad face or narrow edge, because friction depends only on μ and the normal force.

$F=kx$: force (N) = spring constant k (N/m) times deflection x (m) from natural length. The force points back toward natural length, written $F=-kx$. Valid only in the elastic range.

$U=\frac{1}{2}k{x}^2$ in joules (k in N/m, x in m). Doubling the deflection gives 4 times the energy. Equivalent form: $U=\frac{F^2}{2k}$.

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}$, or $k_{eq}=\frac{k_1{k}_2}{k_1+k_2}$ for two. Each carries the full force and stretches add, so the result is softer than the softest spring.

$k_{eq}=k_1+k_2$. The springs share the same deflection and their forces add, so the combination is stiffer than any single spring; n identical springs give nk.

$T=2\pi \sqrt{\frac{m}{k}}$ (s), with $f=1/T$ in hertz. More mass gives a longer period; a stiffer spring a shorter one. Amplitude and gravity do not change T.

The deflection must be in meters: 4 cm = 0.04 m, so $F=500\times 0.04=20$ N. Unconverted centimeters make the answer 100 times too big.

From the natural (unloaded) length, not the hanging equilibrium or installed position. A spring 30 cm relaxed and 34 cm loaded has 4 cm extension, not 34 cm.

Each half has half the coils, so it is twice as stiff (2k). More active coils, thinner wire or a larger coil diameter all make a spring softer.

Temperature measures how hot something is (average particle motion), in °C or K. Heat is energy (joules) transferred because of a temperature difference, always flowing from hot to cold.

$T_K=T_C+273.15$ and $T_F=\frac{9}{5}{T}_C+32$. For changes, 1°C = 1 K and 1°C = 1.8°F (a 10°C change is an 18°F change, no 32 added).

$\Delta L=\alpha L_0\Delta T$: α is the expansion coefficient (1/°C), $L_0$ the original length, $\Delta T$ the temperature change. A 25 m steel rail (α = 12 × 10⁻⁶/°C) warmed 40°C grows 12 mm.

The hole gets bigger. Every dimension grows by the same fraction, like enlarging a photo, so holes, ring bores and gear bores all expand.

Conduction: through a material by particle contact. Convection: carried by a moving fluid (natural or fan-forced). Radiation: electromagnetic waves, which can cross a vacuum.

$Q=mc\Delta T$: Q in joules, m in kg, c specific heat in J/(kg·K), $\Delta T$ in °C or K. Water's c is about 4,200 J/(kg·K), far higher than most metals.

$Q=mL$, with L the latent heat in J/kg; temperature stays constant during the change. Water: fusion about 334,000 J/kg, vaporization about 2,260,000 J/kg.

Stress $\sigma =F/A$ (Pa); strain $\epsilon =\Delta L/L_0$ (no unit); Young's modulus $E=\sigma /\epsilon$ measures stiffness (steel about 200 GPa, aluminum about 70 GPa).