ChapterChapter 5A.5A. TorqueTorque

AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 TorqueTorque isis aa twisttwist oror turnturn thatthat tendstends toto produceproduce rotation.rotation. ** ** ** ApplicationsApplications areare foundfound inin manymany commoncommon toolstools aroundaround thethe homehome oror industryindustry wherewhere itit isis necessarynecessary toto turn,turn, tightentighten oror loosenloosen devices.devices. Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: •• DefineDefine andand givegive examplesexamples ofof thethe termsterms torque,torque, momentmoment arm,arm, axis,axis, andand lineline ofof actionaction ofof aa force.force. •• Draw,Draw, labellabel andand calculatecalculate thethe momentmoment armsarms forfor aa varietyvariety ofof appliedapplied forcesforces givengiven anan axisaxis ofof rotation.rotation. •• CalculateCalculate thethe resultantresultant torquetorque aboutabout anyany axisaxis givengiven thethe magnitudemagnitude andand locationslocations ofof forcesforces onon anan extendedextended objectobject.. •• Optional:Optional: DefineDefine andand applyapply thethe vectorvector crosscross productproduct toto calculatecalculate torque.torque. DefinitionDefinition ofof TorqueTorque

TorqueTorque isis defineddefined asas thethe tendencytendency toto produceproduce aa changechange inin rotationalrotational motion.motion.

Examples: TorqueTorque isis DeterminedDetermined byby ThreeThree Factors:Factors:

••• TheThe The magnitudemagnitudemagnitude ofofof thethethe appliedappliedapplied force.force.force. ••• TheThe The directiondirectiondirection ofofof thethethe appliedappliedapplied force.force.force. ••• TheThe The locationlocationlocation ofofof thethethe appliedappliedapplied force.force.force.

DirectionMagnitudeLocation of of ofForce force force TheEachThe forces 40-Nof thenearer force 20-N the forcesproducesend of has the atwice wrenchdifferent the 20 N  2020 N N torquetorque as due does to the the  20 N have greater torques. 2040 N N direction20-N force. of force. 20 N20 N UnitsUnits forfor TorqueTorque TorqueTorque isis proportionalproportional toto thethe magnitudemagnitude ofof FF andand toto thethe distancedistance rr fromfrom thethe axis.axis. Thus,Thus, aa tentativetentative formulaformula mightmight be:be:

== FrFr Units: Nm or lbft

 = (40 N)(0.60 m) = 24.0 Nm, cw 6 cm == 24.024.0 NNm,m, cwcw 40 N DirectionDirection ofof TorqueTorque

TorqueTorque isis aa vectorvector quantityquantity thatthat hashas directiondirection asas wellwell asas magnitude.magnitude.

Turning the handle of a screwdriver clockwise and then counterclockwise will advance the screw first inward and then outward. SignSign ConventionConvention forfor TorqueTorque By convention, counterclockwise torques are positive and clockwise torques are negative.

Positive torque: ccw Counter-clockwise, out of page cw Negative torque: clockwise, into page LineLine ofof ActionAction ofof aa ForceForce

TheThe lineline ofof actionactionofof aa forceforce isis anan imaginaryimaginary lineline ofof indefiniteindefinite lengthlength drawndrawn alongalong thethe directiondirection ofof thethe force.force.

F2 F1 Line of F3 action TheThe MomentMoment ArmArm

TheThe momentmoment armarmofof aa forceforce isis thethe perpendicularperpendicular distancedistance fromfrom thethe lineline ofof actionaction ofof aa forceforce toto thethe axisaxis ofof rotation.rotation.

F1 r

F r 2 F r 3 CalculatingCalculating TorqueTorque

••• ReadRead Read problemproblemproblem andandand drawdrawdraw aaa roughroughrough figure.figure.figure. ••• ExtendExtend Extend linelineline ofofof actionactionaction ofofof thethethe force.force.force. ••• DrawDraw Draw andandand labellabellabel momentmomentmoment arm.arm.arm. ••• CalculateCalculate Calculate thethethe momentmomentmoment armarmarm ififif necessary.necessary.necessary. ••• ApplyApply Apply definitiondefinitiondefinition ofofof torque:torque:torque:

== FrFr Torque = force x moment arm ExampleExample 1:1: AnAn 8080--NN forceforce actsacts atat thethe endend ofof aa 1212--cmcm wrenchwrench asas shown.shown. FindFind thethe torque.torque.

• Extend line of action, draw, calculate r.

rr == 1212 cmcm sinsin 606000 == (80(80 N)(0.104N)(0.104 m)m) == 10.410.4 cmcm == 8.318.31 NN mm Alternate:Alternate: AnAn 8080--NN forceforce actsacts atat thethe endend ofof aa 1212--cmcm wrenchwrench asas shown.shown. FindFind thethe torque.torque.

positive

12 cm

Resolve 80-N force into components as shown.

Note from figure: rx = 0 and ry = 12 cm

 = (69.3 N)(0.12 m) == 8.318.31 NN mm asas beforebefore CalculatingCalculating ResultantResultant TorqueTorque ••• Read,Read, Read, draw,draw,draw, andandand labellabellabel aaa roughroughrough figure.figure.figure. ••• DrawDraw Draw freefreefree-body--bodybody diagramdiagramdiagram showingshowingshowing allallall forces,forces,forces, distances,distances,distances, andandand axisaxisaxis ofofof rotation.rotation.rotation. ••• ExtendExtend Extend lineslineslines ofofof actionactionaction forforfor eacheacheach force.force.force. ••• CalculateCalculate Calculate momentmomentmoment armsarmsarms ififif necessary.necessary.necessary. ••• CalculateCalculate Calculate torquestorquestorques dueduedue tototo EACHEACHEACH individualindividualindividual forceforceforce affixingaffixingaffixing properproperproper sign.sign.sign. CCWCCWCCW (+)(+)(+) andandand CWCWCW (((-).--).). ••• ResultantResultant Resultant torquetorquetorque isisis sumsumsum ofofof individualindividualindividual torques.torques.torques. ExampleExample 2:2: FindFind resultantresultant torquetorque aboutabout axisaxis AA forfor thethe arrangementarrangement shownshown below:below:

FindFind duedue toto 30 N negative 20 N

each force. 0 r each force. 30 300 Consider 20-N Consider 20-N 6 m 2 m A 4 m forceforce first:first: 40 N r = (4 m) sin 300 The torque about A is = 2.00 m clockwise and negative.  = Fr = (20 N)(2 m) 20 == --4040 NN mm = 40 N m, cw 20 ExampleExample 22 (Cont.):(Cont.): NextNext wewe findfind torquetorque duedue toto 3030--NN forceforce aboutabout samesame axisaxis AA..

r 20 N FindFind duedue toto 30 N negative each force. 0 each force. 30 300 Consider 30-N Consider 30-N 6 m 2 m A 4 m forceforce next.next. 40 N r = (8 m) sin 300 The torque about A is = 4.00 m clockwise and negative.  = Fr = (30 N)(4 m) 30 == -120-120 NN mm = 120 N m, cw 30 ExampleExample 22 (Cont.):(Cont.): Finally,Finally, wewe considerconsider thethe torquetorque duedue toto thethe 4040--NN force.force.

FindFind duedue toto 30 N positive 20 N

each force. 0 each force. 30 r 300 Consider 40-N Consider 40-N 6 m 2 m A 4 m forceforce next:next: 40 N r = (2 m) sin 900 The torque about A is = 2.00 m CCW and positive.  = Fr = (40 N)(2 m)  = +80 N m = 80 N m, ccw 4040 = +80 N m ExampleExample 22 (Conclusion):(Conclusion): FindFind resultantresultant torquetorque aboutabout axisaxis AA forfor thethe arrangementarrangement shownshown below:below:

30 N 20 N ResultantResultant torquetorque isis thethe sumsum ofof 300 300 6 m 2 m individualindividual torques.torques. A 4 m 40 N

R = 20 + 20 + 20 = -40 N m -120 N m + 80 N m

 = - 80 N m Clockwise RR = - 80 N m PartPart II:II: TorqueTorque andand thethe CrossCross ProductProduct oror VectorVector Product.Product. Optional Discussion

This concludes the general treatment of torque. Part II details the use of the vector product in calculating resultant torque. Check with your instructor before studying this section. TheThe VectorVector ProductProduct

Torque can also be found by using the vector product of force F and position vector r. For example, consider the figure below.

F Sin  Torque F The effect of the force F at angle  (torque) r  is to advance the bolt out of the page. Magnitude: Direction = Out of page (+). (FSin)r DefinitionDefinition ofof aa VectorVector ProductProduct The magnitude of the vector (cross) product of two vectors A and B is defined as follows: AxB= l A l l B l Sin  In our example, the cross product of F and r is:

F x r = l F l l r l Sin MagnitudeMagnitude onlyonly

F Sin  F In effect, this becomes simply:  r (F Sin ) r or F (r Sin ) Example:Example: FindFind thethe magnitudemagnitude ofof thethe crosscross productproduct ofof thethe vectorsvectors rr andand FF drawndrawn below:below: 12 lb Torque r x F = l r l l F l Sin  r x F = (6 in.)(12 lb) Sin  600 6 in. r x F = 62.4 lb in.

6 in. r x F = l r l l F l Sin 

 600 r x F= (6 in.)(12 lb) Sin 120 Torque 12 lb r x F = 62.4 lb in. ExplainExplain differencedifference.. Also,Also, whatwhat aboutabout F x r?? DirectionDirection ofof thethe VectorVector Product.Product. C TheThe directiondirection ofof aa vectorvector productproduct isis determineddetermined byby thethe B rightright handhand rule.rule. B A A -C AA xx BB == CC (up)(up) CurlCurl fingersfingers ofof rightright handhand BB xx AA == --CC (Down)(Down) inin directiondirection ofof crosscross propro-- ductduct ((AA toto BB)) oror ((BB toto AA).). WhatWhat isis directiondirection ThumbThumb willwill pointpoint inin thethe ofof AA xx C?C? directiondirection ofof productproduct CC.. Example:Example: WhatWhat areare thethe magnitudemagnitude andand directiondirection ofof thethe crosscross product,product, rr xx F?F? 10 lb Torque r x F = l r l l F l Sin  r x F = (6 in.)(10 lb) Sin  500 6 in. r x F = 38.3 lb in. Magnitude F Direction by right hand rule: r Out of paper (thumb) or +k Out r x F = (38.3 lb in.) k What are magnitude and direction of F x r? CrossCross ProductsProducts UsingUsing ((i,j,ki,j,k)) y Consider 3D axes (x, y, z) j i x Define unit vectors, i, j, k k Consider cross product: i x i z i 0 i i x i = (1)(1) Sin 0 = 0 Magnitudes are j x j = (1)(1) Sin 00 = 0 zero for parallel 0 vector products. k x k = (1)(1)Sin 0 = 0 VectorVector ProductsProducts UsingUsing ((i,j,ki,j,k)) y Consider 3D axes (x, y, z) j i x Define unit vectors, i, j, k k Consider dot product: i x j z j i i x j = (1)(1) Sin 900 = 1 Magnitudes are “1” j x k = (1)(1) Sin 900 = 1 for perpendicular 0 vector products. k x i = (1)(1) Sin 90 = 1 VectorVector ProductProduct (Directions)(Directions) y Directions are given by the j right hand rule. Rotating i x first vector into second. k z j i x j = (1)(1) Sin 900 = +1 k j x k = (1)(1) Sin 900 = +1 i i k x i = (1)(1) Sin 900 = +1 j k VectorVector ProductsProducts PracticePractice ((i,j,ki,j,k)) y Directions are given by the j right hand rule. Rotating i x first vector into second. k i x k = ? -j (down) z j k x j = ? -i (left) j x -i = ? + k (out)

i 2 i x -3 k = ? + 6 j (up) k UsingUsing i,ji,j NotationNotation -- VectorVector ProductsProducts

Consider: A = 2 i -4 j and B = 3 i + 5 j A x B = (2 i - 4 j) x (3 i + 5 j) = 0 k -k 0 (2)(3) ixi + (2)(5) ixj + (-4)(3) jxi + (-4)(5) jxj

A x B = (2)(5) k + (-4)(3)(-k) = +22 k

Alternative: A = 2 i -4 j Evaluate B = 3 i + 5 j determinant A x B = 10 - (-12) = +22 k SummarySummary TorqueTorqueisis thethe productproduct ofof aa forceforce andand itsits momentmoment armarm asas defineddefined below:below:

TheThe momentmoment armarmofof aa forceforce isis thethe perpendicularperpendicular distancedistance fromfrom thethe lineline ofof actionaction ofof aa forceforce toto thethe axisaxis ofof rotation.rotation.

TheThe lineline ofof actionactionofof aa forceforce isis anan imaginaryimaginary lineline ofof indefiniteindefinite lengthlength drawndrawn alongalong thethe directiondirection ofof thethe force.force.

== FrFr TorqueTorque == forceforce xx momentmoment armarm Summary:Summary: ResultantResultant TorqueTorque ••• Read,Read, Read, draw,draw,draw, andandand labellabellabel aaa roughroughrough figure.figure.figure. ••• DrawDraw Draw freefreefree-body--bodybody diagramdiagramdiagram showingshowingshowing allallall forces,forces,forces, distances,distances,distances, andandand axisaxisaxis ofofof rotation.rotation.rotation. ••• ExtendExtend Extend lineslineslines ofofof actionactionaction forforfor eacheacheach force.force.force. ••• CalculateCalculate Calculate momentmomentmoment armsarmsarms ififif necessary.necessary.necessary. ••• CalculateCalculate Calculate torquestorquestorques dueduedue tototo EACHEACHEACH individualindividualindividual forceforceforce affixingaffixingaffixing properproperproper sign.sign.sign. CCWCCWCCW (+)(+)(+) andandand CWCWCW (((-).--).). ••• ResultantResultant Resultant torquetorquetorque isisis sumsumsum ofofof individualindividualindividual torques.torques.torques. CONCLUSION:CONCLUSION: ChapterChapter 5A5A TorqueTorque