Search results for: Convert 1 lb ft to nm

Below are our best 37 results for convert 1 lb ft to nm, updated recently.

transportdiverse.com

Nm to pounds converter. Nm to pounds per inch. Nm to pounds force. Nm to pounds inch. Nm to pounds per foot. nm, or 0.73756214927727 Pantry. Note that rounding errors may occur, then always check the results. Use this page to know how to convert between nm and pounds. Enter your own numbers in the form to convert the units! Blessed Table of ...

motorsteamzena.it

1 hour ago · Convert inches to feet or convert inches to miles to find your final distance measurement. August 5, 2021. 2 - 5/4x6 2-2x6 1 - 2x12. ... The calculator will indicate the number of 80 lb bags of QUIKRETE ® Base Coat Stucco and Finish Coat Stucco you will need to construct your stucco wall using a traditional 3 coat or 2 coat application process ...

snoopergps.it

Tdi injector torque specs

motorsteamzena.it

7 hours ago · 11. Nov 02, 2010 · Two-piece 33 mm flange nuts used with hub-piloted wheels should be tightened to a torque of 450 to 500 foot-pounds. Return Icon. 500 429-0003 Eaton 400 Rear Input 4. They went on to invent the use of oil lubrication for wheel bearings. ft, 2224 Nm @1200 rpm Granite construction, concrete, vocational truck. STEMCO.

unitconverters.net

It is defined as the energy transferred when a force of one pound-force is applied over a linear displacement of one foot. History/origin: The foot-pound is an English Engineering unit that is part of the British Gravitational system, based on the foot-pound-second (FPS) system. The FPS system is built using the units of foot for length, avoirdupois pound for mass or force, and the second for time.

Please provide values below to convert foot-pound [ft*lbf] to newton meter [N*m], or vice versa.


Foot-pound

Definition: A foot-pound (ft·lb or ft·lbf) is a unit of work or energy in the imperial and United States customary systems of units. It is defined as the energy transferred when a force of one pound-force is applied over a linear displacement of one foot.

History/origin: The foot-pound is an English Engineering unit that is part of the British Gravitational system, based on the foot-pound-second (FPS) system. The FPS system is built using the units of foot for length, avoirdupois pound for mass or force, and the second for time.

Current use: The foot-pound is used in countries that have not adopted the International System of Units (SI). The United States is one of the largest countries where the foot-pound is still used in certain applications, though even in the US, the joule is preferred in most scientific contexts.

Newton-meter

Definition: The newton-meter (N·m) is sometimes used as a unit of work or energy and in this context is equal to the joule, the SI unit of energy. In this use, it is defined as the energy transferred to an object by a one newton force acting on that object in the direction of its motion over a distance of one meter.

History/origin: The use of the newton-meter as a unit of work or energy is derived from the fact that the joule, the SI (International System of Units) unit of energy, is dimensionally equivalent to the newton-meter.

Current use: The newton-meter is more commonly used as a unit of torque rather than as a unit of work or energy. As such, the use of the newton-meter as a unit of work or energy is discouraged to avoid confusion regarding whether energy or torque is being expressed.

Foot-pound [ft*lbf]Newton Meter [N*m]
0.01 ft*lbf0.0135581795 N*m
0.1 ft*lbf0.1355817948 N*m
1 ft*lbf1.3558179483 N*m
2 ft*lbf2.7116358966 N*m
3 ft*lbf4.0674538449 N*m
5 ft*lbf6.7790897414 N*m
10 ft*lbf13.5581794829 N*m
20 ft*lbf27.1163589658 N*m
50 ft*lbf67.7908974145 N*m
100 ft*lbf135.581794829 N*m
1000 ft*lbf1355.8179482896 N*m

1 ft*lbf = 1.3558179483 N*m1 N*m = 0.7375621493 ft*lbf

Example: convert 15 ft*lbf to N*m:
15 ft*lbf = 15 × 1.3558179483 N*m = 20.3372692243 N*m


convertunits.com

lb-ft or N-m. The SI derived unit for torque is the newton meter. 1 lb-ft is equal to 1.3558179483314 newton meter. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between pound feet and newton meters. Type in …

Did you mean to convert lb-ft to newton-meter
newton-metre


›› More information from the unit converter

How many lb-ft in 1 N-m? The answer is 0.73756214927727.
We assume you are converting between pound foot and newton-meter. You can view more details on each measurement unit:

lb-ft or N-m


The SI derived unit for torque is the newton meter. 1 lb-ft is equal to 1.3558179483314 newton meter. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between pound feet and newton meters.

Type in your own numbers in the form to convert the units!


›› Quick conversion chart of lb-ft to N-m

1 lb-ft to N-m = 1.35582 N-m

5 lb-ft to N-m = 6.77909 N-m

10 lb-ft to N-m = 13.55818 N-m

15 lb-ft to N-m = 20.33727 N-m

20 lb-ft to N-m = 27.11636 N-m

25 lb-ft to N-m = 33.89545 N-m

30 lb-ft to N-m = 40.67454 N-m

40 lb-ft to N-m = 54.23272 N-m

50 lb-ft to N-m = 67.7909 N-m



You can do the reverse unit conversion from N-m to lb-ft, or enter any two units below:

lb-ft to ounce-foot
lb-ft to poundal-foot
lb-ft to micronewton-meter
lb-ft to millinewton-meter
lb-ft to kilopond-meter
lb-ft to newton metre
lb-ft to foot pound
lb-ft to gram-centimeter
lb-ft to kilogram-meter
lb-ft to kilogram-centimeter


›› Definition: Newton meter

Newton metre is the unit of moment in the SI system. Its symbol is N m or N·m. It is a compound unit of torque corresponding to the torque from a force of 1 newton applied over a distance arm of 1 metre.


›› Metric conversions and more

ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

convertunits.com

We assume you are converting between pound foot and newton-meter. You can view more details on each measurement unit: lb-ft or newton-meter The SI derived unit for torque is the newton meter. 1 lb-ft is equal to 1.3558179483314 newton meter. Note that rounding errors …



How many lb-ft in 1 newton-meter? The answer is 0.73756214927727.
We assume you are converting between pound foot and newton-meter. You can view more details on each measurement unit:

lb-ft or newton-meter


The SI derived unit for torque is the newton meter. 1 lb-ft is equal to 1.3558179483314 newton meter. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between pound feet and newton meters.

Type in your own numbers in the form to convert the units!


›› Quick conversion chart of lb-ft to newton-meter

1 lb-ft to newton-meter = 1.35582 newton-meter

5 lb-ft to newton-meter = 6.77909 newton-meter

10 lb-ft to newton-meter = 13.55818 newton-meter

15 lb-ft to newton-meter = 20.33727 newton-meter

20 lb-ft to newton-meter = 27.11636 newton-meter

25 lb-ft to newton-meter = 33.89545 newton-meter

30 lb-ft to newton-meter = 40.67454 newton-meter

40 lb-ft to newton-meter = 54.23272 newton-meter

50 lb-ft to newton-meter = 67.7909 newton-meter



You can do the reverse unit conversion from newton-meter to lb-ft, or enter any two units below:

lb-ft to kilogram-meter
lb-ft to foot pound
lb-ft to gram-centimeter
lb-ft to pound-inch
lb-ft to millinewton-meter
lb-ft to newton-centimeter
lb-ft to poundal-foot
lb-ft to kilonewton-meter
lb-ft to newton metre
lb-ft to ton meter


›› Definition: Newton meter

Newton metre is the unit of moment in the SI system. Its symbol is N m or N·m. It is a compound unit of torque corresponding to the torque from a force of 1 newton applied over a distance arm of 1 metre.


›› Metric conversions and more

ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

metric-conversions.org

Foot-pounds Newton-meters; 0 ft-lb: 0.00 Nm: 1 ft-lb: 1.36 Nm: 2 ft-lb: 2.71 Nm: 3 ft-lb: 4.07 Nm: 4 ft-lb: 5.42 Nm: 5 ft-lb: 6.78 Nm: 6 ft-lb: 8.13 Nm: 7 ft-lb: 9.49 Nm: 8 ft-lb: 10.85 Nm: 9 ft-lb: 12.20 Nm: 10 ft-lb: 13.56 Nm: 11 ft-lb: 14.91 Nm: 12 ft-lb: 16.27 Nm: 13 ft-lb: 17.63 Nm: 14 ft-lb: 18.98 Nm: 15 ft-lb: 20.34 Nm: 16 ft-lb: 21.69 Nm: 17 ft-lb: 23.05 Nm: 18 ft-lb: 24.40 Nm: 19 ft-lb: 25.76 Nm

Convert Pound Foot to Newton Meters (lb ft to Nm ...

1 Pound Foot to common torque units; 1 lb ft = 1.35581795 newton meters (Nm) 1 lb ft = 0.13825495454615 kilogram force meters (kgf m) 1 lb ft = 13.825495454615 kilogram force centimeters (kgf cm) 1 lb ft = 1 foot pounds force (lbf路ft) 1 lb ft = 1 pound foot (lb ft) 1 lb ft = 13558179.5 dyne centimeters (dyn cm) 1 lb ft = 32.174048596027 foot poundals (ft pdl) 1 lb ft ...


Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)

Pound FootSymbol/abbreviation: lb ftUnit of: TORQUETORQUE's base unit: newton meters (SI Unit)In relation to the base unit (newton meters), 1 Pound Foot = 1.35581795 newton meters.
1 lb ft= 13558179.5 dyne centimeters (dyn cm)
1 lb ft= 32.174048596027 foot poundals (ft pdl)
1 lb ft= 1 foot pounds force (lbf路ft)
1 lb ft= 12.000000017701 inch pounds force (in lbf)
1 lb ft= 13.825495454615 kilogram force centimeters (kgf cm)
1 lb ft= 0.13825495454615 kilogram force meters (kgf m)
1 lb ft= 0.00135581795 kilonewton meters (kN m)
1 lb ft= 1355817.95 meganewton meters (MN m)
1 lb ft= 1355817.95 micronewton meters (碌N m)
1 lb ft= 1355.81795 millinewton meters (mN m)
1 lb ft= 135.581795 newton centimeters (N cm)
1 lb ft= 1.35581795 newton meters (Nm)
1 lb ft= 1 pound foot (lb ft)
1 lb ft= 1 pound force foot (lbf ft)
1 lb ft= 12.000000017701 pound force inch (lbf in)
wrenchturn.com

Convert lb-ft to Nm Pound-feet (lb-ft), sometimes called "foot-pounds" (ft-lbs), is a unit of torque mainly used in the United States. The Newton-meter is an internationally recognized unit of torque commonly used in countries such as Japan and Germany.

advancedconverter.com

Conversion Pound foot to Newton meter. A pound-foot (lb·ft or lbf·ft) is a unit of torque. One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point. This tool converts pound foot to newton meter (lb·ft to nm) and vice versa. 1 pound foot ≈ 1.3558 newton meter. The user must fill one of the two fields and the conversion will ...

A pound-foot (lb·ft or lbf·ft) is a unit of torque. One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point.

This tool converts pound foot to newton meter (lb·ft to nm) and vice versa. 1 pound foot ≈ 1.3558 newton meter. The user must fill one of the two fields and the conversion will become automatically.


1 pound foot = 1.3558 newton meter Formula pound foot in newton meter (lb·ft in nm). Nm = pound foot*1.3558179483314004 ≈ pound foot*1.3558

Conversions pound foot to other units

Table pound foot to nm

1 pound foot = 1.3558 nm11 pound foot = 14.914 nm21 pound foot = 28.4722 nm
2 pound foot = 2.7116 nm12 pound foot = 16.2698 nm22 pound foot = 29.828 nm
3 pound foot = 4.0675 nm13 pound foot = 17.6256 nm23 pound foot = 31.1838 nm
4 pound foot = 5.4233 nm14 pound foot = 18.9815 nm24 pound foot = 32.5396 nm
5 pound foot = 6.7791 nm15 pound foot = 20.3373 nm25 pound foot = 33.8954 nm
6 pound foot = 8.1349 nm16 pound foot = 21.6931 nm26 pound foot = 35.2513 nm
7 pound foot = 9.4907 nm17 pound foot = 23.0489 nm27 pound foot = 36.6071 nm
8 pound foot = 10.8465 nm18 pound foot = 24.4047 nm28 pound foot = 37.9629 nm
9 pound foot = 12.2024 nm19 pound foot = 25.7605 nm29 pound foot = 39.3187 nm
10 pound foot = 13.5582 nm20 pound foot = 27.1164 nm30 pound foot = 40.6745 nm
40 pound foot = 54.2327 nm70 pound foot = 94.9073 nm100 pound foot = 135.582 nm
50 pound foot = 67.7909 nm80 pound foot = 108.465 nm110 pound foot = 149.14 nm
60 pound foot = 81.3491 nm90 pound foot = 122.024 nm120 pound foot = 162.698 nm
200 pound foot = 271.164 nm500 pound foot = 677.909 nm800 pound foot = 1084.65 nm
300 pound foot = 406.745 nm600 pound foot = 813.491 nm900 pound foot = 1220.24 nm
400 pound foot = 542.327 nm700 pound foot = 949.073 nm1000 pound foot = 1355.82 nm

Torque Conversions

easyunitconverter.com

One newton force applied to an object through a distance of one meter in a force direction. One newton meter is equal to 0.7375621493 foot-pounds. Foot Pounds to Newton Meters (ft-lb to nm): It is a free online foot-pound to newton meters (ft-lb to nm) energy converter. A foot-pound is a measuring unit of energy and denoted as "ft.lb or ft.lbf".

calculator-online.net

Add Foot-pound to Newton meter converter to your website to use this unit converter directly. Feel hassle-free to account this widget as it is 100% free. Get Code! Available on App. Try Unit Converter App for your Mobile to get the ease of converting thousands of units. It’s 100% free with ample of features! Convert Foot-pound to Other Energy ...

unitconverters.net

Instant free online tool for newton meter to foot-pound conversion or vice versa. The newton meter [N*m] to foot-pound [ft*lbf] conversion table and conversion steps are also listed. Also, explore tools to convert newton meter or foot-pound to other energy units or learn more about energy conversions.

Please provide values below to convert newton meter [N*m] to foot-pound [ft*lbf], or vice versa.


Newton-meter

Definition: The newton-meter (N·m) is sometimes used as a unit of work or energy and in this context is equal to the joule, the SI unit of energy. In this use, it is defined as the energy transferred to an object by a one newton force acting on that object in the direction of its motion over a distance of one meter.

History/origin: The use of the newton-meter as a unit of work or energy is derived from the fact that the joule, the SI (International System of Units) unit of energy, is dimensionally equivalent to the newton-meter.

Current use: The newton-meter is more commonly used as a unit of torque rather than as a unit of work or energy. As such, the use of the newton-meter as a unit of work or energy is discouraged to avoid confusion regarding whether energy or torque is being expressed.

Foot-pound

Definition: A foot-pound (ft·lb or ft·lbf) is a unit of work or energy in the imperial and United States customary systems of units. It is defined as the energy transferred when a force of one pound-force is applied over a linear displacement of one foot.

History/origin: The foot-pound is an English Engineering unit that is part of the British Gravitational system, based on the foot-pound-second (FPS) system. The FPS system is built using the units of foot for length, avoirdupois pound for mass or force, and the second for time.

Current use: The foot-pound is used in countries that have not adopted the International System of Units (SI). The United States is one of the largest countries where the foot-pound is still used in certain applications, though even in the US, the joule is preferred in most scientific contexts.

Newton Meter [N*m]Foot-pound [ft*lbf]
0.01 N*m0.0073756215 ft*lbf
0.1 N*m0.0737562149 ft*lbf
1 N*m0.7375621493 ft*lbf
2 N*m1.4751242986 ft*lbf
3 N*m2.2126864479 ft*lbf
5 N*m3.6878107465 ft*lbf
10 N*m7.375621493 ft*lbf
20 N*m14.751242986 ft*lbf
50 N*m36.878107465 ft*lbf
100 N*m73.75621493 ft*lbf
1000 N*m737.5621493 ft*lbf

1 N*m = 0.7375621493 ft*lbf1 ft*lbf = 1.3558179483 N*m

Example: convert 15 N*m to ft*lbf:
15 N*m = 15 × 0.7375621493 ft*lbf = 11.0634322395 ft*lbf


omnicalculator.com

The conversion from Nm to ft-lbs is very simple. You can, of course, use our torque conversion calculator, but it's straightforward to calculate it by hand, too! All you need to remember is that 1 Nm is equal to 0.7376 ft-lbs. The conversion table below might come in handy, should you need to convert the most fundamental values of torque. If your value doesn't appear in the table, you might …

Torque - Wikipedia

20-12-2021 · A force applied perpendicularly to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque.A force of three newtons applied two meters from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if …

20-12-2021
Physics concept
For other uses, see Torque (disambiguation).
Torque animation.gifTorque
Relationship between force F, torque τ, linear momentum p, and angular momentum L in a system which has rotation constrained to only one plane (forces and moments due to gravity and friction not considered).
Common symbols
τ{\displaystyle \tau }\tau, MSI unitN⋅m
Other units
pound-force-feet, lbf⋅inch, ozf⋅inIn SI base unitskg⋅m2⋅s−2DimensionM L2T−2
Part of a series on
Classical mechanics
F=ddt(mv){\displaystyle {\textbf }={\frac {d}{dt}}(m{\textbf {v}})}{\displaystyle {\textbf }={\frac {d}{dt}}(m{\textbf {v}})}
Second law of motion
  • History
  • Timeline
  • Textbooks
Branches
  • Applied
  • Celestial
  • Continuum
  • Dynamics
  • Kinematics
  • Kinetics
  • Statics
  • Statistical
Fundamentals
  • Acceleration
  • Angular momentum
  • Couple
  • D'Alembert's principle
  • Energy
    • kinetic
    • potential
  • Force
  • Frame of reference
  • Inertial frame of reference
  • Impulse
  • Inertia / Moment of inertia
  • Mass

  • Mechanical power
  • Mechanical work

  • Moment
  • Momentum
  • Space
  • Speed
  • Time
  • Torque
  • Velocity
  • Virtual work
Formulations
  • Newton's laws of motion
  • Analytical mechanics
    • Lagrangian mechanics
    • Hamiltonian mechanics
    • Routhian mechanics
    • Hamilton–Jacobi equation
    • Appell's equation of motion
    • Koopman–von Neumann mechanics
Core topics
  • Damping ratio
  • Displacement
  • Equations of motion
  • Euler's laws of motion
  • Fictitious force
  • Friction
  • Harmonic oscillator
  • Inertial / Non-inertial reference frame
  • Mechanics of planar particle motion
  • Motion (linear)
  • Newton's law of universal gravitation
  • Newton's laws of motion
  • Relative velocity
  • Rigid body
    • dynamics
    • Euler's equations
  • Simple harmonic motion
  • Vibration
Rotation
  • Circular motion
  • Rotating reference frame
  • Centripetal force
  • Centrifugal force
    • reactive
  • Coriolis force
  • Pendulum
  • Tangential speed
  • Rotational speed
  • Angular acceleration / displacement / frequency / velocity
Scientists
  • Kepler
  • Galileo
  • Huygens
  • Newton
  • Horrocks
  • Halley
  • Daniel Bernoulli
  • Johann Bernoulli
  • Euler
  • d'Alembert
  • Clairaut
  • Lagrange
  • Laplace
  • Hamilton
  • Poisson
  • Cauchy
  • Routh
  • Liouville
  • Appell
  • Gibbs
  • Koopman
  • von Neumann
  • Stylised atom with three Bohr model orbits and stylised nucleus.svg Physics portal
  • Category Category
  • v
  • t
  • e

In physics and mechanics, torque is the rotational equivalent of linear force.[1] It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically τ{\displaystyle {\boldsymbol {\tau }}}{\boldsymbol {\tau }} or τ, the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M.

In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector[2] connecting the point about which the torque is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols:

τ=r×F{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf \,\!}{\boldsymbol {\tau }}=\mathbf {r} \times \mathbf  \,\!
τ=‖r‖‖F‖sin⁡θ{\displaystyle \tau =\|\mathbf {r} \|\,\|\mathbf \|\sin \theta \,\!}\tau =\|\mathbf {r} \|\,\|\mathbf  \|\sin \theta \,\!

where

  • τ{\displaystyle {\boldsymbol {\tau }}}{\boldsymbol {\tau }} is the torque vector and τ{\displaystyle \tau }\tau is the magnitude of the torque,
  • r{\displaystyle \mathbf {r} }\mathbf {r} is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied),
  • F{\displaystyle \mathbf }\mathbf is the force vector,
  • ×{\displaystyle \times }\times denotes the cross product, which produces a vector that is perpendicular to both r and F following the right-hand rule,
  • θ{\displaystyle \theta }\theta is the angle between the force vector and the lever arm vector.

The SI unit for torque is the newton-metre (N⋅m). For more on the units of torque, see § Units.

Defining terminology

See also: Couple (mechanics)

James Thomson, the brother of Lord Kelvin, introduced the term torque into English scientific literature in 1884.[3] However, torque is referred to using different vocabulary depending on geographical location and field of study. This article follows the definition used in US physics in its usage of the word torque.[4] In the UK and in US mechanical engineering, torque is referred to as moment of force, usually shortened to moment.[5] These terms are interchangeable in US physics[4] and UK physics terminology, unlike in US mechanical engineering, where the term torque is used for the closely related "resultant moment of a couple".[5][contradictory]

Torque and moment in the US mechanical engineering terminology

In US mechanical engineering, torque is defined mathematically as the rate of change of angular momentum of an object (in physics it is called "net torque"). The definition of torque states that one or both of the angular velocity or the moment of inertia of an object are changing. Moment is the general term used for the tendency of one or more applied forces to rotate an object about an axis, but not necessarily to change the angular momentum of the object (the concept which is called torque in physics).[5] For example, a rotational force applied to a shaft causing acceleration, such as a drill bit accelerating from rest, results in a moment called a torque. By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the angular momentum of the beam is not changing, this bending moment is not called a torque. Similarly with any force couple on an object that has no change to its angular momentum, such moment is also not called a torque.

Definition and relation to angular momentum

A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F produces a torque. This torque τ = r × F has magnitude τ = |r| |F| = |r| |F| sin θ and is directed outward from the page.

A force applied perpendicularly to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two meters from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand are curled from the direction of the lever arm to the direction of the force, then the thumb points in the direction of the torque.[6]

More generally, the torque on a point particle (which has the position r in some reference frame) can be defined as the cross product:

τ=r×F,{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf ,}{\boldsymbol {\tau }}=\mathbf {r} \times \mathbf  ,

where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by

τ=rFsin⁡θ,{\displaystyle \tau =rF\sin \theta ,}{\displaystyle \tau =rF\sin \theta ,}

where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,

τ=rF⊥,{\displaystyle \tau =rF_{\perp },}\tau =rF_{\perp },

where F is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[7][8]

It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. Conversely, the torque vector defines the plane in which the position and force vectors lie. The resulting torque vector direction is determined by the right-hand rule.[7]

The net torque on a body determines the rate of change of the body's angular momentum,

τ=dLdt{\displaystyle {\boldsymbol {\tau }}={\frac {\mathrm {d} \mathbf }{\mathrm {d} t}}}{\boldsymbol {\tau }}={\frac {\mathrm {d} \mathbf  }{\mathrm {d} t}}

where L is the angular momentum vector and t is time.

For the motion of a point particle,

L=Iω,{\displaystyle \mathbf =I{\boldsymbol {\omega }},}\mathbf  =I{\boldsymbol {\omega }},

where I is the moment of inertia and ω is the orbital angular velocity pseudovector. It follows that

τnet=dLdt=d(Iω)dt=Idωdt dIdtω=Iα d(mr2)dtω=Iα 2rp||ω,{\displaystyle {\boldsymbol {\tau }}_{\mathrm {net} }={\frac {\mathrm {d} \mathbf }{\mathrm {d} t}}={\frac {\mathrm {d} (I{\boldsymbol {\omega }})}{\mathrm {d} t}}=I{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}} {\frac {\mathrm {d} I}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }} {\frac {\mathrm {d} (mr^)}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }} 2rp_{||}{\boldsymbol {\omega }},}{\displaystyle {\boldsymbol {\tau }}_{\mathrm {net} }={\frac {\mathrm {d} \mathbf  }{\mathrm {d} t}}={\frac {\mathrm {d} (I{\boldsymbol {\omega }})}{\mathrm {d} t}}=I{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}} {\frac {\mathrm {d} I}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }} {\frac {\mathrm {d} (mr^)}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }} 2rp_{||}{\boldsymbol {\omega }},}

where α is the angular acceleration of the particle, and p|| is the radial component of its linear momentum. This equation is the rotational analogue of Newton's Second Law for point particles, and is valid for any type of trajectory. Note that although force and acceleration are always parallel and directly proportional, the torque τ need not be parallel or directly proportional to the angular acceleration α. This arises from the fact that although mass is always conserved, the moment of inertia in general is not.

Proof of the equivalence of definitions

The definition of angular momentum for a single point particle is:

L=r×p{\displaystyle \mathbf =\mathbf {r} \times {\boldsymbol {p}}}\mathbf  =\mathbf {r} \times {\boldsymbol {p}}

where p is the particle's linear momentum and r is the position vector from the origin. The time-derivative of this is:

dLdt=r×dpdt drdt×p.{\displaystyle {\frac {\mathrm {d} \mathbf }{\mathrm {d} t}}=\mathbf {r} \times {\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}} {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\times {\boldsymbol {p}}.}{\displaystyle {\frac {\mathrm {d} \mathbf  }{\mathrm {d} t}}=\mathbf {r} \times {\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}} {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\times {\boldsymbol {p}}.}

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definition of force F=dpdt{\displaystyle \mathbf ={\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}}{\displaystyle \mathbf  ={\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}} (whether or not mass is constant) and the definition of velocity drdt=v{\displaystyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\mathbf {v} }{\displaystyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\mathbf {v} }

dLdt=r×F v×p.{\displaystyle {\frac {\mathrm {d} \mathbf }{\mathrm {d} t}}=\mathbf {r} \times \mathbf \mathbf {v} \times {\boldsymbol {p}}.}{\displaystyle {\frac {\mathrm {d} \mathbf  }{\mathrm {d} t}}=\mathbf {r} \times \mathbf   \mathbf {v} \times {\boldsymbol {p}}.}

The cross product of momentum p{\displaystyle {\boldsymbol {p}}}{\boldsymbol {p}} with its associated velocity v{\displaystyle \mathbf {v} }\mathbf {v} is zero because velocity and momentum are parallel, so the second term vanishes.

By definition, torque τ = r × F. Therefore, torque on a particle is equal to the first derivative of its angular momentum with respect to time.

If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that

dLdt=r×Fnet=τnet.{\displaystyle {\frac {\mathrm {d} \mathbf }{\mathrm {d} t}}=\mathbf {r} \times \mathbf _{\mathrm {net} }={\boldsymbol {\tau }}_{\mathrm {net} }.}{\displaystyle {\frac {\mathrm {d} \mathbf  }{\mathrm {d} t}}=\mathbf {r} \times \mathbf  _{\mathrm {net} }={\boldsymbol {\tau }}_{\mathrm {net} }.}

This is a general proof for point particles.

The proof can be generalized to a system of point particles by applying the above proof to each of the point particles and then summing over all the point particles. Similarly, the proof can be generalized to a continuous mass by applying the above proof to each point within the mass, and then integrating over the entire mass.

Units

Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule.[9][10] The unit newton metre is properly denoted N⋅m.[10]

The traditional Imperial and U.S. customary units for torque are the pound foot (lbf-ft), or for small values the pound inch (lbf-in). In the US, torque is most commonly referred to as the foot-pound (denoted as either lb-ft or ft-lb) and the inch-pound (denoted as in-lb).[11][12] Practitioners depend on context and the hyphen in the abbreviation to know that these refer to torque and not to energy or moment of mass (as the symbolism ft-lb would properly imply).

Special cases and other facts

Moment arm formula

Moment arm diagram

A very useful special case, often given as the definition of torque in fields other than physics, is as follows:

τ=(moment arm)(force).{\displaystyle \tau =({\text{moment arm}})({\text{force}}).}\tau =({\text{moment arm}})({\text{force}}).

The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in three-dimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:

τ=(distance to centre)(force).{\displaystyle \tau =({\text{distance to centre}})({\text{force}}).}\tau =({\text{distance to centre}})({\text{force}}).

For example, if a person places a force of 10 N at the terminal end of a wrench that is 0.5 m long (or a force of 10 N exactly 0.5 m from the twist point of a wrench of any length), the torque will be 5 N⋅m – assuming that the person moves the wrench by applying force in the plane of movement and perpendicular to the wrench.

The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.

Static equilibrium

For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in two-dimensions, three equations are used.

Net force versus torque

When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a current-carrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force F{\displaystyle \mathbf }\mathbf is not zero, and τ1{\displaystyle {\boldsymbol {\tau }}_}{\boldsymbol {\tau }}_ is the torque measured from r1{\displaystyle \mathbf {r} _}\mathbf {r} _, then the torque measured from r2{\displaystyle \mathbf {r} _}\mathbf {r} _ is

τ2=τ1 (r1−r2)×F{\displaystyle {\boldsymbol {\tau }}_={\boldsymbol {\tau }}_ (\mathbf {r} _-\mathbf {r} _)\times \mathbf }
{\displaystyle {\boldsymbol {\tau }}_={\boldsymbol {\tau }}_ (\mathbf {r} _-\mathbf {r} _)\times \mathbf  }

Machine torque

Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis shows the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in newton metres) that the engine is capable of providing at that speed.

Torque forms part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internal-combustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). One can measure the varying torque output over that range with a dynamometer, and show it as a torque curve.

Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam-engines and electric motors can start heavy loads from zero rpm without a clutch.

Relationship between torque, power, and energy

If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass, the work W can be expressed as

W=∫θ1θ2τ dθ,{\displaystyle W=\int _{\theta _}^{\theta _}\tau \ \mathrm {d} \theta ,}W=\int _{\theta _}^{\theta _}\tau \ \mathrm {d} \theta ,

where τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[13]

Proof

The work done by a variable force acting over a finite linear displacement s{\displaystyle s}s is given by integrating the force with respect to an elemental linear displacement ds{\displaystyle \mathrm {d} \mathbf {s} }{\displaystyle \mathrm {d} \mathbf {s} }

W=∫s1s2F⋅ds{\displaystyle W=\int _{s_}^{s_}\mathbf \cdot \mathrm {d} \mathbf {s} }{\displaystyle W=\int _{s_}^{s_}\mathbf  \cdot \mathrm {d} \mathbf {s} }

However, the infinitesimal linear displacement ds{\displaystyle \mathrm {d} \mathbf {s} }{\displaystyle \mathrm {d} \mathbf {s} } is related to a corresponding angular displacement dθ{\displaystyle \mathrm {d} {\boldsymbol {\theta }}}{\displaystyle \mathrm {d} {\boldsymbol {\theta }}} and the radius vector r{\displaystyle \mathbf {r} }\mathbf {r} as

ds=dθ×r{\displaystyle \mathrm {d} \mathbf {s} =\mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }{\displaystyle \mathrm {d} \mathbf {s} =\mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }

Substitution in the above expression for work gives

W=∫s1s2F⋅dθ×r{\displaystyle W=\int _{s_}^{s_}\mathbf \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }{\displaystyle W=\int _{s_}^{s_}\mathbf  \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }

The expression F⋅dθ×r{\displaystyle \mathbf \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }{\displaystyle \mathbf  \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} } is a scalar triple product given by [Fdθr]{\displaystyle \left[\mathbf \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]}{\displaystyle \left[\mathbf  \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]}. An alternate expression for the same scalar triple product is

[Fdθr]=r×F⋅dθ{\displaystyle \left[\mathbf \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]=\mathbf {r} \times \mathbf \cdot \mathrm {d} {\boldsymbol {\theta }}}{\displaystyle \left[\mathbf  \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]=\mathbf {r} \times \mathbf  \cdot \mathrm {d} {\boldsymbol {\theta }}}

But as per the definition of torque,

τ=r×F{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf }{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf  }

Corresponding substitution in the expression of work gives,

W=∫s1s2τ⋅dθ{\displaystyle W=\int _{s_}^{s_}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}{\displaystyle W=\int _{s_}^{s_}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}

Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving

W=∫θ1θ2τ⋅dθ{\displaystyle W=\int _{\theta _}^{\theta _}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}{\displaystyle W=\int _{\theta _}^{\theta _}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}

If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e., τ⋅dθ=|τ||dθ|cos⁡0=τdθ{\displaystyle {\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}=\left|{\boldsymbol {\tau }}\right|\left|\mathrm {d} {\boldsymbol {\theta }}\right|\cos 0=\tau \,\mathrm {d} \theta }{\displaystyle {\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}=\left|{\boldsymbol {\tau }}\right|\left|\mathrm {d} {\boldsymbol {\theta }}\right|\cos 0=\tau \,\mathrm {d} \theta } giving

W=∫θ1θ2τdθ{\displaystyle W=\int _{\theta _}^{\theta _}\tau \,\mathrm {d} \theta }{\displaystyle W=\int _{\theta _}^{\theta _}\tau \,\mathrm {d} \theta }

It follows from the work-energy theorem that W also represents the change in the rotational kinetic energy Er of the body, given by

Er=12Iω2,{\displaystyle E_{\mathrm {r} }={\tfrac }I\omega ^,}E_{\mathrm {r} }={\tfrac }I\omega ^,

where I is the moment of inertia of the body and ω is its angular speed.[13]

Power is the work per unit time, given by

P=τ⋅ω,{\displaystyle P={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }},}{\displaystyle P={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }},}

where P is power, τ is torque, ω is the angular velocity, and ⋅{\displaystyle \cdot }\cdot represents the scalar product.

Algebraically, the equation may be rearranged to compute torque for a given angular speed and power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).

In practice, this relationship can be observed in bicycles: Bicycles are typically composed of two road wheels, front and rear gears (referred to as sprockets) meshing with a circular chain, and a derailleur mechanism if the bicycle's transmission system allows multiple gear ratios to be used (i.e. multi-speed bicycle), all of which attached to the frame. A cyclist, the person who rides the bicycle, provides the input power by turning pedals, thereby cranking the front sprocket (commonly referred to as chainring). The input power provided by the cyclist is equal to the product of cadence (i.e. the number of pedal revolutions per minute) and the torque on spindle of the bicycle's crankset. The bicycle's drivetrain transmits the input power to the road wheel, which in turn conveys the received power to the road as the output power of the bicycle. Depending on the gear ratio of the bicycle, a (torque, rpm)input pair is converted to a (torque, rpm)output pair. By using a larger rear gear, or by switching to a lower gear in multi-speed bicycles, angular speed of the road wheels is decreased while the torque is increased, product of which (i.e. power) does not change.

Consistent units must be used. For metric SI units, power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).

Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar. This means that the dimensional equivalence of the newton metre and the joule may be applied in the former, but not in the latter case. This problem is addressed in orientational analysis which treats radians as a base unit rather than a dimensionless unit.[14]

Conversion to other units

A conversion factor may be necessary when using different units of power or torque. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution. In the following formulas, P is power, τ is torque, and ν (Greek letter nu) is rotational speed.

P=τ⋅2π⋅ν{\displaystyle P=\tau \cdot 2\pi \cdot \nu }{\displaystyle P=\tau \cdot 2\pi \cdot \nu }

Showing units:

P(W)=τ(N⋅m)⋅2π(rad/rev)⋅ν(rev/sec){\displaystyle P({\rm })=\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rev/sec)}}}{\displaystyle P({\rm })=\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rev/sec)}}}

Dividing by 60 seconds per minute gives us the following.

P(W)=τ(N⋅m)⋅2π(rad/rev)⋅ν(rpm)60{\displaystyle P({\rm })={\frac {\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rpm)}}}}}{\displaystyle P({\rm })={\frac {\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rpm)}}}}}

where rotational speed is in revolutions per minute (rpm).

Some people (e.g., American automotive engineers) use horsepower (mechanical) for power, foot-pounds (lbf⋅ft) for torque and rpm for rotational speed. This results in the formula changing to:

P(hp)=τ(lbf⋅ft)⋅2π(rad/rev)⋅ν(rpm)33,000.{\displaystyle P({\rm {hp}})={\frac {\tau {\rm {(lbf\cdot ft)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu ({\rm {rpm}})}{33,000}}.}{\displaystyle P({\rm {hp}})={\frac {\tau {\rm {(lbf\cdot ft)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu ({\rm {rpm}})}{33,000}}.}

The constant below (in foot-pounds per minute) changes with the definition of the horsepower; for example, using metric horsepower, it becomes approximately 32,550.

The use of other units (e.g., BTU per hour for power) would require a different custom conversion factor.

Derivation

For a rotating object, the linear distance covered at the circumference of rotation is the product of the radius with the angle covered. That is: linear distance = radius × angular distance. And by definition, linear distance = linear speed × time = radius × angular speed × time.

By the definition of torque: torque = radius × force. We can rearrange this to determine force = torque ÷ radius. These two values can be substituted into the definition of power:

power=force⋅linear distancetime=(torquer)⋅(r⋅angular speed⋅t)t=torque⋅angular speed.{\displaystyle {\begin{aligned}{\text{power}}&={\frac {{\text{force}}\cdot {\text{linear distance}}}{\text{time}}}\[6pt]&={\frac {\left({\dfrac {\text{torque}}{r}}\right)\cdot (r\cdot {\text{angular speed}}\cdot t)}{t}}\[6pt]&={\text{torque}}\cdot {\text{angular speed}}.\end{aligned}}}{\displaystyle {\begin{aligned}{\text{power}}&={\frac {{\text{force}}\cdot {\text{linear distance}}}{\text{time}}}\[6pt]&={\frac {\left({\dfrac {\text{torque}}{r}}\right)\cdot (r\cdot {\text{angular speed}}\cdot t)}{t}}\[6pt]&={\text{torque}}\cdot {\text{angular speed}}.\end{aligned}}}

The radius r and time t have dropped out of the equation. However, angular speed must be in radians per unit of time, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:

power=torque⋅2π⋅rotational speed.{\displaystyle {\text{power}}={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}.\,}{\displaystyle {\text{power}}={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}.\,}

If torque is in newton metres and rotational speed in revolutions per second, the above equation gives power in newton metres per second or watts. If Imperial units are used, and if torque is in pounds-force feet and rotational speed in revolutions per minute, the above equation gives power in foot pounds-force per minute. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower:

power=torque⋅2π⋅rotational speed⋅ft⋅lbfmin⋅horsepower33,000⋅ft⋅lbfmin≈torque⋅RPM5,252{\displaystyle {\begin{aligned}{\text{power}}&={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}\cdot {\frac {\text{horsepower}}{33,000\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}}}\[6pt]&\approx {\frac {{\text{torque}}\cdot {\text}}{5,252}}\end{aligned}}}{\displaystyle {\begin{aligned}{\text{power}}&={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}\cdot {\frac {\text{horsepower}}{33,000\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}}}\[6pt]&\approx {\frac {{\text{torque}}\cdot {\text}}{5,252}}\end{aligned}}}

because 5252.113122≈33,0002π.{\displaystyle 5252.113122\approx {\frac {33,000}{2\pi }}.\,}5252.113122\approx {\frac {33,000}{2\pi }}.\,

Principle of moments

The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:

(r×F1) (r×F2) ⋯=r×(F1 F2 ⋯).{\displaystyle (\mathbf {r} \times \mathbf _) (\mathbf {r} \times \mathbf _) \cdots =\mathbf {r} \times (\mathbf _ \mathbf _ \cdots ).}(\mathbf {r} \times \mathbf  _) (\mathbf {r} \times \mathbf  _) \cdots =\mathbf {r} \times (\mathbf  _ \mathbf  _ \cdots ).

From this it follows that if a pivoted beam of zero mass is balanced with two opposed forces then:

(r×F1)=(r×F2).{\displaystyle (\mathbf {r} \times \mathbf _)=(\mathbf {r} \times \mathbf _).}{\displaystyle (\mathbf {r} \times \mathbf  _)=(\mathbf {r} \times \mathbf  _).}

Torque multiplier

Main article: Torque multiplier

Torque can be multiplied via three methods: by locating the fulcrum such that the length of a lever is increased; by using a longer lever; or by the use of a speed reducing gearset or gear box. Such a mechanism multiplies torque, as rotation rate is reduced.

See also

  • Moment
  • Conversion of units
  • Friction torque
  • Mechanical equilibrium
  • Rigid body dynamics
  • Statics
  • Torque converter
  • Torque limiter
  • Torque screwdriver
  • Torque tester
  • Torque wrench
  • Torsion (mechanics)

References

  1. ^ Serway, R. A. and Jewett, Jr. J.W. (2003). Physics for Scientists and Engineers. 6th Ed. Brooks Cole. ISBN 0-534-40842-7.
  2. ^ Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4.
  3. ^ Thomson, James; Larmor, Joseph (1912). Collected Papers in Physics and Engineering. University Press. p. civ.
  4. ^ a b Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link
  5. ^ a b c Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download.
  6. ^ "Right Hand Rule for Torque". Retrieved 2007-09-08.
  7. ^ a b Halliday, David; Resnick, Robert (1970). Fundamentals of Physics. John Wiley & Sons, Inc. pp. 184–85.
  8. ^ Knight, Randall; Jones, Brian; Field, Stuart (2016). College Physics: A Strategic Approach. Jones, Brian, 1960-, Field, Stuart, 1958- (Third edition, technology update ed.). Boston: Pearson. p. 199. ISBN 9780134143323. OCLC 922464227.
  9. ^ From the official SI website: "...For example, the quantity torque is the cross product of a position vector and a force vector. The SI unit is newton metre. Even though torque has the same dimension as energy (SI unit joule), the joule is never used for expressing torque."
  10. ^ a b "SI brochure Ed. 9, Section 2.3.4" (PDF). Bureau International des Poids et Mesures. 2019. Retrieved 2020-05-29.
  11. ^ "Dial Torque Wrenches from Grainger". Grainger. 2020. Demonstration that, as in most US industrial settings, the torque ranges are given in ft-lb rather than lbf-ft.
  12. ^ Erjavec, Jack (22 January 2010). Manual Transmissions & Transaxles: Classroom manual. p. 38. ISBN 978-1-4354-3933-7.
  13. ^ a b Kleppner, Daniel; Kolenkow, Robert (1973). An Introduction to Mechanics. McGraw-Hill. pp. 267–68. ISBN 9780070350489.
  14. ^ Page, Chester H. (1979). "Rebuttal to de Boer's "Group properties of quantities and units"". American Journal of Physics. 47 (9): 820. Bibcode:1979AmJPh..47..820P. doi:10.1119/1.11704.

External links

Look up torque in Wiktionary, the free dictionary.
Wikimedia Commons has media related to Torque.
  • "Horsepower and Torque" An article showing how power, torque, and gearing affect a vehicle's performance.
  • "Torque vs. Horsepower: Yet Another Argument" An automotive perspective
  • Torque and Angular Momentum in Circular Motion on Project PHYSNET.
  • An interactive simulation of torque
  • Torque Unit Converter
  • A feel for torque An order-of-magnitude interactive.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Torque&oldid=1061290529"
convertunits.com

The answer is 1.3558179483314. We assume you are converting between newton-meter and pound foot. You can view more details on each measurement unit: newton meters or pound feet. The SI derived unit for torque is the newton meter. 1 …

wrenchturn.com

Convert Nm to lb-ft The Newton-meter is an internationally recognized unit of torque commonly used in countries such as Japan and Germany. It is derived from the SI system using the metric units Newton (force) and meter (distance).

Convert Pound-force feet to Newton metres (lbf·ft → Nm)

1 Pound-force feet = 1.3558 Newton metres: 10 Pound-force feet = 13.5582 Newton metres: 2500 Pound-force feet = 3389.54 Newton metres: 2 Pound-force feet = 2.7116 Newton metres: 20 Pound-force feet = 27.1164 Newton metres: 5000 Pound-force feet = 6779.09 Newton metres: 3 Pound-force feet = 4.0675 Newton metres: 30 Pound-force feet = 40.6745 Newton metres: …

Precision: decimal digits

Convert from Pound-force feet to Newton metres. Type in the amount you want to convert and press the Convert button.

Belongs in category
Torque

  • To other units
  • Conversion table
  • For your website
1 Pound-force feet = 1.3558 Newton metres 10 Pound-force feet = 13.5582 Newton metres 2500 Pound-force feet = 3389.54 Newton metres
2 Pound-force feet = 2.7116 Newton metres 20 Pound-force feet = 27.1164 Newton metres 5000 Pound-force feet = 6779.09 Newton metres
3 Pound-force feet = 4.0675 Newton metres 30 Pound-force feet = 40.6745 Newton metres 10000 Pound-force feet = 13558.18 Newton metres
4 Pound-force feet = 5.4233 Newton metres 40 Pound-force feet = 54.2327 Newton metres 25000 Pound-force feet = 33895.45 Newton metres
5 Pound-force feet = 6.7791 Newton metres 50 Pound-force feet = 67.7909 Newton metres 50000 Pound-force feet = 67790.9 Newton metres
6 Pound-force feet = 8.1349 Newton metres 100 Pound-force feet = 135.58 Newton metres 100000 Pound-force feet = 135581.79 Newton metres
7 Pound-force feet = 9.4907 Newton metres 250 Pound-force feet = 338.95 Newton metres 250000 Pound-force feet = 338954.49 Newton metres
8 Pound-force feet = 10.8465 Newton metres 500 Pound-force feet = 677.91 Newton metres 500000 Pound-force feet = 677908.97 Newton metres
9 Pound-force feet = 12.2024 Newton metres 1000 Pound-force feet = 1355.82 Newton metres 1000000 Pound-force feet = 1355817.95 Newton metres

Embed this unit converter in your page or blog, by copying the following HTML code:

Your browser does not support iframes. convertlive.

convertlive

Convert Pound Force Foot to Newton Meters (lbf ft to Nm ...

1 lbf ft. = 12.000000017701 inch pounds force (in lbf) 1 lbf ft. = 0.00135581795 kilonewton meters (kN m) 1 lbf ft. = 1355817.95 meganewton meters (MN m) Pound Force Foot. to Newton Meters (table conversion) 1 lbf ft.


Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)

Pound Force FootSymbol/abbreviation: lbf ftUnit of: TORQUETORQUE's base unit: newton meters (SI Unit)In relation to the base unit (newton meters), 1 Pound Force Foot = 1.35581795 newton meters.
1 lbf ft= 13558179.5 dyne centimeters (dyn cm)
1 lbf ft= 32.174048596027 foot poundals (ft pdl)
1 lbf ft= 1 foot pounds force (lbf路ft)
1 lbf ft= 12.000000017701 inch pounds force (in lbf)
1 lbf ft= 13.825495454615 kilogram force centimeters (kgf cm)
1 lbf ft= 0.13825495454615 kilogram force meters (kgf m)
1 lbf ft= 0.00135581795 kilonewton meters (kN m)
1 lbf ft= 1355817.95 meganewton meters (MN m)
1 lbf ft= 1355817.95 micronewton meters (碌N m)
1 lbf ft= 1355.81795 millinewton meters (mN m)
1 lbf ft= 135.581795 newton centimeters (N cm)
1 lbf ft= 1.35581795 newton meters (Nm)
1 lbf ft= 1 pound foot (lb ft)
1 lbf ft= 1 pound force foot (lbf ft)
1 lbf ft= 12.000000017701 pound force inch (lbf in)
fastunitconverter.com

Convert ft lb to Nm. Please provide values below to convert foot-pound [ft*lbf] to newton meter [N*m], or vice versa.

1library.net

Nm To Ft Lbs Conversion Table . 17 0 Download (0) 0 Download (0)

daprofitclub.com

Why used the ft lbs to nm calculator for conversion? Energy conversion Calculator by DaProfitClub helps you Convert ft lbs to nm easily and efficiently. You do not need any knowledge or technical skills to use this calculator. This is free online energy conversion tool. How does this conversion calculator work? * this value by 1.356, to see the ...

domiciliotrieste.it

3 hours ago · Aug 11, 2021 · Instead, we're 5 feet 10 inches. 81 in. Many Other Conversions. Convert Meter Meters. 3700787. 110. 905 13/16 0. 3/64 inches to decimal. If you have a volume measuring 11 cm by 5 cm by 2 cm, the volume is 110 cubic centimeters (11 x 5 x 2). When measured diagonally as a rectangle, the iPad Pro 12. 75" 13/16" =.

sparkyexpress.ca

One pound (force) = 4.448 222 newtons One foot = 0.3048 m This gives the conversion factor: One pound-foot = 1.35582 newton metres. The name "pound-foot", intended to minimize confusion with the foot-pound as a unit of work, was apparently first proposed by British physicist Arthur Mason Worthington.

flightpedia.org

2214 Newton Meters is equal to 1,627.94 Foot Pound Force. Formula to convert 2214 Nm to ft lbf is 2214 / 1.36 Q: How many Newton Meters in 2214 Foot Pounds Force?

kylesconverter.com

20 Foot-pounds Force to Newton Meters = 27.1164: 900 Foot-pounds Force to Newton Meters = 1220.2362: 30 Foot-pounds Force to Newton Meters = 40.6745: 1,000 Foot-pounds Force to Newton Meters = 1355.8179: 40 Foot-pounds Force to Newton Meters = 54.2327: 10,000 Foot-pounds Force to Newton Meters = 13558.1795: 50 Foot-pounds Force to Newton Meters ...

Reverse conversion?
Newton Meters to Foot-pounds Force
(or just enter a value in the "to" field)
Unit Descriptions
1 Foot-Pound Force:
g x 1 lb x 1 ft
1 Newton Meter:
kg·m2/s2. SI unit.
Conversions Table
1 Foot-pounds Force to Newton Meters = 1.355870 Foot-pounds Force to Newton Meters = 94.9073
2 Foot-pounds Force to Newton Meters = 2.711680 Foot-pounds Force to Newton Meters = 108.4654
3 Foot-pounds Force to Newton Meters = 4.067590 Foot-pounds Force to Newton Meters = 122.0236
4 Foot-pounds Force to Newton Meters = 5.4233100 Foot-pounds Force to Newton Meters = 135.5818
5 Foot-pounds Force to Newton Meters = 6.7791200 Foot-pounds Force to Newton Meters = 271.1636
6 Foot-pounds Force to Newton Meters = 8.1349300 Foot-pounds Force to Newton Meters = 406.7454
7 Foot-pounds Force to Newton Meters = 9.4907400 Foot-pounds Force to Newton Meters = 542.3272
8 Foot-pounds Force to Newton Meters = 10.8465500 Foot-pounds Force to Newton Meters = 677.909
9 Foot-pounds Force to Newton Meters = 12.2024600 Foot-pounds Force to Newton Meters = 813.4908
10 Foot-pounds Force to Newton Meters = 13.5582800 Foot-pounds Force to Newton Meters = 1084.6544
20 Foot-pounds Force to Newton Meters = 27.1164900 Foot-pounds Force to Newton Meters = 1220.2362
30 Foot-pounds Force to Newton Meters = 40.67451,000 Foot-pounds Force to Newton Meters = 1355.8179
40 Foot-pounds Force to Newton Meters = 54.232710,000 Foot-pounds Force to Newton Meters = 13558.1795
50 Foot-pounds Force to Newton Meters = 67.7909100,000 Foot-pounds Force to Newton Meters = 135581.7948
60 Foot-pounds Force to Newton Meters = 81.34911,000,000 Foot-pounds Force to Newton Meters = 1355817.9483

Similar Torque Units

Common Units

Newton-Meters to Foot-Pounds Converter - The Calculator Site

How to convert newton-meters to foot-pounds. 1 newton-meter is equal to 0.7376 ft-lbs, rounded to 4 decimal places. To convert newton-meters to foot-pounds, simply multiply your figure by 0.7376. The conversion chart below can be used to convert some of the most common torque figures for newton-meters (Nm) and foot-pounds (ft-lb).

  • Nm to foot-pounds
  • Nm to inch-pounds

Use this conversion tool to convert between newton-meters (Nm) and foot-pounds (ft-lb), both of which are units of work, energy or torque (pound-foot).

Advertisements

Please help me spread the word by sharing this with friends or on your website/blog. Thank you.

Disclaimer: Whilst every effort has been made in building this calculator, we are not to be held liable for any damages or monetary losses arising out of or in connection with the use of it. This tool is here purely as a service to you, please use it at your own risk. Full disclaimer. Do not use calculations for anything where loss of life, money, property, etc could result from inaccurate calculations.

1 newton-meter is equal to 0.7376 ft-lbs, rounded to 4 decimal places. To convert newton-meters to foot-pounds, simply multiply your figure by 0.7376.

The conversion chart below can be used to convert some of the most common torque figures for newton-meters (Nm) and foot-pounds (ft-lb).

Newton-meters to foot-pounds chart

If you wish to convert between inch-pounds and foot-pounds, take a look at the inch-pounds and foot-pounds converter.

Advertisement

For any figures not represented in our Nm and ft-lb chart, please use the converter at the top of the page.

protoindustrial.com

100. 73.76. 1 Nm = 0.73756 ft/lb. NOTE: The International Metric System measures torque in Newton meters (Nm). A Newton is defined as the force necessary to move one kilogram one meter per second.

buildingclub.info

How to convert from lb/ft to kN/m? Correctly measure in lbf/ft (pound-force per foot). 1 lbf/ft= 1 lb/ft. 1000 lb/ft = 1000 lbf/ft =14.59390 kN/m (gravitational acceleration (free fall) g = 9.80665 m/s 2 ≈ 32.174048 ft/s 2 ). 1 lb/ft = 1 lbf/ft = 0.01459399 kN/m

convert-to.com

Exchange reading in foot pounds unit ft lb into newton meters unit N-m as in an equivalent measurement result (two different units but the same identical physical total value, which is also equal to their proportional parts when divided or multiplied). One foot pound converted into newton meter equals = 1.36 N-m 1 ft lb = 1.36 N-m

coolstuffshub.com

Newton meters to foot pounds (Nm to ft-lb) conversion calculator with tables and formula. Energy conversion has never been so easy! = Home Length Area Volume Weight Speed. Energy converter > Newton meters conversion > Newton meters to foot pounds. Newton meters to foot pounds. 1 Nm = 0.73757890546509 ft-lb.

metric-conversions.org

Newton-meters to Foot-pounds formula. ft-lb = Nm * 0.73756 . Foot-pounds. One foot pound is the work done by a force of one pounl acting through a distance of one foot, in the direction of the force. It equates to 1.355 817 948 331 4004 joules . Newton-meters to Foot-pounds table. Start Increments Accuracy Format Print table < Smaller Values Larger Values > Newton-meters Foot-pounds; 0 Nm: 0 ...

henryfilms.com

torque = (moment of inertia) (angular acceleration) τ = Iα. The first formula is for metric units, whereas the second for imperial ones. Torque Conversion . Micrometer "Click" T

rollerbearings.pl

37 minutes ago · Features a durable 22" (56 cm) steel deck and a 21" (53. 74"] Comes with gaskets Manual choke lever Fuel shut off lever Idle adjustment screw 16100/168FB JF168-I-04 JF168FLH-8B. A simple liters to cubic inch displacement converter (also converts to cubic centimeters) Mar 12, 2018 · Solved: Hello from Germany, after installing Windows 10 Home ...

graeberroad.com

/a > convert /a. 1 m 2 = 1 cm * 0.000001. cubic meter = cubic cm * 1 *. Equal to one hundredth of a meter really just have to remind ourselves what the.... 1Cm / 2.54cm/in = 0.393