How To Use A Torque Multiplier

Physics concept

Torque
Relationship between force F, torque τ, linear momentum p, and angular momentum 50 in a system which has rotation constrained to only one plane (forces and moments due to gravity and friction not considered).
Common symbols	${\displaystyle \tau }$ , Yard
SI unit of measurement	Northward⋅m
Other units	pound-force-feet, lbf⋅inch, ozf⋅in
In SI base units	kg⋅m2⋅south−2
Dimension	Thousand 50 2 T −2

In physics and mechanics, torque is the rotational equivalent of linear force.^[1] It is too referred to as the moment, moment of force, rotational force or turning event, depending on the discipline. It represents the capability of a force to produce alter in the rotational motion of the trunk. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push button or a pull, a torque can be idea of every bit a twist to an object effectually a specific axis. Torque is defined as the production of the magnitude of the force and the perpendicular altitude of the line of activeness of a forcefulness from the axis of rotation. The symbol for torque is typically ${\displaystyle {\boldsymbol {\tau }}}$ , the lowercase Greek letter tau. When beingness referred to as moment of force, it is commonly denoted past Thousand.

In three dimensions, the torque is a pseudovector; for signal particles, it is given by the cross production of the position vector (distance vector) and the forcefulness vector. The magnitude of torque of a rigid torso depends on 3 quantities: the force applied, the lever arm vector ^[ii] connecting the point about which the torque is beingness measured to the betoken of forcefulness application, and the angle betwixt the force and lever arm vectors. In symbols:

${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \,\!}$

${\displaystyle \tau =\|\mathbf {r} \|\,\|\mathbf {F} \|\sin \theta \,\!}$

where

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

Defining terminology [edit]

The term torque (from Latin torquēre "to twist") is said to have been suggested by James Thomson and appeared in print in Apr, 1884.^{[iii] [4] [five]} Usage is attested the same year past Silvanus P. Thompson in the first edition of Dynamo-Electrical Machinery.^[v] Thompson motivates the term every bit follows:^[4]

"Just as the Newtonian definition of force is that which produces or tends to produce move (along a line), so torque may be defined as that which produces or tends to produce torsion (around an axis). It is better to utilise a term which treats this action as a single definite entity than to use terms like "couple" and "moment," which propose more complex ideas. The single notion of a twist practical to plow a shaft is meliorate than the more complex notion of applying a linear force (or a pair of forces) with a sure leverage."

Today, torque is referred to using unlike 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.^{[half dozen]} In the United kingdom and in US mechanical applied science, torque is referred to as moment of force, commonly shortened to moment.^[7] These terms are interchangeable in US physics^[six] and U.k. physics terminology, unlike in US mechanical engineering, where the term torque is used for the closely related "resultant moment of a couple".^{[7] [ contradictory ]}

Torque and moment in the US mechanical engineering terminology [edit]

In The states mechanical engineering, torque is defined mathematically every bit the rate of change of angular momentum of an object (in physics it is called "net torque"). The definition of torque states that 1 or both of the angular velocity or the moment of inertia of an object are irresolute. Moment is the general term used for the tendency of 1 or more practical forces to rotate an object virtually an axis, but not necessarily to change the athwart momentum of the object (the concept which is chosen torque in physics).^[7] For case, 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 (chosen a bending moment), but since the athwart momentum of the beam is not irresolute, this bending moment is not called a torque. Similarly with whatever strength couple on an object that has no modify to its athwart momentum, such moment is also not called a torque.

Definition and relation to angular momentum [edit]

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 past its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons practical two metres 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 tin be adamant past using the right manus grip rule: if the fingers of the right hand are curled from the direction of the lever arm to the direction of the strength, then the thumb points in the direction of the torque.^[8]

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

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

where F is the force acting on the particle. The magnitude τ of the torque is given by

${\displaystyle \tau =rF\sin \theta ,}$

where F is the magnitude of the force applied, and θ is the bending between the position and forcefulness vectors. Alternatively,

${\displaystyle \tau =rF_{\perp },}$

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 non produce a torque.^{[9] [10]}

Information technology 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 adamant by the right-hand rule.^[9]

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

${\displaystyle {\boldsymbol {\tau }}={\frac {\mathrm {d} \mathbf {Fifty} }{\mathrm {d} t}}}$

where Fifty is the athwart momentum vector and t is time.

For the motion of a signal particle,

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

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

${\displaystyle {\boldsymbol {\tau }}_{\mathrm {net} }={\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}={\frac {\mathrm {d} (I{\boldsymbol {\omega }})}{\mathrm {d} t}}=I{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}}+{\frac {\mathrm {d} I}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }}+{\frac {\mathrm {d} (mr^{ii})}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\blastoff }}+2rp_{||}{\boldsymbol {\omega }},}$

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 whatever blazon of trajectory. Notation that although strength and dispatch are always parallel and directly proportional, the torque τ need not exist parallel or direct proportional to the angular dispatch 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 [edit]

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

$${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$$

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

$${\displaystyle {\frac {\mathrm {d} \mathbf {Fifty} }{\mathrm {d} t}}=\mathbf {r} \times {\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}+{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\times \mathbf {p} .}$$

This result tin can easily exist proven by splitting the vectors into components and applying the production rule. Now using the definition of force ${\textstyle \mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}}$ (whether or not mass is constant) and the definition of velocity ${\textstyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\mathbf {5} }$

$${\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} +\mathbf {5} \times \mathbf {p} .}$$

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

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

If multiple forces are applied, Newton'south 2nd law instead reads F _internet = m a , and it follows that

$${\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} _{\mathrm {cyberspace} }={\boldsymbol {\tau }}_{\mathrm {net} }.}$$

This is a general proof for signal particles.

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

Units [edit]

Torque has the dimension of force times distance, symbolically T ⁻ⁱⁱ Fifty ² One thousand. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit of measurement newton metre (North⋅chiliad) and never the joule.^{[11] [12]} The unit newton metre is properly denoted Northward⋅grand.^[12]

The traditional Imperial and U.S. customary units for torque are the pound foot (lbf-ft), or for pocket-size values the pound inch (lbf-in). In the U.s.a., torque is about commonly referred to equally the human foot-pound (denoted as either lb-ft or ft-lb) and the inch-pound (denoted equally in-lb).^{[thirteen] [xiv]} 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 [edit]

Moment arm formula [edit]

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

${\displaystyle \tau =({\text{moment arm}})({\text{strength}}).}$

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 hard to utilize in iii-dimensional cases. If the forcefulness is perpendicular to the displacement vector r, the moment arm volition exist equal to the distance to the centre, and torque will be a maximum for the given forcefulness. The equation for the magnitude of a torque, arising from a perpendicular force:

${\displaystyle \tau =({\text{altitude to centre}})({\text{force}}).}$

For example, if a person places a strength of 10 N at the terminal end of a wrench that is 0.5 m long (or a forcefulness of 10 N exactly 0.v k from the twist point of a wrench of any length), the torque will be v N⋅m – bold that the person moves the wrench by applying force in the aeroplane of motility and perpendicular to the wrench.

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

Static equilibrium [edit]

For an object to be in static equilibrium, non only must the sum of the forces be zero, but too the sum of the torques (moments) near 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 2-dimensions, three equations are used.

Net forcefulness versus torque [edit]

When the net force on the arrangement is nada, the torque measured from whatever indicate in space is the same. For example, the torque on a electric current-conveying loop in a compatible magnetic field is the same regardless of the point of reference. If the net force ${\displaystyle \mathbf {F} }$ is not zilch, and ${\displaystyle {\boldsymbol {\tau }}_{i}}$ is the torque measured from ${\displaystyle \mathbf {r} _{i}}$ , then the torque measured from ${\displaystyle \mathbf {r} _{two}}$ is

$${\displaystyle {\boldsymbol {\tau }}_{2}={\boldsymbol {\tau }}_{1}+(\mathbf {r} _{1}-\mathbf {r} _{2})\times \mathbf {F} }$$

Car torque [edit]

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

Torque forms role of the basic specification of an engine: the ability output of an engine is expressed as its torque multiplied past its rotational speed of the axis. Internal-combustion engines produce useful torque only over a express range of rotational speeds (typically from around one,000–6,000 rpm for a small automobile). One can measure the varying torque output over that range with a dynamometer, and prove it as a torque curve.

Steam engines and electric motors tend to produce maximum torque shut to naught rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam-engines and electric motors can showtime heavy loads from nothing rpm without a clutch.

Relationship between torque, ability, and energy [edit]

If a force is allowed to deed through a altitude, it is doing mechanical work. Similarly, if torque is allowed to deed through a rotational distance, information technology is doing work. Mathematically, for rotation about a fixed centrality through the heart of mass, the work W tin be expressed as

${\displaystyle Westward=\int _{\theta _{1}}^{\theta _{2}}\tau \ \mathrm {d} \theta ,}$

where τ is torque, and θ ₁ and θ ₂ stand for (respectively) the initial and final athwart positions of the body.^[fifteen]

Proof [edit]

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

${\displaystyle Due west=\int _{s_{1}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} \mathbf {south} }$

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

${\displaystyle \mathrm {d} \mathbf {due south} =\mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }$

Substitution in the in a higher place expression for work gives

${\displaystyle W=\int _{s_{i}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }$

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

${\displaystyle \left[\mathbf {F} \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]=\mathbf {r} \times \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}}$

Only as per the definition of torque,

${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} }$

Respective exchange in the expression of piece of work gives,

${\displaystyle W=\int _{s_{i}}^{s_{2}}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}$

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

${\displaystyle West=\int _{\theta _{1}}^{\theta _{ii}}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}$

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

${\displaystyle Westward=\int _{\theta _{i}}^{\theta _{2}}\tau \,\mathrm {d} \theta }$

It follows from the work–free energy principle that Westward besides represents the change in the rotational kinetic free energy E _r of the torso, given by

${\displaystyle E_{\mathrm {r} }={\tfrac {1}{2}}I\omega ^{2},}$

where I is the moment of inertia of the torso and ω is its athwart speed.^[15]

Power is the piece of work per unit of measurement time, given by

${\displaystyle P={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }},}$

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 practical (this is equivalent to the linear instance where the power injected by a force depends just on the instantaneous speed – not on the resulting acceleration, if any).

In practise, this human 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 bike's transmission system allows multiple gear ratios to be used (i.east. multi-speed bicycle), all of which fastened to the frame. A cyclist, the person who rides the bike, provides the input ability past turning pedals, thereby cranking the front sprocket (commonly referred to as chainring). The input ability provided past the cyclist is equal to the production of cadence (i.e. the number of pedal revolutions per minute) and the torque on spindle of the bicycle's crankset. The wheel'south drivetrain transmits the input power to the route wheel, which in plough conveys the received power to the route 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, production of which (i.e. power) does not alter.

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 non revolutions per second).

Likewise, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. Nonetheless, in the case of torque, the unit of measurement 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 non in the latter instance. This trouble is addressed in orientational analysis which treats radians as a base of operations unit rather than a dimensionless unit.^[sixteen]

Conversion to other units [edit]

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

${\displaystyle P=\tau \cdot 2\pi \cdot \nu }$

Showing units:

${\displaystyle P({\rm {West}})=\tau {\rm {(North\cdot yard)}}\cdot ii\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rev/sec)}}}$

Dividing by 60 seconds per minute gives u.s. the following.

${\displaystyle P({\rm {West}})={\frac {\tau {\rm {(N\cdot yard)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rpm)}}}{lx}}}$

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

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

${\displaystyle P({\rm {hp}})={\frac {\tau {\rm {(lbf\cdot ft)}}\cdot two\pi {\rm {(rad/rev)}}\cdot \nu ({\rm {rpm}})}{33,000}}.}$

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

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

Derivation [edit]

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 altitude. And by definition, linear altitude = linear speed × fourth dimension = radius × angular speed × time.

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

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

The radius r and time t take dropped out of the equation. Yet, angular speed must be in radians per unit of time, past the causeless directly human relationship between linear speed and athwart 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 iiπ in the in a higher place derivation to give:

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

If torque is in newton metres and rotational speed in revolutions per second, the above equation gives power in newton metres per 2nd or watts. If Imperial units are used, and if torque is in pounds-force feet and rotational speed in revolutions per minute, the to a higher place equation gives power in foot pounds-strength per infinitesimal. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower:

${\displaystyle {\begin{aligned}{\text{power}}&={\text{torque}}\cdot two\pi \cdot {\text{rotational speed}}\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}\cdot {\frac {\text{horsepower}}{33,000\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}}}\\[6pt]&\approx {\frac {{\text{torque}}\cdot {\text{RPM}}}{5,252}}\end{aligned}}}$

because ${\displaystyle 5252.113122\approx {\frac {33,000}{2\pi }}.\,}$

Principle of moments [edit]

The principle of moments, also known equally Varignon'due south theorem (non to exist confused with the geometrical theorem of the aforementioned name) states that the resultant torques due to several forces practical to about a indicate is equal to the sum of the contributing torques:

${\displaystyle \tau =\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\ldots +\mathbf {r} _{N}\times \mathbf {F} _{N}.}$

From this information technology follows that the torques resulting from ii forces acting effectually a pivot on an object are balanced when

${\displaystyle \mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{two}=\mathbf {0} .}$

Torque multiplier [edit]

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

See also [edit]

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

References [edit]

1. ^ Serway, R. A. and Jewett, Jr. J.W. (2003). Physics for Scientists and Engineers. 6th Ed. Brooks Cole. ISBN 0-534-40842-seven.
2. ^ Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (fifth ed.). Westward. H. Freeman. ISBN0-7167-0809-4.
3. ^ Thomson, James; Larmor, Joseph (1912). Collected Papers in Physics and Engineering science. University Press. p. civ.
4. ^ ^{a b} Thompson, Silvanus Phillips (1893). Dynamo-electrical machinery: A Manual For Students Of Electrotechnics (4th ed.). New York, Harvard publishing co. p. 108.
5. ^ ^{a b} "torque". Oxford English language Dictionary. 1933.
6. ^ ^{a b} Physics for Technology by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link Archived 2017-07-eleven at the Wayback Car
7. ^ ^{a b c} Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download Archived 2015-06-nineteen at the Wayback Machine.
8. ^ "Correct Hand Rule for Torque". Archived from the original on 2007-08-19. Retrieved 2007-09-08 .
9. ^ ^{a b} Halliday, David; Resnick, Robert (1970). Fundamentals of Physics. John Wiley & Sons, Inc. pp. 184–85.
10. ^ Knight, Randall; Jones, Brian; Field, Stuart (2016). College Physics: A Strategic Arroyo. Jones, Brian, 1960-, Field, Stuart, 1958- (Third edition, technology update ed.). Boston: Pearson. p. 199. ISBN9780134143323. OCLC 922464227.
11. ^ From the official SI website Archived 2021-04-xix at the Wayback Motorcar, The International System of Units – 9th edition – Text in English Department 2.3.4: "...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 equally energy (SI unit joule), the joule is never used for expressing torque."
12. ^ ^{a b} "SI brochure Ed. 9, Department 2.iii.4" (PDF). Bureau International des Poids et Mesures. 2019. Archived (PDF) from the original on 2020-07-26. Retrieved 2020-05-29 .
13. ^ "Punch Torque Wrenches from Grainger". Grainger. 2020. Demonstration that, as in most United states industrial settings, the torque ranges are given in ft-lb rather than lbf-ft.
14. ^ Erjavec, Jack (22 January 2010). Manual Transmissions & Transaxles: Classroom transmission. p. 38. ISBN978-ane-4354-3933-vii.
15. ^ ^{a b} Kleppner, Daniel; Kolenkow, Robert (1973). An Introduction to Mechanics . McGraw-Hill. pp. 267–68. ISBN9780070350489.
16. ^ Page, Chester H. (1979). "Rebuttal to de Boer's "Group properties of quantities and units"". American Journal of Physics. 47 (nine): 820. Bibcode:1979AmJPh..47..820P. doi:x.1119/1.11704.

External links [edit]

Await upwardly 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: All the same Some other Argument" An automotive perspective
• Torque and Athwart Momentum in Circular Motion on Project PHYSNET.
• An interactive simulation of torque
• Torque Unit Converter
• A feel for torque An order-of-magnitude interactive.

How To Use A Torque Multiplier,

Source: https://en.wikipedia.org/wiki/Torque

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