torque
TRANSCRIPT
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TORQUE
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Think of these!• Why is it easier to open the tight lid of a
can of paint using a long screw driver than a 5-peso coin?
• Why are door knobs placed at the other end of the door opposite the hinges and not at the middle?
• Why is it easier to maneuver a bicycle with handle bar than one without?
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•The answer to all is
“torque”.
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OUR TARGET• Define torque;• Illustrate by example the understanding
of the term moment arm;• Calculate the resultant torque about
any axis when given the magnitude and position of forces on an extended object;
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Let’s recall:•What does the first condition of equilibrium tell us about the condition of an object in equilibrium?
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The First Condition of Equilibrium
• “If all forces acting on a body intersect at a single point and their vector sum is zero, the system must be in equilibrium.”
• Mathematically: ΣF = 0
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LET’S ANALYZE THE
FOLLOWING
SITUATIONS
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F
F
N
S
F F
Does equilibrium exist if you exert a pair of equal opposing forces F to the right and to the left of the lug wrench?
Situation 1
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FF
EW
F
F
If the same two forces are applied as shown in the figure, does equilibrium exist?
Situation 2
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Questions
• How would you compare the vector sum of the forces in the two situations?
• How would you compare the effect of the pair of equal forces applied on the lug wrench in the two situations?
• What is the difference in the application of the two equal forces in situation 1 and 2?
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N
S
F
FEW
F
F
VECTOR DIAGRAMS
SITUATION 1 SITUATION 2
How would you compare the vector sum of the forces in the two situations?
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N
S
F
FNW
F
F
VECTOR DIAGRAMS
SITUATION 1 SITUATION 2
How would you compare the effect of the pair of equal forces applied on the lug wrench in the two situations?
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N
S
FF NW
F
F
VECTOR DIAGRAMS
SITUATION 1 SITUATION 2
What is the difference in the application of the two equal forces in situations 1 and 2?
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THE LINE OF ACTION• The line of action of a force is an
imaginary line extended indefinitely along the vector in both directions.
F
F
F
F
SΛ
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THE MOMENT ARM• The moment arm of a force is the
perpendicular distance from the line of action of the force to the axis of rotation.
WAxis of rotation
Line of action
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THE MOMENT ARM
Line of action
r = L sinθ
Axis of rotation
Length L
• The moment arm of a force is the perpendicular distance from the line of action of the force to the axis of rotation.
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LET’S ANALYZE
THE MEANING OF
MOMENT ARM
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A B C
F F F
Given three unbalanced forces A, B and C
1. What is the moment arm of A? B? C?
2. How would you compare the rotational effect of A, B and C?
3. Which of the three unbalanced forces A, B, and C is the most effective in producing rotational motion? Why? Least effective? Why?
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A B C
F F F
1. What is the moment arm of A? B? C?2. How would you compare the rotational effect of A, B and C?3. Which of the three unbalanced forces A, B, and C is the most effective in producing rotational motion? Why? Least effective? Why?
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TORQUEis the tendency to produce a change in rotational motion. It is also called moment of force.
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Rotational motion is affected by both the magnitude of force F and its moment arm (also called torque arm) r.
Thus we define torque as the product of a force and its moment arm.
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Mathematically:
τ = Fr Where:
τ (tau) = torque( in Newton-meter, N-m or pound-foot, lb-ft)
F = force (in Newton, N or pound, lb)
r = moment arm (in meter, m or foot, ft)
Torque = Force x moment arm
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Sign ConventionThe directions of torque depends
on whether it tends to produce clockwise or counterclockwise rotation about an axis.
We take counterclockwise rotation positive and
clockwise rotation negative.
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F
F(-) (+)
(+)(-)FClockwise torque Counterclockwise
torque
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Illustrative Example
1. Draw and label the moment arm of force F about an axis at point A and point B in the figure below? What is the magnitude of the moment arm? What is the resultant torque about axis A and B?
A
F
B2 m 3 m
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Solution
The moment arm about point A is the perpendicular distance from point A to the line of action of the force F
A
F
B2 m 3 m
The moment arm about point B is the perpendicular distance from point B to the line of action of the force F
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Illustrative Example
2. Draw and label the moment arm of force F about an axis at point A in the figure below? What is the magnitude of the moment arm?
A
F
B2 m 3 m
30°
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2. SolutionThe moment arm about point A is the
perpendicular distance from point A to the line of action of force F.
A
F
B2 m 3 m
30°
Its magnitude is :
r = L sin 30°
= 2m (0.500) = 1.0 m
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Illustrative Example
3. Find the moment arm about axis B.
A
F
B2 m 3 m
30°
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3. SolutionThe moment arm about point B is the
perpendicular distance from point B to the line of action of force F.
A
F
B2 m 3 m
30°
Its magnitude is :
r = L sin 30°
= 3m (0.500) = 1.5 m
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Illustrative Examples
4. If force F is 100 N, what is the resultant torque about an axis A? What is the resultant torque about axis B?
A
F
B2 m 3 m
30°
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4. Solution
A
F
B2 m 3 m
30°
Since rA = 1.0 m and F = 100 N
TA = FrA = 100 N (1.0 m) = 100N-m
r = 1.0 m
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4. Solution
Since rB = 1.5 m and F = 100 N
TB = FrB = 100 N (1.5 m) = 150N-m
A
F
B2 m 3 m
30°
r = 1.5 m
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Illustrative Example
5. What if force F is applied directly towards point B, what would be the resultant torque about axis A and B?
AF
B2 m 3 m
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Sum it up.If you are to explain torque to someone
who does not study physics, how would you explain it so that she/he can readily understand it?
How would you explain to him/her the importance of the moment arm in producing torque?
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Check your understanding.
1. Why is it easier to open a tight lid of a can of paint using a long screw driver than with a coin?
2. Why is the door knob always located at the opposite end of the door hinge? Why is it not placed at middle of the door?
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Which of the following will produce the greatest torque?
D= 1000 N
C=600 N
A= 1000 N
B=150 N
2.0 m 3.5 m 2.5 m
8.0 m
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Which of the following will produce the greatest torque?
D= 1000 N
C=600 N
A= 1000 N
B=450 N
2.0 m 3.5 m 2.5 m
8.0 m
20°
60°
T = 1000N (2.0m)sin60°
= 1732 N-m
T = 450N (5.50 m) = 2475 N-mT = 600N (5.50m)sin 20° = 1129 N-mT = 1000N (0.0m) = 0
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Assignment
A. Answer these in your assignment notebook.
• What is the 2nd Condition of Equilibrium?
• How is it different from the 1st condition of Equilibrium?
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B. Figure 1 shows an object with several forces acting on it. The pivot point is at O.
If F1 = 10 N, and is at a distance of 0.25 m from O, where Θ1 = 30° F2 = 7.0 N, acting perpendicular to the object, at a distance of 1.25 m from O F3 = 12 N, is 0.60 m from O, and acts at Θ3 = 40° from the horizontal
Find the total (net) torque on the object.
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Torque is greater with longer moment arm. How could this idea be applied to you when you are in trouble? Or when someone is in need?
Did MCS ever extend a helping hand to you when you are in need?
At your end, how do you support the school in helping others in need?
Think about this.
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THE END
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Figure 1 object with several torques. The net total torque on the object is: A. -1.7 N m B. -15.8 N m C. -6.6 N m D. 18 N m
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Solution
Τ1 = F1 r1 =
Τ2 = F2 r2 =
Τ3 = F3 r3 =
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Calculate for moment arm and torque
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r1r2
r3
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