Torque► Consider force required to

open door. Is it easier to open the door by pushing/pulling away from hinge or close to hinge?

close to hinge

away from hinge

Farther from from hinge, larger rotational effect!

Physics concept: torque

Torque►Torque, , is the tendency of a force to

rotate an object about some axist

Torque

§ is the torque§ d is the lever arm (or moment arm)§ F is the force

t

Door example:

Torque is proportional to the magnitude of F and to the distance from the axis. Thus, a tentative formula might be:

Fd=t

Torque units

A torque of 1 N�m applied on an angle of 1 radian requires 1 Joule (1 N�m) of energy.

E = t q1J = 1 N�m�rad1 J/rad = 1 N�m of torqueE/q = t Note: When we studied WORK, we learned that

a N�m is a Joule. However, for TORQUE we never call a N�m a Joule, since technically, it’s a J/rad. Therefore we simply say “Newton-meter"

Another way of thinking about it:

E = t q is analogous to W = F d

Units:

TorqueTorque is proportional to the magnitude of F and to the distance r from the axis of rotation. Thus, a tentative formula might be:

t = Fr Units: N×m or lb×ft

6 cm40 N

t = (40 N)(0.06 m)

= 2.4 N×m

t = 2.4 N×m

Calculate a torque

§ A force of 50 newtons is applied to a wrench that is 30 centimeters long.

§ Calculate the torque if the force is applied perpendicular to the wrench so the lever arm is 30 cm.

§ Torque = 50N (0.30 m) = 15 N-m

§ Sometimes the force is notapplied perpendicular to the length of the object being rotated.

TorqueHere, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.

Torque

The torque is defined as:

OR

t = r F

Two ways of looking at the definition of torque

Lever Arm• The lever arm, r , is

the shortest perpendicular distance from the axis of rotation to a line drawn along the the direction of the force

r = r sin Φ

• It is not necessarily the distance between the axis of rotation and point where the force is applied

Note: The important angle to consider is the angle between F and r

An Alternative Look at Torque► The force could also be resolved into its x- and y-components

§ The x-component, F cos Φ, produces 0 torque§ The y-component, F sin Φ, produces a non-zero torque

ϕ

F

= F sin f = F r sin ft = r

Another example:

Alternately…

= r sin f t = F = F r sin f

ϕ

In both diagrams

t = F r sin f

where f is the angle between F and r

Concept Check #1You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?

1. No2. Yes

Response: The torque is the same, because the force is the same, and the perpendicular distance isn't changed by lengthening the line of force. It's defined as the distance from the axis to the LINE of force, no matter where on that line the force acts!

You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased?

Concept Check #1

Concept Check #1You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?

1. No2. Yes

Concept Check #2You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:

Concept Check #2 You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:

2, 1, 4, 3or

2, 4, 1, 3 Since 4 & 1 are equivalent

Direction of TorqueTorque 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 TorqueBy convention, counterclockwise torques are positive and clockwise torques are negative.

Positive torque: Counter-clockwise,

out of pagecw

ccw

Negative torque: clockwise, into page

.

Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque.

Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque.

Resolve 80-N force into components as shown.

t = (69.3 N)(0.12 m)

t = 8.31 N m

positive12 cm

t = F(sin60) r

Alternately: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque.

• Extend line of action, draw, calculate r⊥

t = (80 N)(0.104 m)

= 8.31 N m, as beforer⊥= 12 cm sin 600

= 10.4 cm

Example #2: Calculate a torque

§ It takes 50 newtons to loosen the bolt when the force is applied perpendicular to the wrench.

§ How much force would it take if the force was applied at a 30-degree angle from perpendicular?

§ A 20-centimeter wrench is used to loosen a bolt.

§ The force is applied 0.20 m from the bolt.

t1 = F1 r where f is 60o, = t2 = F2 r sin f⁄ ⁄

= 57.7 N

the angle between F and r is not 30o !

What if two or more different forces act on a lever arm?

Net Torque

►The net torque is the sum of all the torques produced by all the forces

§ Remember to account for the direction of the tendency for rotation

►Counterclockwise torques are positive►Clockwise torques are negative

Example #3: Finish labeling the diagram on your paper, according to the following…

Example 3: Find resultant torque about axis A for the arrangement shown below:

300300

6 m 2 m 4 m

20 N30 N

40 NA

Find t due to each force.

Consider 20-Nforce first:

r = (4 m) sin 300

= 2.00 mt = Fr = (20 N)(2 m)

= 40 N m, cw

The torque about A is clockwise and negative.

t20 = -40 N m

r

negative

Example 3 (Cont.): Next we find torque due to 30-N force about same axis A.

300300

6 m 2 m 4 m

20 N30 N

40 NA

Find t due to each force.

Consider 30-Nforce next.

r = (8 m) sin 300

= 4.00 mt = Fr = (30 N)(4 m)

= 120 N m, cw

The torque about A is clockwise and negative.

t30 = -120 N m

rnegative

Example 3 (Cont.): Finally, we consider the torque due to the 40-N force.

Find t due to each force.

Consider 40-N force next:

r = (2 m) sin 900

= 2.00 mt = Fr = (40 N)(2 m)

= 80 N m, ccw

The torque about A is CCW and positive.

t40 = +80 N m

300300

6 m 2 m 4 m

20 N30 N

40 NA

r

positive

Example 3 (Conclusion): Find resultant torque about axis A for the arrangement shown below:

300300

6 m 2 m 4 m

20 N30 N

40 NA

Resultant torque is the sum of individual torques.

tR = - 80 N m Clockwise

tR = t20 + t30 + t40 = -40 N m -120 N m + 80 N m

Example 4:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d1=4 md2=2 m

Find:

St = ?

1. Draw all applicable forces

ü

2. Consider CCW rotation to be positive

500 N 800 N

4 m 2 m

Rotation would be CCW

N’

Determine the net torque:

(500 )(4 ) ( )(800 )(2 )2000 1600400

N m N mN m N mN m

t = + -

= + × - ×= + ×

å

Example 5: Where would the 500 N person have to be relative to fulcrum for

zero torque?

Example 5:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d2=2 mSt = 0

Find:

d1= ?

1. Draw all applicable forces and moment arms

500 N 800 N

2 md1

N’ y

2

2 2

(800 )(2 )

(500 )( )800 2 [ ] 500 [ ] 0 3.2

RHS

LHS

N m

N d mN m d N m d m

t

t

=-

=

- × × + × × = Þ =

åå

Example 5:

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d2=2 mSt = 0

Find:

d1= ?

1. Draw all applicable forces and moment arms

ü

500 N 800 N

2 md1 m

According to our understanding of torque there would be no rotation and no motion!

N’ y

What does it say about acceleration and force?

Thus, according to 2nd Newton’s law SF=0 and a=0! NNNNNFi

1300'0)800(')500(

=

=-++-=å

Torque and Equilibrium► First Condition of Equilibrium

► The net external force must be zero

§ This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium

§ This is a statement of translational equilibrium

► Second Condition of Equilibrium► The net external torque must be zero

►S t = 0►This is a statement of rotational equilibrium

S F = 0S Fx = 0 and S Fy = 0

Calculate using equilibrium

§ A boy and his cat sit on a seesaw.

§ The cat has a mass of 4 kg and sits 2 m from the center of rotation.

§ If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?

#6.

Calculate using equilibrium

§ A boy and his cat sit on a seesaw.

§ The cat has a mass of 4 kg and sits 2 m from the center of rotation.

§ If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance? (0.16m)

#6.

Axis of Rotation

► So far we have chosen an obvious axis of rotation► If the object is in equilibrium, it does not

matter where you put the axis of rotation for calculating the net torque§ The location of the axis of rotation is completely

arbitrary§ Often the nature of the problem will suggest a

convenient location for the axis§ When solving a problem, you must specify an axis

of rotation►Once you have chosen an axis, you must maintain that

choice consistently throughout the problem

Rotational Equilibrium§ When an object is in rotational equilibrium, the net torque

applied to it is zero.§ Rotational equilibrium is often used to determine unknown

forces.§ What are the forces (FA , FB) holding the bridge up at either

end?

Since it doesn’t matter where we define our point of rotation, let’s make it at point A.

Now set the sum of all torques = 0Solve for FB

Plug value of FB into first equation to solve for FA:

Rotational Equilibrium