pt notes unit 1 - force unit 1 - subunit 1 mechanical force

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PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

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Page 1: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 1 - Force

Unit 1 - Subunit 1 Mechanical Force

Page 2: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Linear

Force = Mass x Acceleration

F = m x aUnitsAm.St. [lb] = [slugs] x [ft/s2]

S.I. [N] = [kg] x [m/s2]

Page 3: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

acceleration due to gravitygravity = a

a = 32 ft/s2 Am. St.

a = 9.8 m/s2 S.I.

Page 4: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Torque = Force x Lever Arm

T = F x L Units

Am. St. [lb·ft] = [lb] x [ft]

S.I. [N·m] = [N] x [m]

Page 5: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 1 - Force

Unit 1 - Subunit 2 Fluid Force

Page 6: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Mass Density = _Mass_

Volume

D = _m_

VUnits Am. St. [slugs/ft3] = [slugs] / [ft3]S.I. [g/cm3] = [g] / [cm3] [kg/m3] = [kg] / [m3] [g/mL] = [g] / [mL]

Page 7: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Weight Density = _Weight_

Volume

*ρ w = __W__

VUnits Am. St. [lbs/ft3] = [lbs] / [ft3]S.I. [N/cm3] = [N] / [cm3] [N/m3] = [N] / [m3]

* “ρ” - Rho is a Greek letter

Page 8: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Specific Gravity = Density of "stuff"

Density of H2O

sp. gr. = Dstuff

Dwater

****NO UNITS FOR SPECIFIC GRAVITY****Density of H2O = 1 g/cm3 = 1,000 kg/m3

= 62.4 lb/ft3

1 cc = 1 cm3 = 1 mL

Page 9: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Pressure = Force

Area

P = F_ A

Units Am. St. [lb/ft2] = [lb] / [ft2] [p.s.i.] = [lb] / [in2] S.I. [Pa*] = [N] / [m2]

*Pa = Pascal

Page 10: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Pressure = weight density x height

P = ρw* x h

* weight density Units

Am. St. [lb/ft2] = [lb/ft3] x[ft] S.I. [N/m2] = [N/m3] x [m]

Page 11: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

1 atm (atmosphere) = 14.7 lb/in2 (psi)= 2117 lb/ft2

= 1.013 x 105 N/m2 or Pascal (Pa)

= 33.92 ft. of H2O= 760 mm of Hg (mercury) (Chem

- torr)

= 29.92 in of Hg

Units of Atmospheric Pressure (at sea level)

Page 12: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Absolute Pressure =

Total Pressure =

GaugePres. + AtmosphericPres.

Page 13: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Pascal’s Principle

PLarge = Psmall

_FL_ = _FS_

AL AS

Page 14: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Buoyant Volume weight Force = displaced X

density

FB = Vdisplaced x ρw Units

Am. St. [lb] = [ft3] x [lb/ft3]

S. I. [N] = [m3] x [N/m3]

Page 15: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 1 - Force

Unit 1 - Subunit 3 Electrical Force

Page 17: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Voltage –Prime Mover

Parallel Circuit:•Brighter•1 out others stay on•Vsource = VL1 = VL2

Page 18: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 1- Force

Unit 1 – Subunit 4Thermal Force

Page 19: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Temperature – Molecular Motion

F = 9/5 C + 32

C = 5/9 (F – 32)

Page 20: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 2 - Work

Unit 2 – Subunit 1Mechanical Work

Page 21: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Linear

Work = Force x Distance

W = F x dUnitsAm. St. [ft·lb] = [lb] x [ft]S.I. [J] = [N] x [m]

J = Joule = N·m

Page 22: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Torque Work = Torque x radians

WT = T x θ*

UnitsAm. St. [ft·lb] = [lb·ft] x radians (unitless)

S.I. [J] = [N·m] x radians (unitless)

* θ = (theta) is a Greek letter used to label angles

Angular(rotational)

Page 23: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

1 rotation = 360 = 2 radians

2 = 6.28

1 radian = 57.3

Page 24: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Efficiency = Workout

WorkinNote: UNITLESS the units

cancel!!

Efficiency is usually

given as a percent

Multiply by 100 and add a “%” sign

Page 25: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Unit 2 - Subunit 2

Fluid Work

Page 26: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Fluid Work = Volume X Pressure Change Change

WF = Δ V x Δ P

UnitsAm.St. [ft·lb] = [ft3] x [lb/ft2]S.I. [J] = [m3] x [N/m2]

Page 27: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Formulas

Area of circle = r2

Volume of cylinder = hr2

= h(area of circle)

Page 28: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Unit 2 - Subunit 3

Electrical Work

Page 29: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Electrical = Change x quantity

Work in Voltage of charge

WE = ΔV x q

Units Am.St. [J] = [V] x [C] & S.I.

C = coulomb 1 coulomb = 6.25 x 1018 electrons = 1

A·sec

Page 30: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Charge = Current x Time

q = I x tUnitsAm.St. [C] = [A] x [sec]& S.I.

A = Amperes = Amps

Page 31: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Electrical=change in x Current x Time Work Voltage

WE = Δ V x I x tUnitsAm.St. [J] = [V] x [A] x [sec]& S.I.

1 J = 1 V·A·sec = V·C

Page 32: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 3 – Rate

Unit 3 - Subunit 1 Mechanical - Rate

Page 33: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Linear Rate

Velocity = distance_ time

v = l__ t

Units Am.St. [mi/hr or mph] = [mi] /

[hr] [ft/sec] = [ft] / [sec]

S.I. [km/hr or kph] = [km] / [hr]

[m/sec] = [m] / [sec]

Page 34: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

velocity - has magnitude and direction(vector)

speed - has magnitude only(scalar)

average velocity = displacement/time

average speed = total dist. traveled

/time

Page 35: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Acceleration= final velocity – initial velocity

time = Vf - Vi

tUnits

Am. St. [ft/sec 2] = [ft/sec] – [ft/sec]

[sec]

S.I. [m/sec2 ] = [m/sec] – [m/sec]

[sec]

Page 36: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Angular Rate

Angular Rate = number of rotations time ω = θ tUnitsAm. St. [rev/min] or rpm= [rev] / [min]& S.I. [rot/sec] = [rot] / [sec] [rad/sec] = [rad] / [sec]

Page 37: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Angular Acceleration angular = final rate – initial

rate acceleration time

= ωf - ωi

tUnitsAm. St. & S.I. [rev/min2] = [rev/min] / [min]

[rot/sec2] = [rot/sec] / [sec] [rad/sec2] = [rad/sec] / [sec]

Page 38: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 3 – Rate

Unit 3 - Subunit 2

Fluid - Rate

Page 39: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Volume Flow Rate = Volume Time

QV = V t

UnitsAm. St. [gal/min] = [gal] / [min]

[ft3/sec] = [ft3] / [sec]S.I. [L/min] = [L] / [min]

[m3/hr] = [m3] / [hr]

Page 40: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Mass Flow Rate = Mass

Time QM = m

tUnitsAm. St. [lb/hr] = [lb] / [hr]

S.I. [kg/hr] = [kg] / [hr]

Page 41: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Area of TrapezoidArea = 1/2( base 1 + base 2 ) x

heightBase 2

Height

Base 1

Page 42: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Volume of a Trapezoid

Volume = 1/2( base 1 + base 2) x height x distance

Base 1

Distance

Base 2

Height

Page 43: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 3 – Rate

Unit 3 - Subunit 3 Electrical - Rate

Page 44: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Current = Quantity of Charge time

I = q t

UnitsAm.St. [A] = [Coulombs] & S.I. [sec]

*this is an old formula from Unit 2 rearranged q = I x t

Page 45: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Frequency = number of cycles time f = # cycles tUnitsAm. St. [Hz] = [cycles]& S.I [sec]

Page 46: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Period = time # of Cycles

T = t # of cyclesUnitsAm. St. [sec/cycle] = [sec]& S.I. [cycle]

Page 47: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

f = 1 / T frequency & period are

T = 1 / f inverses of each other

1 sec = 1,000 milliseconds [msec]

1 sec = 1,000,000 microseconds [μsec]

Page 48: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Oscilloscope sine waves square waves triangle waves saw-tooth waves

Page 49: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Vertical - measures voltage

Page 50: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Horizontal - measures period

Page 51: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force
Page 52: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

PT NotesUnit 3 – Rate

Unit 3 - Subunit

4 Thermal - Rate

Page 53: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Heat Flow = Heat Energy Transferred Rate Elapsed Time

QH = H t

UnitsAm. St. [Btu/hr] = [Btu] / [hr] S.I. [cal/min] = [cal] / [min] [J/sec] = [J] / [sec]

Page 54: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Do not confuse Heat with

Temperature

Heat is Energy!!!

Page 55: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

1 calorie = the amount of heat

required to raise temperature

of 1 gram of water 1° C

1 British Thermal Unit (Btu) =

the amt of heat required to raise the temperature of 1 lb. of water 1 F

Page 56: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

1 Btu = 252 cal

1 cal = 4.18 J

1 kcal = 1,000 cal

1kcal = 1 Cal

Big “C” is food calories

Page 57: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

SpecificHeat = Mass * Heat * Δ Temp Constant

H = m * c * Δ TUnitsAm. St. [Btu] = [lb] * [Btu/lb·F°] * [F°]S.I. [cal] = [g] * [cal/g·C°] * [C°]

Page 58: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Heat Thermal Flow Conductivity Rate = constant * Area * ΔTemp

Thickness

QH = k * A * ΔT lUnitsAm. St. [Btu/hr] = [(Btu·in) / (hr·ft2·F°)] * [ft2] *

[F°][in]

S.I. [cal/sec] = [(cal·cm) / (sec·cm2·C°)] *[cm2] * [C°]

[cm]

Page 59: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Lab book p. 96 has table of specific heat

constants (“c”)

Lab book p. 99 has table of thermal conductivity

constants (“k”)

Page 60: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force
Page 61: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

This is the only method of heat transfer in opaque solids. If the temperature at one end of a metal rod is raised by heating, heat is conducted to the colder end, but the exact mechanism of heat conduction in solids is not entirely understood. It is believed, however, to be partially due to the motion of free electrons in the solid matter, which transport energy if a temperature difference is applied. This theory helps to explain why good electrical conductors also tend to be good heat conductors (see Conductor, Electrical). Although the phenomenon of heat conduction had been observed for centuries, it was not until 1882 that the French mathematician Jean Baptiste Joseph Fourier gave it precise mathematical expression in what is now regarded as Fourier's law of heat conduction. This physical law states that the rate at which heat is conducted through a body per unit cross-sectional area is proportional to the negative of the temperature gradient existing in the body. The proportionality factor is called the thermal conductivity of the material. Materials such as gold, silver, and copper have high thermal conductivities and conduct heat readily, but materials such as glass and asbestos have values of thermal conductivity hundreds and thousands of times smaller, conduct heat poorly, and are referred to as insulators.

Page 62: PT Notes Unit 1 - Force Unit 1 - Subunit 1 Mechanical Force

Temperature Change Versus Heat Added: Water