Chapter 5A. Torque A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern...

Chapter 5A. TorqueChapter 5A. Torque

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Torque is a Torque is a twist or turn twist or turn that tends to that tends to produce produce rotation. * * * rotation. * * * Applications Applications are found in are found in many common many common tools around tools around the home or the home or industry where industry where it is necessary it is necessary to turn, tighten to turn, tighten or loosen or loosen devices.devices.

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to:• Define and give examples ofDefine and give examples of the terms the terms torque, torque,

moment arm, axis, moment arm, axis, andand line of action line of action of a force. of a force.

• Draw, label and calculate the Draw, label and calculate the moment armsmoment arms for a for a variety of applied forces given an axis of rotation.variety of applied forces given an axis of rotation.

• Calculate the Calculate the resultant torqueresultant torque about any axis about any axis given the magnitude and locations of forces on given the magnitude and locations of forces on an extended object.an extended object.

• Optional:Optional: Define and apply the Define and apply the vector cross vector cross productproduct to calculate torque. to calculate torque.

Definition of TorqueDefinition of Torque

Torque is defined as the tendency to produce a change in rotational motion.

Torque is defined as the tendency to produce a change in rotational motion.

Examples:

Torque is Determined by Three Torque is Determined by Three Factors:Factors:

• The The magnitudemagnitude of the applied force. of the applied force.

• The The directiondirection of the applied force. of the applied force.

• The The locationlocation of the applied force. of the applied force.

• The The magnitudemagnitude of the applied force. of the applied force.

• The The directiondirection of the applied force. of the applied force.

• The The locationlocation of the applied force. of the applied force.

20 N

Magnitude of force

40 N

The 40-N force produces twice the torque as does the

20-N force.

Each of the 20-N forces has a different

torque due to the direction of force. 20 N

Direction of Force

20 N

20 N20 N

Location of forceThe forces nearer the

end of the wrench have greater torques.

20 N20 N

Units for TorqueUnits for TorqueTorque is proportional to the magnitude of F and to the distance r from the axis. Thus, a tentative formula might be:

Torque is proportional to the magnitude of F and to the distance r from the axis. Thus, a tentative formula might be:

= Fr = Fr Units: Nm or lbft

6 cm

40 N

= (40 N)(0.60 m)

= 24.0 Nm, cw

= 24.0 Nm, cw = 24.0 Nm, cw

Direction of TorqueDirection of Torque

Torque is a vector quantity that has direction as well as magnitude.

Torque is a vector quantity that has direction as well as magnitude.

Turning the handle of a screwdriver clockwise

and then counterclockwise will

advance the screw first inward and then outward.

Sign Convention for Sign Convention for TorqueTorque

By convention, counterclockwise torques are positive and clockwise

torques are negative.Positive torque:

Counter-clockwise, out of

pagecw

ccw

Negative torque: clockwise, into page

Line of Action of a ForceLine of Action of a Force

The line of action of a force is an imaginary line of indefinite length drawn along the direction of the force.


F1

F2

F3Line of action

The Moment ArmThe Moment Arm

The moment arm of a force is the perpendicular distance from the line of action of a force to the axis of rotation.


F2

F1

F3

r

rr

Calculating TorqueCalculating Torque

• Read problem and draw a rough figure.Read problem and draw a rough figure.

• Extend line of action of the force.Extend line of action of the force.

• Draw and label moment arm.Draw and label moment arm.

• Calculate the moment arm if necessary.Calculate the moment arm if necessary.

• Apply definition of torque:Apply definition of torque:

• Read problem and draw a rough figure.Read problem and draw a rough figure.

• Extend line of action of the force.Extend line of action of the force.

• Draw and label moment arm.Draw and label moment arm.

• Calculate the moment arm if necessary.Calculate the moment arm if necessary.

• Apply definition of torque:Apply definition of torque:

= Fr = Fr Torque = force x moment arm

Example 1:Example 1: An An 80-N80-N force acts at the end of force acts at the end of a a 12-cm12-cm wrench as shown. Find the torque. wrench as shown. Find the torque.

• Extend line of action, draw, calculate r.

= (80 N)(0.104 m) = 8.31 N m

= (80 N)(0.104 m) = 8.31 N m

r = 12 cm sin 600 = 10.4 cm

r = 12 cm sin 600 = 10.4 cm

Alternate:Alternate: An An 80-N80-N force acts at the end of force acts at the end of a a 12-cm12-cm wrench as shown. Find the torque. wrench as shown. Find the torque.

Resolve 80-N force into components as shown.

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

= (69.3 N)(0.12 m) = 8.31 N m as before = 8.31 N m as before

positive

12 cm

Calculating Resultant Calculating Resultant TorqueTorque• Read, draw, and label a rough figure.Read, draw, and label a rough figure.

• Draw free-body diagram showing all Draw free-body diagram showing all forces, distances, and axis of rotation.forces, distances, and axis of rotation.

• Extend lines of action for each force.Extend lines of action for each force.

• Calculate moment arms if necessary.Calculate moment arms if necessary.

• Calculate torques due to EACH individual Calculate torques due to EACH individual force affixing proper sign. CCW (+) and force affixing proper sign. CCW (+) and CW (-).CW (-).

• Resultant torque is sum of individual Resultant torque is sum of individual torques.torques.

• Read, draw, and label a rough figure.Read, draw, and label a rough figure.






Example 2:Example 2: Find resultant torque Find resultant torque about axis about axis AA for the arrangement for the arrangement shown below:shown below:

300300

6 m 2 m4 m

20 N30 N

40 NA

Find due to each force.

Consider 20-N force first:


Consider 20-N force first:

r = (4 m) sin 300 = 2.00 m

= Fr = (20 N)(2 m) = 40 N m, cw

The torque about A is clockwise and

negative.

20 = -40 N m20 = -40 N m

r

negative

Example 2 (Cont.):Example 2 (Cont.): Next we find Next we find torque due to torque due to 30-N30-N force about force about same axis same axis AA..

300300

6 m 2 m4 m

20 N30 N

40 NA


Consider 30-N force next.


Consider 30-N force next.

r = (8 m) sin 300 = 4.00 m

= Fr = (30 N)(4 m) = 120 N m, cw

The torque about A is clockwise and

negative.

30 = -120 N m30 = -120 N m

rnegative

Example 2 (Cont.):Example 2 (Cont.): Finally, we Finally, we consider the torque due to the consider the torque due to the 40-N40-N force.force.


Consider 40-N force next:


Consider 40-N force next:

r = (2 m) sin 900 = 2.00 m

= Fr = (40 N)(2 m) = 80 N m, ccw

The torque about A is CCW and positive.

40 = +80 N m40 = +80 N m

300300

6 m 2 m4 m

20 N30 N

40 NA

r

positive

Example 2 (Conclusion):Example 2 (Conclusion): Find Find resultant torque about axis resultant torque about axis AA for for the arrangement shown below:the arrangement shown below:

300300

6 m 2 m4 m

20 N30 N

40 NA

Resultant torque is the sum of individual torques.

Resultant torque is the sum of individual torques.

R = - 80 N mR = - 80 N m Clockwise

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

Part II: Torque and the Part II: Torque and the Cross Product or Vector Cross Product or Vector

Product.Product.Optional

DiscussionThis 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.

The Vector ProductThe Vector Product

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

F

r

F Sin The effect of the

force F at angle (torque) is to advance the bolt out of the page.

Torque

Magnitude:

(F Sin )r

Direction = Out of page (+).

Definition of a Vector Definition of a Vector ProductProduct

The magnitude of the vector (cross) product of two vectors A and B is defined as follows:

A x B = l A l l B l Sin

F x r = l F l l r l Sin Magnitude onlyMagnitude only

F

(F Sin ) r or F (r Sin )

In our example, the cross product of F and r is:

In effect, this becomes simply:

r

F Sin

Example:Example: Find the Find the magnitudemagnitude of of the cross product of the vectors r the cross product of the vectors r and F drawn below:and F drawn below:

r x F = l r l l F l Sin r x F = (6 in.)(12 lb) Sin

r x F = l r l l F l Sin

r x F = (6 in.)(12 lb) Sin 120

Explain Explain differencedifference.. Also, what aboutAlso, what about F x r??

12 lb

r x F = 62.4 lb in.

Torque

600

6 in.

Torque

600

6 in.

12 lb r x F = 62.4 lb in.

Direction of the Vector Direction of the Vector Product.Product.

The The directiondirection of of a vector product a vector product is determined by is determined by the the right hand right hand rule.rule. A

C

BB

-CA

A x B = C (up)A x B = C (up)

B x A = -C B x A = -C (Down)(Down)

Curl fingers of right Curl fingers of right hand in direction of hand in direction of cross pro-duct (cross pro-duct (AA to to BB) ) or (or (BB to to AA). ). ThumbThumb will will point in the direction of point in the direction of product product CC..

What is What is direction of A x direction of A x

C?C?

Example:Example: What are the magnitude What are the magnitude and direction of the cross product, and direction of the cross product, r x F?r x F?

r x F = l r l l F l Sin r x F = (6 in.)(10 lb) Sin

r x F = 38.3 lb in.

10 lbTorqu

e

500

6 in. Magnitude

Out

r

FDirection by right hand rule:Out of paper (thumb) or +kr x F = (38.3 lb in.) k

What are magnitude and direction of F x r?

Cross Products Using Cross Products Using ((i,j,ki,j,k))

x

z

yConsider 3D axes (x, y, z)Define unit vectors, i, j, k

ij

kConsider cross product: i x i

i x i = (1)(1) Sin 00 = 0

i i

j x j = (1)(1) Sin 00 = 0

k x k = (1)(1)Sin 00= 0

Magnitudes are zero for parallel vector products.

Vector Products Using Vector Products Using ((i,j,ki,j,k))

Consider 3D axes (x, y, z)Define unit vectors, i, j, k

x

z

y

ij

k Consider dot product: i x j

i x j = (1)(1) Sin 900 = 1j x k = (1)(1) Sin 900 = 1

k x i = (1)(1) Sin 900 = 1

j i

Magnitudes are “1” for perpendicular vector products.

Vector Product Vector Product (Directions)(Directions)

x

z

y

ij

k

i x j = (1)(1) Sin 900 = +1 kj x k = (1)(1) Sin 900 = +1 i

k x i = (1)(1) Sin 900 = +1 j

Directions are given by the right hand rule. Rotating first vector into second.

k

j

i

Vector Products Practice Vector Products Practice ((i,j,ki,j,k))

x

z

y

ij

ki x k = ?k x j = ?

Directions are given by the right hand rule. Rotating first vector into second.

k

j

i 2 i x -3 k = ?

- j (down)

- i (left)

+ 6 j (up)

j x -i = ?

+ k (out)

Using i,j Notation - Vector Using i,j Notation - Vector 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) =

(2)(3) ixi + (2)(5) ixj + (-4)(3) jxi + (-4)(5) jxj

k -k0 0

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

Alternative:

A = 2 i - 4 j B = 3 i + 5 jA x B = 10 - (-12) = +22

k

Evaluate determinant

SummarySummaryTorque is the product of a force and its

moment arm as defined below:Torque is the product of a force and its

moment arm as defined below:





= Fr = Fr Torque = force x moment arm

Torque = force x moment arm

Summary: Resultant Summary: Resultant TorqueTorque













CONCLUSION: Chapter 5ACONCLUSION: Chapter 5ATorque Torque

Chapter 5A. Torque A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern...

Documents

Transcript of Chapter 5A. Torque A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern...