Post on 16-Dec-2015
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.
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.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.
• 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.
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:
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 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
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 Finally, we consider the torque due to the consider the torque due to the 40-N40-N 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 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:
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 arm
Torque = force x moment arm
Summary: Resultant Summary: 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.
• 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.
CONCLUSION: Chapter 5ACONCLUSION: Chapter 5ATorque Torque