Chapter 5A. TorqueChapter 5A. TorqueA PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

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

Objectives: After completing this Objectives: After completing this module, you should be able to:module, you should be 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 for a variety of applied forces given an axis of a variety of applied forces given an axis of rotation.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 objectan 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 Factors:Torque is Determined by Three Factors:

• The magnitude of the applied force.

• The direction of the applied force.

• The location 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 cm40 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 TorqueSign Convention for TorqueBy 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.

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.

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.

• Extend line of action of the force.

• Draw and label moment arm.

• Calculate the moment arm if necessary.

• 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 8080--NN force acts at the end of force acts at the end of a a 1212--cmcm 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 8080--NN force acts at the end of force acts at the end of a a 1212--cmcm 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 TorqueCalculating Resultant Torque• Read, draw, and label a rough figure.

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

• Extend lines of action for each force.

• Calculate moment arms if necessary.

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

• Resultant torque is sum of individual torques.

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

•• Draw freeDraw free--body diagram showing all forces, body diagram showing all forces, distances, and axis of rotation.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 force Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (affixing proper sign. CCW (+) and CW (--).).

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

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

300300

6 m 2 m4 m

20 N30 N

40 NA

Find due to each force.

Consider 20-N force first:

Find due to each force.

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 torque Next we find torque due to due to 3030--NN force about same axis force about same axis AA..

300300

6 m 2 m4 m

20 N30 N

40 NA

Find due to each force.

Consider 30-N force next.

Find due to each force.

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 consider Finally, we consider the torque due to the the torque due to the 4040--NN force.force.

Find due to each force.

Consider 40-N force next:

Find due to each force.

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 resultant Find resultant torque about axis torque about axis AA for the arrangement for the arrangement shown below: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 Cross Part II: Torque and the Cross Product or Vector Product.Product or Vector 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.

The Vector ProductThe Vector ProductTorque 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 )rDirection = Out of page (+).

Definition of a Vector ProductDefinition of a Vector ProductThe 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 the of the cross product of the vectors r and F cross product of the vectors r and F drawn below: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.

Torque600

6 in.

12 lb r x F = 62.4 lb in.

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

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

A

C

BB

-CA

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

B x A = B x A = --C (Down)C (Down)Curl fingers of right hand Curl fingers of right hand in direction of cross proin direction of cross pro-- duct (duct (AA to to BB) or () or (BB to to AA). ). ThumbThumb will point in the will point in the direction of product direction of product CC..

What is direction What is direction of A x C?of A x C?

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

500

6 in. Magnitude

Out

r

F Direction by right hand rule:Out of paper (thumb) or +k

r 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, kij

k Consider cross product: i x i

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

z

y

ij

k Consider dot product: i x j

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

j 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 (Directions)Vector Product (Directions)

x

z

y

ij

k

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

j 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

k i 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 Using i,j Notation -- Vector ProductsVector Products

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) jxjk -k0 0

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

Alternative: A = 2 i - 4 jB = 3 i + 5 j

A 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:

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

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

= Fr = Fr Torque = force x moment armTorque = force x moment arm

Summary: Resultant TorqueSummary: Resultant Torque• Read, draw, and label a rough figure.

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

• Extend lines of action for each force.

• Calculate moment arms if necessary.

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

• Resultant torque is sum of individual torques.

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

•• Draw freeDraw free--body diagram showing all forces, body diagram showing all forces, distances, and axis of rotation.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 force Calculate torques due to EACH individual force affixing proper sign. CCW (+) and CW (affixing proper sign. CCW (+) and CW (--).).

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

CONCLUSION: Chapter 5ACONCLUSION: Chapter 5A Torque Torque