TORQUE

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?

•The answer to all is

“torque”.

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;

Let’s recall:•What does the first condition of equilibrium tell us about the condition of an object in equilibrium?

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

LET’S ANALYZE THE

FOLLOWING

SITUATIONS

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

FF

EW

F

F

If the same two forces are applied as shown in the figure, does equilibrium exist?

Situation 2

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?

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?

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?

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?

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

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

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.

LET’S ANALYZE

THE MEANING OF

MOMENT ARM

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?

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?

TORQUEis the tendency to produce a change in rotational motion. It is also called moment of force.

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.

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

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.

F

F(-) (+)

(+)(-)FClockwise torque Counterclockwise

torque

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

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

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°

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

Illustrative Example

3. Find the moment arm about axis B.

A

F

B2 m 3 m

30°

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

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°

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

-

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

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

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?

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?

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

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

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?

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.

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.

THE END

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

Solution

Τ1 = F1 r1 =

Τ2 = F2 r2 =

Τ3 = F3 r3 =

Calculate for moment arm and torque

r1r2

r3