Chapter 5A. Torque

Chapter 5A. TorqueAA PowerPointPowerPoint PresentationPresentation bybyPaulPaul E.E. Tippens,Tippens, ProfessorProfessor ofof PhysicsPhysicsSouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity 2007

Torque is a twistor turn that tendsto producerotation. * * *Applications arefound in manycommon toolsaround the homeor industry whereit is necessary toturn, tighten orloosen devices.

Objectives: After completing thismodule, you should be able to: Define and give examples of the terms torque,moment arm, axis, and line of action of a force. Draw, label and calculate the moment arms fora variety of applied forces given an axis ofrotation. Calculate the resultant torque about any axisgiven the magnitude and locations of forces onan extended object. Optional: Define and apply the vector crossproduct to calculate torque.

Definition of TorqueTorqueTorque isis defineddefined asas thethe tendencytendency totoproduceproduce aa changechange inin rotationalrotational motion.motion.Examples:

Torque is Determined by Three Factors: TheThe magnitudemagnitude ofof thethe appliedapplied force.force. TheThe directiondirection ofof thethe appliedapplied force.force. TheThe locationlocation ofof thethe appliedapplied force.force.EachThe40-Nof enttheend ofhastheatwicewrenchtorquetorqueasduedoesto thethehavegreatertorques.direction20-N force.of rce20 N 2020N20NN 2040NN20 N20 N

Units for TorqueTorqueTorque isis proportionalproportional toto thethe magnitudemagnitude ofofFF andand toto thethe distancedistance rr fromfrom thethe axis.axis. Thus,Thus,aa tentativetentative formulaformula mightmight be:be: FrFrUnits: N m or lb ft (40 N)(0.60 m) 24.0 N m, cw 24.024.0 N m,N m, cwcw6 cm40 N

Direction of TorqueTorqueTorque isis aa vectorvector quantityquantity thatthat hashasdirectiondirection asas wellwell asas magnitude.magnitude.Turning the handle of ascrewdriver clockwise andthen counterclockwise willadvance the screw firstinward and then outward.

Sign Convention for TorqueBy convention, counterclockwise torques arepositive and clockwise torques are negative.Positive torque:Counter-clockwise,out of pagecwccwNegative torque:clockwise, into page

Line of Action of a ForceTheThe lineline ofof actionaction ofof aa forceforce isis anan imaginaryimaginarylineline ofof indefiniteindefinite lengthlength drawndrawn alongalong thethedirectiondirection ofof thethe force.force.F1F2Line ofactionF3

The Moment ArmTheThe momentmoment armarm ofof aa forceforce isis thethe perpendicularperpendiculardistancedistance fromfrom thethe lineline ofof actionaction ofof aa forceforce toto thetheaxisaxis ofof rotation.rotation.F1F2rrrF3

Calculating Torque ReadRead problemproblem andand drawdraw aa roughrough figure.figure. ExtendExtend lineline ofof actionaction ofof thethe force.force. DrawDraw andand labellabel momentmoment arm.arm. CalculateCalculate thethe momentmoment armarm ifif necessary.necessary. ApplyApply definitiondefinition ofof torque:torque: FrFrTorque force x moment arm

Example 1: An 80-N force acts at the end ofa 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r.rr 1212 cmcm sinsin 606000 10.410.4 cmcm (80(80 N)(0.104N)(0.104 m)m) 8.318.31 NNmm

Alternate: An 80-N force acts at the end ofa 12-cm wrench as shown. Find the torque.positive12 cmResolve 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 NNmm asas beforebefore

Calculating Resultant Torque Read,Read, draw,draw, andand labellabel aa roughrough figure.figure. Draw-body diagramDraw freefree-bodydiagram showingshowing allall forces,forces,distances,distances, andand axisaxis ofof rotation.rotation. ExtendExtend lineslines ofof actionaction forfor eacheach force.force. CalculateCalculate momentmoment armsarms ifif necessary.necessary. CalculateCalculate torquestorques duedue toto EACHEACH individualindividual forceforceaffixing-).affixing properproper sign.sign. CCWCCW ( )( ) andand CWCW ((-). ResultantResultant torquetorque isis sumsum ofof individualindividual torques.torques.

Example 2: Find resultant torque aboutaxis A for the arrangement shown below:FindFind duedue totoeacheach force.force.ConsiderConsider 20-N20-Nforceforce first:first:negative30 Nr3002m6m40 N20 N300A4mr (4 m) sin 300The torque about A isclockwise and negative. Fr (20 N)(2 m) 2020 -40-40 NNmm 2.00 m 40 N m, cw

Example 2 (Cont.): Next we find torquedue to 30-N force about same axis A.FindFind duedue totoeacheach force.force.ConsiderConsider 30-N30-Nforceforce next.next.rnegative30 N30020 N3002m6m40 NA4mr (8 m) sin 300The torque about A isclockwise and negative. Fr (30 N)(4 m) 3030 -120-120 NNmm 4.00 m 120 N m, cw

Example 2 (Cont.): Finally, we considerthe torque due to the 40-N force.FindFind duedue totoeacheach force.force.ConsiderConsider 40-N40-Nforceforce next:next:r (2 m) sin 900 2.00 m Fr (40 N)(2 m) 80 N m, ccwpositive30 Nr3002m6m40 N20 N300A4mThe torque about A isCCW and positive. 4040 80 80 NNmm

Example 2 (Conclusion): Find resultanttorque about axis A for the arrangementshown below:ResultantResultant torquetorqueisis thethe sumsum ofofindividualindividual torques.torques.20 N30 N3003002m6m40 NA4m R 20 20 20 -40 N m -120 N m 80 N m RR -- 8080 NNmmClockwise

Part II: Torque and the CrossProduct or Vector Product.Optional DiscussionThis concludes the general treatmentof torque. Part II details the use ofthe vector product in calculatingresultant torque. Check with yourinstructor before studying thissection.

The Vector ProductTorque can also be found by using the vectorproduct of force F and position vector r. Forexample, consider the figure below.F Sin TorquerMagnitude:(F Sin )rF The effect of the forceF at angle (torque)is to advance the boltout of the page.Direction Out of page ( ).

Definition of a Vector ProductThe magnitude of the vector (cross) productof two vectors A and B is defined as follows:A x B 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 Magnitude onlyF Sin rFIn effect, this becomes simply:(F Sin ) r or F (r Sin )

Example: Find the magnitude of thecross product of the vectors r and Fdrawn below:Torque12 lb6006 in.6 in.Torque12 lb600r x F l r l l F l Sin r x F (6 in.)(12 lb) Sin r x F 62.4 lb in.r x F l r l l F l Sin r x F (6 in.)(12 lb) Sin 120 r x F 62.4 lb in.Explain difference. Also, what about F x r?

Direction of the Vector Product.The direction of avector product isdetermined by theright hand rule.A x B C (up)B x A -C (Down)What is directionof A x C?CABAB-CCurl fingers of right handin direction of cross product (A to B) or (B to A).Thumb will point in thedirection of product C.

Example: What are the magnitude anddirection of the cross product, r x F?10 lbTorque5006 in.FrOutr x F l r l l F l Sin r x F (6 in.)(10 lb) Sin r x F 38.3 lb in.MagnitudeDirection by right hand rule:Out of paper (thumb) or kr x F (38.3 lb in.) kWhat are magnitude and direction of F x r?

Cross Products Using (i,j,k)yjConsider 3D axes (x, y, z)ikziiMagnitudes arezero for parallelvector products.xDefine unit vectors, i, j, kConsider cross product: i x ii x i (1)(1) Sin 00 0j x j (1)(1) Sin 00 0k x k (1)(1)Sin 00 0

Vector Products Using (i,j,k)yjConsider 3D axes (x, y, z)ixkzDefine unit vectors, i, j, kConsider dot product: i x jjiMagnitudes are “1”for perpendicularvector products.i x j (1)(1) Sin 900 1j x k (1)(1) Sin 900 1k x i (1)(1) Sin 900 1

Vector Product (Directions)yjixDirections are given by theright hand rule. Rotatingfirst vector into second.kzji x j (1)(1) Sin 900 1 kj x k (1)(1) Sin 900 1 ikik x i (1)(1) Sin 900 1 j

Vector Products Practice (i,j,k)yjikzkjixDirections are given by theright hand rule. Rotatingfirst vector into second.ixk ?- j (down)kxj ?- i (left)j x -i ? k (out)2 i x -3 k ? 6 j (up)

Using i,j Notation - Vector ProductsConsider: A 2 i - 4 j and B 3 i 5 jA x B (2 i - 4 j) x (3 i 5 j) 0k-k(2)(3) ixi (2)(5) ixj (-4)(3) jxi (-4)(5) jxjA x B (2)(5) k (-4)(3)(-k) 22 kAlternative: A 2 i - 4 jB 3i 5jEvaluatedeterminantA x B 10 - (-12) 22 k0

SummaryTorqueTorque isis thethe productproduct ofof aa forceforce andand itsitsmomentmoment armarm asas defineddefined below:below:TheThe momentmoment armarm ofof aa forceforce isis thethe perpendicularperpendicular distancedistancefromfrom thethe lineline ofof actionaction ofof aa forceforce toto thethe axisaxis ofof rotation.rotation.TheThe lineline ofof actionaction ofof aa forceforce isis anan imaginaryimaginary lineline ofofindefiniteindefinite lengthlength drawndrawn alongalong thethe directiondirection ofof thethe force.force. FrFrTorqueTorque forceforce xx momentmoment armarm

Summary: Resultant Torque Read,Read, draw,draw, andand labellabel aa roughrough figure.figure. Draw-body diagramDraw freefree-bodydiagram showingshowing allall forces,forces,distances,distances, andand axisaxis ofof rotation.rotation. ExtendExtend lineslines ofof actionaction forfor eacheach force.force. CalculateCalculate momentmoment armsarms ifif necessary.necessary. CalculateCalculate torquestorques duedue toto EACHEACH individualindividual forceforceaffixing-).affixing properproper sign.sign. CCWCCW ( )( ) andand CWCW ((-). ResultantResultant torquetorque isis sumsum ofof individualindividual torques.torques.

CONCLUSION: Chapter 5ATorque

moment arm, axis, and and line of action of a force. Draw, label and calculate the moment arms moment arms for a variety of applied forces given an axis of rotation. Calculate the resultant torque resultant torque about any axis