chapter5a_torque.ppt

8/14/2019 Chapter5A_Torque.ppt

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Chapter 5A. Torque

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

2007


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Torque is a twist

or turn that tendsto producerotation. * * *

Applications are

found in manycommon toolsaround the home

or industry whereit is necessary toturn, tighten orloosen devices.


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Objectives: After completing this

module, you should be able to: Define and give examples ofthe terms torque,

moment arm, axis, andline of action of a force.

Draw, label and calculate the moment armsfora variety of applied forces given an axis ofrotation.

Calculate the resultant torqueabout any axis

given the magnitude and locations of forces onan extended object.

Optional:Define and apply the vector crossproductto calculate torque.


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Definition of Torque

Torqueis defined as the tendency toproduce a change in rotational motion.

Examples:


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Torque is Determined by Three Factors:

The magnitudeof the applied force.

The directionof the applied force.

The locationof the applied force.

20 N

Magnitude of force

40 N

The 40-Nforce

produces twice thetorque as does the

20-Nforce.

Each of the 20-N

forces has a differenttorque due to the

direction of force. 20 N

Direction of Force

20 Nq

q20 N

20 N

Location of forceThe forces nearer the

end of the wrenchhave greater torques.

20 N

20 N


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Units for Torque

Torque is proportional to the magnitude ofF and to the distance r from the axis. Thus,a tentativeformula might be:

t= Fr Units: Nmor lbft

6cm

40 N

t= (40 N)(0.60 m)

= 24.0 Nm, cw

t= 24.0 Nm, cw


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Direction of Torque

Torque is a vector quantity that hasdirection as well as magnitude.

Turning the handle of ascrewdriver clockwiseand

then counterclockwisewilladvance the screw firstinward and then outward.


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Sign Convention for Torque

By convention, counterclockwise torques arepositive and clockwise torques are negative.

Positivetorque:Counter-clockwise,

out of page

cw

ccw

Negativetorque:clockwise, into page


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Line of Action of a Force

The line of actionof a force is an imaginaryline of indefinite length drawn along thedirection of the force.

F1

F2

F3Line ofaction


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The Moment Arm

The moment armof a force is the perpendiculardistance from the line of action of a force to theaxis of rotation.

F2

F1

F3

r

rr


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Calculating 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:

t= Fr Torque = force x moment arm


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Example 1: An 80-Nforce acts at the end ofa 12-cmwrench as shown. Find the torque.

Extend line of action, draw, calculate r.

t= (80 N)(0.104 m)

= 8.31 N m

r = 12cm sin 600

= 10.4 cm


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Alternate: An 80-Nforce acts at the end ofa 12-cmwrench as shown. Find the torque.

Resolve 80-N force into components as shown.

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

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

positive

12 cm


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Calculating 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 forceaffixing proper sign. CCW (+) and CW (-).

Resultant torque is sum of individual torques.


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Example 2: Find resultant torque aboutaxisAfor the arrangement shown below:

300

300

6 m 2 m

4 m

20 N30 N

40 NA

Find t due to

each force.

Consider 20-N

force first:

r = (4 m) sin 300

= 2.00 m

t= Fr = (20 N)(2 m)

= 40 N m, cw

The torque about A isclockwise and negative.

t20= -40 N m

r

negative


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Example 2 (Cont.): Next we find torquedue to 30-Nforce about same axisA.

300

300

6 m 2 m

4 m

20 N30 N

40 NA

Find t due to

each force.

Consider 30-N

force next.

r = (8 m) sin 300

= 4.00 m

t= Fr = (30 N)(4 m)

= 120 N m, cw

The torque about A isclockwise and negative.

t30= -120N m

rnegative


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Example 2 (Cont.): Finally, we considerthe torque due to the 40-Nforce.

Find t due to

each force.

Consider 40-N

force next:

r = (2 m) sin 900

= 2.00 m

t= Fr = (40 N)(2 m)

= 80 N m, ccw

The torque about A isCCW and positive.

t40= +80N m

300

300

6 m 2 m

4 m

20 N30 N

40 NA

r

positive


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Example 2 (Conclusion): Find resultanttorque about axisAfor the arrangement

shown below:

300300

6 m 2 m4 m

20 N30 N

40 NA

Resultant torque

is the sum ofindividual torques.

tR= - 80N m Clockwise

tR=

t20 +

t20+

t20= -40 N m -120 N m + 80 N m


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Part II: Torque and the Cross

Product or Vector Product.Optional Discussion

This concludes the general treatmentof torque. Part II details the use ofthe vector product in calculating

resultant torque. Check with yourinstructor before studying this section.


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The Vector Product

Torque can also be found by using the vectorproduct of force Fand position vector r. Forexample, consider the figure below.

F

qr

F Sin qThe effect of the forceFat angle q(torque)

is to advance the boltoutof the page.

Torque

Magnitude:

(FSinq)r

Direction = Out of page (+).


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Definition of a Vector Product

The magnitude of the vector (cross) productof two vectorsAand Bis defined as follows:

AxB= l A l l B l Sin q

F xr = l F l l r l Sin q Magnitude only

F

(F Sinq) r or F (r Sinq)

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

In effect, this becomes simply:q

r

F Sin


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Example: Find the magnitude of thecross product of the vectors r and Fdrawn below:

r xF = l r l l F l Sin q

r xF = (6 in.)(12 lb) Sin 600


r xF= (6 in.)(12 lb) Sin 1200

Explain difference. Also, what about Fx r?

12 lb

r xF = 62.4 lb in.

Torque

600

6 in.

Torque600

6 in.

12 lb r xF = 62.4 lb in.


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Direction of the Vector Product.

The directionof avector product isdetermined by the

right hand rule. A

C

BB

-CA

A x B = C (up)

B x A = -C (Down)

Curl fingers of right handin direction of cross pro-duct (Ato B) or (BtoA).Thumbwill point in thedirection of product C.

What is directionof A x C?


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


r xF = (6 in.)(10 lb) Sin 500

r xF = 38.3 lb in.

10 lbTorque

500

6 in. Magnitude

Out

r

FDirection by right hand rule:

Out of paper (thumb) or +k

r xF = (38.3 lb in.) k

What are magnitude and direction of F x r?


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Cross Products Using (i,j,k)

x

z

yConsider 3D axes (x, y, z)

Define unit vectors, i, j, kij

kConsider cross product: i x i

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

= 0

i

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

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

Magnitudes arezero for parallelvector products.


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Vector Products Using (i,j,k)

Consider 3D axes (x, y, z)

Define unit vectors, i, j, kx

z

y

ij

k Consider dot product: i xj

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

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

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

ji

Magnitudes are1for perpendicularvector products.


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Vector Product (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= +1j

Directions are given by theright hand rule. Rotatingfirst vector into second.

k

j

i


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Vector Products Practice (i,j,k)

x

z

y

ij

ki xk= ?

k xj = ?

Directions are given by theright hand rule. Rotatingfirst vector into second.

k

j

i 2ix -3 k= ?

- j (down)

- i (left)

+ 6j (up)

j x-i = ? + k (out)


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Using i,j Notation - Vector Products

Consider: A= 2 i- 4j and B= 3 i + 5j

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

Alternative: A= 2 i- 4j

B= 3 i + 5j

A xB= 10 - (-12) = +22 k

Evaluatedeterminant


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Summary

Torqueis the product of a forceand itsmoment armas defined below:

The moment armof a force is the perpendicular distance

from the line of action of a force to the axis of rotation.

The line of actionof a force is an imaginary line ofindefinite length drawn along the direction of the force.

t= Fr Torque = force x moment arm


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Summary: 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 forceaffixing proper sign. CCW (+) and CW (-).

Resultant torque is sum of individual torques.


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CONCLUSION: Chapter 5A

Torque

chapter5a_torque.ppt

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