Search results for: Convert 1 lb ft to nm
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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 ...
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1 hour ago · Convert inches to feet or convert inches to miles to find your final distance measurement. August 5, 2021. 2  5/4x6 22x6 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 ...
Tdi injector torque specs
7 hours ago · 11. Nov 02, 2010 · Twopiece 33 mm flange nuts used with hubpiloted wheels should be tightened to a torque of 450 to 500 footpounds. Return Icon. 500 4290003 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.
It is defined as the energy transferred when a force of one poundforce is applied over a linear displacement of one foot. History/origin: The footpound is an English Engineering unit that is part of the British Gravitational system, based on the footpoundsecond (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 footpound [ft*lbf] to newton meter [N*m], or vice versa.
Footpound
Definition: A footpound (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 poundforce is applied over a linear displacement of one foot.
History/origin: The footpound is an English Engineering unit that is part of the British Gravitational system, based on the footpoundsecond (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 footpound 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 footpound is still used in certain applications, though even in the US, the joule is preferred in most scientific contexts.
Newtonmeter
Definition: The newtonmeter (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 newtonmeter 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 newtonmeter.
Current use: The newtonmeter is more commonly used as a unit of torque rather than as a unit of work or energy. As such, the use of the newtonmeter as a unit of work or energy is discouraged to avoid confusion regarding whether energy or torque is being expressed.
Footpound [ft*lbf]  Newton Meter [N*m] 

0.01 ft*lbf  0.0135581795 N*m 
0.1 ft*lbf  0.1355817948 N*m 
1 ft*lbf  1.3558179483 N*m 
2 ft*lbf  2.7116358966 N*m 
3 ft*lbf  4.0674538449 N*m 
5 ft*lbf  6.7790897414 N*m 
10 ft*lbf  13.5581794829 N*m 
20 ft*lbf  27.1163589658 N*m 
50 ft*lbf  67.7908974145 N*m 
100 ft*lbf  135.581794829 N*m 
1000 ft*lbf  1355.8179482896 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
lbft or Nm. The SI derived unit for torque is the newton meter. 1 lbft 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  lbft  to  newtonmeter newtonmetre 
›› More information from the unit converter
How many lbft in 1 Nm?
The answer is 0.73756214927727.
We assume you are converting between pound foot and newtonmeter.
You can view more details on each measurement unit:
lbft or Nm
The SI derived unit for torque is the newton meter. 1 lbft 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 lbft to Nm
1 lbft to Nm = 1.35582 Nm
5 lbft to Nm = 6.77909 Nm
10 lbft to Nm = 13.55818 Nm
15 lbft to Nm = 20.33727 Nm
20 lbft to Nm = 27.11636 Nm
25 lbft to Nm = 33.89545 Nm
30 lbft to Nm = 40.67454 Nm
40 lbft to Nm = 54.23272 Nm
50 lbft to Nm = 67.7909 Nm
You can do the reverse unit conversion from Nm to lbft, or enter any two units below:
lbft to ouncefoot
lbft to poundalfoot
lbft to micronewtonmeter
lbft to millinewtonmeter
lbft to kilopondmeter
lbft to newton metre
lbft to foot pound
lbft to gramcentimeter
lbft to kilogrammeter
lbft to kilogramcentimeter
›› 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!
We assume you are converting between pound foot and newtonmeter. You can view more details on each measurement unit: lbft or newtonmeter The SI derived unit for torque is the newton meter. 1 lbft is equal to 1.3558179483314 newton meter. Note that rounding errors …
How many lbft in 1 newtonmeter?
The answer is 0.73756214927727.
We assume you are converting between pound foot and newtonmeter.
You can view more details on each measurement unit:
lbft or newtonmeter
The SI derived unit for torque is the newton meter. 1 lbft 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 lbft to newtonmeter
1 lbft to newtonmeter = 1.35582 newtonmeter
5 lbft to newtonmeter = 6.77909 newtonmeter
10 lbft to newtonmeter = 13.55818 newtonmeter
15 lbft to newtonmeter = 20.33727 newtonmeter
20 lbft to newtonmeter = 27.11636 newtonmeter
25 lbft to newtonmeter = 33.89545 newtonmeter
30 lbft to newtonmeter = 40.67454 newtonmeter
40 lbft to newtonmeter = 54.23272 newtonmeter
50 lbft to newtonmeter = 67.7909 newtonmeter
You can do the reverse unit conversion from newtonmeter to lbft, or enter any two units below:
lbft to kilogrammeter
lbft to foot pound
lbft to gramcentimeter
lbft to poundinch
lbft to millinewtonmeter
lbft to newtoncentimeter
lbft to poundalfoot
lbft to kilonewtonmeter
lbft to newton metre
lbft 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!
Footpounds Newtonmeters; 0 ftlb: 0.00 Nm: 1 ftlb: 1.36 Nm: 2 ftlb: 2.71 Nm: 3 ftlb: 4.07 Nm: 4 ftlb: 5.42 Nm: 5 ftlb: 6.78 Nm: 6 ftlb: 8.13 Nm: 7 ftlb: 9.49 Nm: 8 ftlb: 10.85 Nm: 9 ftlb: 12.20 Nm: 10 ftlb: 13.56 Nm: 11 ftlb: 14.91 Nm: 12 ftlb: 16.27 Nm: 13 ftlb: 17.63 Nm: 14 ftlb: 18.98 Nm: 15 ftlb: 20.34 Nm: 16 ftlb: 21.69 Nm: 17 ftlb: 23.05 Nm: 18 ftlb: 24.40 Nm: 19 ftlb: 25.76 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 ...
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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)  ⇛ 
Convert lbft to Nm Poundfeet (lbft), sometimes called "footpounds" (ftlbs), is a unit of torque mainly used in the United States. The Newtonmeter is an internationally recognized unit of torque commonly used in countries such as Japan and Germany.
Conversion Pound foot to Newton meter. A poundfoot (lb·ft or lbf·ft) is a unit of torque. One poundfoot 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 poundfoot (lb·ft or lbf·ft) is a unit of torque. One poundfoot 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 nm  11 pound foot = 14.914 nm  21 pound foot = 28.4722 nm 
2 pound foot = 2.7116 nm  12 pound foot = 16.2698 nm  22 pound foot = 29.828 nm 
3 pound foot = 4.0675 nm  13 pound foot = 17.6256 nm  23 pound foot = 31.1838 nm 
4 pound foot = 5.4233 nm  14 pound foot = 18.9815 nm  24 pound foot = 32.5396 nm 
5 pound foot = 6.7791 nm  15 pound foot = 20.3373 nm  25 pound foot = 33.8954 nm 
6 pound foot = 8.1349 nm  16 pound foot = 21.6931 nm  26 pound foot = 35.2513 nm 
7 pound foot = 9.4907 nm  17 pound foot = 23.0489 nm  27 pound foot = 36.6071 nm 
8 pound foot = 10.8465 nm  18 pound foot = 24.4047 nm  28 pound foot = 37.9629 nm 
9 pound foot = 12.2024 nm  19 pound foot = 25.7605 nm  29 pound foot = 39.3187 nm 
10 pound foot = 13.5582 nm  20 pound foot = 27.1164 nm  30 pound foot = 40.6745 nm 
40 pound foot = 54.2327 nm  70 pound foot = 94.9073 nm  100 pound foot = 135.582 nm 
50 pound foot = 67.7909 nm  80 pound foot = 108.465 nm  110 pound foot = 149.14 nm 
60 pound foot = 81.3491 nm  90 pound foot = 122.024 nm  120 pound foot = 162.698 nm 
200 pound foot = 271.164 nm  500 pound foot = 677.909 nm  800 pound foot = 1084.65 nm 
300 pound foot = 406.745 nm  600 pound foot = 813.491 nm  900 pound foot = 1220.24 nm 
400 pound foot = 542.327 nm  700 pound foot = 949.073 nm  1000 pound foot = 1355.82 nm 
Torque Conversions
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 footpounds. Foot Pounds to Newton Meters (ftlb to nm): It is a free online footpound to newton meters (ftlb to nm) energy converter. A footpound is a measuring unit of energy and denoted as "ft.lb or ft.lbf".
Add Footpound to Newton meter converter to your website to use this unit converter directly. Feel hasslefree 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 Footpound to Other Energy ...
Instant free online tool for newton meter to footpound conversion or vice versa. The newton meter [N*m] to footpound [ft*lbf] conversion table and conversion steps are also listed. Also, explore tools to convert newton meter or footpound to other energy units or learn more about energy conversions.
Please provide values below to convert newton meter [N*m] to footpound [ft*lbf], or vice versa.
Newtonmeter
Definition: The newtonmeter (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 newtonmeter 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 newtonmeter.
Current use: The newtonmeter is more commonly used as a unit of torque rather than as a unit of work or energy. As such, the use of the newtonmeter as a unit of work or energy is discouraged to avoid confusion regarding whether energy or torque is being expressed.
Footpound
Definition: A footpound (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 poundforce is applied over a linear displacement of one foot.
History/origin: The footpound is an English Engineering unit that is part of the British Gravitational system, based on the footpoundsecond (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 footpound 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 footpound is still used in certain applications, though even in the US, the joule is preferred in most scientific contexts.
Newton Meter [N*m]  Footpound [ft*lbf] 

0.01 N*m  0.0073756215 ft*lbf 
0.1 N*m  0.0737562149 ft*lbf 
1 N*m  0.7375621493 ft*lbf 
2 N*m  1.4751242986 ft*lbf 
3 N*m  2.2126864479 ft*lbf 
5 N*m  3.6878107465 ft*lbf 
10 N*m  7.375621493 ft*lbf 
20 N*m  14.751242986 ft*lbf 
50 N*m  36.878107465 ft*lbf 
100 N*m  73.75621493 ft*lbf 
1000 N*m  737.5621493 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
The conversion from Nm to ftlbs 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 ftlbs. 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 …
20122021 · 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 …
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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 }}} 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 \,\!}
 τ=‖r‖‖F‖sinθ{\displaystyle \tau =\\mathbf {r} \\,\\mathbf \\sin \theta \,\!}
where
 τ{\displaystyle {\boldsymbol {\tau }}} is the torque vector and τ{\displaystyle \tau } is the magnitude of the torque,
 r{\displaystyle \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 } is the force vector,
 ×{\displaystyle \times } denotes the cross product, which produces a vector that is perpendicular to both r and F following the righthand rule,
 θ{\displaystyle \theta } is the angle between the force vector and the lever arm vector.
The SI unit for torque is the newtonmetre (N⋅m). For more on the units of torque, see § Units.
Defining terminology
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 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 ,}
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=rFsinθ,{\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 },}
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 righthand 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}}}
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 }},}
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 2rpω,{\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}}}
where p is the particle's linear momentum and r is the position vector from the origin. The timederivative 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}}.}
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}}} (whether or not mass is constant) and the definition of velocity drdt=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}}.}
The cross product of momentum p{\displaystyle {\boldsymbol {p}}} with its associated velocity v{\displaystyle \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 F_{net} = 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} }.}
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^{−2}L^{2}M. 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 (lbfft), or for small values the pound inch (lbfin). In the US, torque is most commonly referred to as the footpound (denoted as either lbft or ftlb) and the inchpound (denoted as inlb).^{[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 ftlb would properly imply).
Special cases and other facts
Moment arm formula
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}}).}
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 threedimensional 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}}).}
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.
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 twodimensional 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 twodimensions, 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 currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force F{\displaystyle \mathbf } is not zero, and τ1{\displaystyle {\boldsymbol {\tau }}_} is the torque measured from r1{\displaystyle \mathbf {r} _}, then the torque measured from r2{\displaystyle \mathbf {r} _} is
Machine torque
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. Internalcombustion 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 steamengines 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 ,}
where τ is torque, and θ_{1} and θ_{2} 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} is given by integrating the force with respect to an elemental linear displacement ds{\displaystyle \mathrm {d} \mathbf {s} }
 W=∫s1s2F⋅ds{\displaystyle W=\int _{s_}^{s_}\mathbf \cdot \mathrm {d} \mathbf {s} }
However, the infinitesimal linear displacement ds{\displaystyle \mathrm {d} \mathbf {s} } is related to a corresponding angular displacement dθ{\displaystyle \mathrm {d} {\boldsymbol {\theta }}} and the radius vector r{\displaystyle \mathbf {r} } as
 ds=dθ×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} }
The expression F⋅dθ×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]}. 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 }}}
But as per the definition of torque,
 τ=r×F{\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 }}}
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 }}}
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=τdθcos0=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 } giving
 W=∫θ1θ2τdθ{\displaystyle W=\int _{\theta _}^{\theta _}\tau \,\mathrm {d} \theta }
It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy E_{r} of the body, given by
 Er=12Iω2,{\displaystyle 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 }},}
where P is power, τ is torque, ω is the angular velocity, and ⋅{\displaystyle \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. multispeed 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 multispeed 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 }
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)}}}
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)}}}}}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g., American automotive engineers) use horsepower (mechanical) for power, footpounds (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}}.}
The constant below (in footpounds 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}}}
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}}.\,}
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 poundsforce feet and rotational speed in revolutions per minute, the above equation gives power in foot poundsforce 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}}}
because 5252.113122≈33,0002π.{\displaystyle 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 ).}
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 _).}
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
 ^ Serway, R. A. and Jewett, Jr. J.W. (2003). Physics for Scientists and Engineers. 6th Ed. Brooks Cole. ISBN 0534408427.
 ^ Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0716708094.
 ^ Thomson, James; Larmor, Joseph (1912). Collected Papers in Physics and Engineering. University Press. p. civ.
 ^ ^{a} ^{b} Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link
 ^ ^{a} ^{b} ^{c} Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download.
 ^ "Right Hand Rule for Torque". Retrieved 20070908.
 ^ ^{a} ^{b} Halliday, David; Resnick, Robert (1970). Fundamentals of Physics. John Wiley & Sons, Inc. pp. 184–85.
 ^ 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.
 ^ 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."
 ^ ^{a} ^{b} "SI brochure Ed. 9, Section 2.3.4" (PDF). Bureau International des Poids et Mesures. 2019. Retrieved 20200529.
 ^ "Dial Torque Wrenches from Grainger". Grainger. 2020. Demonstration that, as in most US industrial settings, the torque ranges are given in ftlb rather than lbfft.
 ^ Erjavec, Jack (22 January 2010). Manual Transmissions & Transaxles: Classroom manual. p. 38. ISBN 9781435439337.
 ^ ^{a} ^{b} Kleppner, Daniel; Kolenkow, Robert (1973). An Introduction to Mechanics. McGrawHill. pp. 267–68. ISBN 9780070350489.
 ^ 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 orderofmagnitude interactive.
The answer is 1.3558179483314. We assume you are converting between newtonmeter 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 …
Convert Nm to lbft The Newtonmeter 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).
1 Poundforce feet = 1.3558 Newton metres: 10 Poundforce feet = 13.5582 Newton metres: 2500 Poundforce feet = 3389.54 Newton metres: 2 Poundforce feet = 2.7116 Newton metres: 20 Poundforce feet = 27.1164 Newton metres: 5000 Poundforce feet = 6779.09 Newton metres: 3 Poundforce feet = 4.0675 Newton metres: 30 Poundforce feet = 40.6745 Newton metres: …
Convert from Poundforce feet to Newton metres. Type in the amount you want to convert and press the Convert button.
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1 Poundforce feet = 1.3558 Newton metres  10 Poundforce feet = 13.5582 Newton metres  2500 Poundforce feet = 3389.54 Newton metres 
2 Poundforce feet = 2.7116 Newton metres  20 Poundforce feet = 27.1164 Newton metres  5000 Poundforce feet = 6779.09 Newton metres 
3 Poundforce feet = 4.0675 Newton metres  30 Poundforce feet = 40.6745 Newton metres  10000 Poundforce feet = 13558.18 Newton metres 
4 Poundforce feet = 5.4233 Newton metres  40 Poundforce feet = 54.2327 Newton metres  25000 Poundforce feet = 33895.45 Newton metres 
5 Poundforce feet = 6.7791 Newton metres  50 Poundforce feet = 67.7909 Newton metres  50000 Poundforce feet = 67790.9 Newton metres 
6 Poundforce feet = 8.1349 Newton metres  100 Poundforce feet = 135.58 Newton metres  100000 Poundforce feet = 135581.79 Newton metres 
7 Poundforce feet = 9.4907 Newton metres  250 Poundforce feet = 338.95 Newton metres  250000 Poundforce feet = 338954.49 Newton metres 
8 Poundforce feet = 10.8465 Newton metres  500 Poundforce feet = 677.91 Newton metres  500000 Poundforce feet = 677908.97 Newton metres 
9 Poundforce feet = 12.2024 Newton metres  1000 Poundforce feet = 1355.82 Newton metres  1000000 Poundforce feet = 1355817.95 Newton metres 
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convertlive1 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.
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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)  ⇛ 
Convert ft lb to Nm. Please provide values below to convert footpound [ft*lbf] to newton meter [N*m], or vice versa.
Nm To Ft Lbs Conversion Table . 17 0 Download (0) 0 Download (0)
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 ...
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" =.
One pound (force) = 4.448 222 newtons One foot = 0.3048 m This gives the conversion factor: One poundfoot = 1.35582 newton metres. The name "poundfoot", intended to minimize confusion with the footpound as a unit of work, was apparently first proposed by British physicist Arthur Mason Worthington.
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?
20 Footpounds Force to Newton Meters = 27.1164: 900 Footpounds Force to Newton Meters = 1220.2362: 30 Footpounds Force to Newton Meters = 40.6745: 1,000 Footpounds Force to Newton Meters = 1355.8179: 40 Footpounds Force to Newton Meters = 54.2327: 10,000 Footpounds Force to Newton Meters = 13558.1795: 50 Footpounds Force to Newton Meters ...
Newton Meters to Footpounds Force (or just enter a value in the "to" field)
Unit Descriptions  



Conversions Table  

1 Footpounds Force to Newton Meters = 1.3558  70 Footpounds Force to Newton Meters = 94.9073 
2 Footpounds Force to Newton Meters = 2.7116  80 Footpounds Force to Newton Meters = 108.4654 
3 Footpounds Force to Newton Meters = 4.0675  90 Footpounds Force to Newton Meters = 122.0236 
4 Footpounds Force to Newton Meters = 5.4233  100 Footpounds Force to Newton Meters = 135.5818 
5 Footpounds Force to Newton Meters = 6.7791  200 Footpounds Force to Newton Meters = 271.1636 
6 Footpounds Force to Newton Meters = 8.1349  300 Footpounds Force to Newton Meters = 406.7454 
7 Footpounds Force to Newton Meters = 9.4907  400 Footpounds Force to Newton Meters = 542.3272 
8 Footpounds Force to Newton Meters = 10.8465  500 Footpounds Force to Newton Meters = 677.909 
9 Footpounds Force to Newton Meters = 12.2024  600 Footpounds Force to Newton Meters = 813.4908 
10 Footpounds Force to Newton Meters = 13.5582  800 Footpounds Force to Newton Meters = 1084.6544 
20 Footpounds Force to Newton Meters = 27.1164  900 Footpounds Force to Newton Meters = 1220.2362 
30 Footpounds Force to Newton Meters = 40.6745  1,000 Footpounds Force to Newton Meters = 1355.8179 
40 Footpounds Force to Newton Meters = 54.2327  10,000 Footpounds Force to Newton Meters = 13558.1795 
50 Footpounds Force to Newton Meters = 67.7909  100,000 Footpounds Force to Newton Meters = 135581.7948 
60 Footpounds Force to Newton Meters = 81.3491  1,000,000 Footpounds Force to Newton Meters = 1355817.9483 
Similar Torque Units
Common Units
How to convert newtonmeters to footpounds. 1 newtonmeter is equal to 0.7376 ftlbs, rounded to 4 decimal places. To convert newtonmeters to footpounds, simply multiply your figure by 0.7376. The conversion chart below can be used to convert some of the most common torque figures for newtonmeters (Nm) and footpounds (ftlb).
 Nm to footpounds
 Nm to inchpounds
Use this conversion tool to convert between newtonmeters (Nm) and footpounds (ftlb), both of which are units of work, energy or torque (poundfoot).
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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 newtonmeter is equal to 0.7376 ftlbs, rounded to 4 decimal places. To convert newtonmeters to footpounds, simply multiply your figure by 0.7376.
The conversion chart below can be used to convert some of the most common torque figures for newtonmeters (Nm) and footpounds (ftlb).
Newtonmeters to footpounds chart
If you wish to convert between inchpounds and footpounds, take a look at the inchpounds and footpounds converter.
For any figures not represented in our Nm and ftlb chart, please use the converter at the top of the page.
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.
How to convert from lb/ft to kN/m? Correctly measure in lbf/ft (poundforce 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
Exchange reading in foot pounds unit ft lb into newton meters unit Nm 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 Nm 1 ft lb = 1.36 Nm
Newton meters to foot pounds (Nm to ftlb) 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 ftlb.
Newtonmeters to Footpounds formula. ftlb = Nm * 0.73756 . Footpounds. 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 . Newtonmeters to Footpounds table. Start Increments Accuracy Format Print table < Smaller Values Larger Values > Newtonmeters Footpounds; 0 Nm: 0 ...
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
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 JF168I04 JF168FLH8B. 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 ...
/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