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
, is the tendency of a force to
t
rotate an object about some axis
► Torque,
Torque
Torque is proportional to the magnitude of
F and to the distance from the axis. Thus, a
tentative formula might be:
Door example:
t = Fd
§ t is the torque
§ d is the lever arm (or moment arm)
§ F is the force
Torque units
A torque of 1 NŸm applied on an angle of
1 radian requires 1 Joule (1 NŸm) of energy.
E =tq
1J = 1 NŸmŸrad
1 J/rad = 1 NŸm of torque
When we studied WORK, we learned that
E/q = t Note:
a NŸm is a Joule. However, for TORQUE we
never call a NŸm a Joule, since technically, it’s a
J/rad. Therefore we simply say “Newton-meter"
Another way of thinking about it:
E =tq
is analogous to W
Units:
=Fd
Torque
Torque 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
t = (40 N)(0.06 m)
= 2.4 N×m
t = 2.4 N×m
6 cm
40 N
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 not
applied perpendicular to
the length of the object
being rotated.
Torque
Here, 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 = rF
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
Another example:
ϕ
F
= F sin f
t=
r = F r sin f
Alternately…
ϕ
= r sin f
t=F
= F r sin f
In both diagrams
t = F r sin f
where
f is the angle between F and r
Concept Check #1
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? Will it
be easier to open the door?
1. No
2. Yes
Concept Check #1
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?
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!
Concept Check #1
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? Will it
be easier to open the door?
1. No
2. Yes
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:
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, 3
or
2, 4, 1, 3
Since 4 & 1 are equivalent
Direction of Torque
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 Torque
By convention, counterclockwise torques are
positive and clockwise torques are negative.
Positive torque:
Counter-clockwise,
out of page
cw
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.
positive
12 cm
Resolve 80-N force into components as shown.
t = F(sin60) r
t = (69.3 N)(0.12 m)
t = 8.31 N m
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
r = 12 cm sin 600
= 10.4 cm
t = (80 N)(0.104 m)
= 8.31 N m, as before
Example #2: Calculate a torque
§ A 20-centimeter wrench is
used to loosen a bolt.
§ The force is applied 0.20 m
from the bolt.
§ 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?
t1 = F1 r⁄ = t2
= F2 r⁄ sin f where f is
= 57.7 N
60o,
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:
Find t due to
each force.
Consider 20-N
force first:
negative
30 N
r
300
2m
6m
40 N
20 N
300
A
4m
r = (4 m) sin 300
= 2.00 m
The torque about A is
clockwise and negative.
t = Fr = (20 N)(2 m)
= 40 N m, cw
t20 = -40 N m
Example 3 (Cont.): Next we find torque
due to 30-N force about same axis A.
Find t due to
each force.
Consider 30-N
force next.
r
negative
30 N
300
20 N
300
2m
6m
40 N
A
4m
r = (8 m) sin 300
= 4.00 m
The torque about A is
clockwise and negative.
t = Fr = (30 N)(4 m)
= 120 N m, cw
t30 = -120 N m
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:
positive
30 N
r
300
2m
6m
40 N
20 N
300
A
4m
r = (2 m) sin 900
= 2.00 m
The torque about A is
CCW and positive.
t = Fr = (40 N)(2 m)
= 80 N m, ccw
t40 = +80 N m
Example 3 (Conclusion): Find resultant
torque about axis A for the arrangement
shown below:
Resultant torque
is the sum of
individual torques.
20 N
30 N
300
300
2m
6m
40 N
A
4m
tR = t20 + t30 + t40 = -40 N m -120 N m + 80 N m
tR = - 80 N m
Clockwise
Example 4:
N’
Determine the net torque:
4m
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d1=4 m
d2=2 m
Find:
St = ?
500 N
800 N
1. Draw all applicable forces
2. Consider CCW rotation to be positive
åt = (500 N )(4 m) + (-)(800 N )(2 m)
= +2000 N × m - 1600 N × m
= +400 N × m
Rotation would be CCW
ü
Example 5: Where would the 500 N
person have to be relative to fulcrum for
zero torque?
Example 5:
N’
d1
y
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d2=2 m
St = 0
Find:
d1= ?
500 N
800 N
1. Draw all applicable forces and moment arms
Example 5:
N’
d1 m
y
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d2=2 m
St = 0
Find:
d1= ?
500 N
800 N
1. Draw all applicable forces and moment arms
åt
åt
RHS
= - (800 N )(2 m)
LHS
= (500 N )(d 2 m)
-800 × 2 [ N × m] + 500 × d 2 [ N × m] = 0
Þ d 2 = 3.2 m
ü
According to our understanding of torque
there would be no rotation and no motion!
What does it say about acceleration and force?
Thus, according to 2nd Newton’s law SF=0 and a=0!
å F =(-500 N ) + N '+(-800 N ) = 0
i
N ' = 1300 N
Torque and Equilibrium
►
First Condition of Equilibrium
► The
net external force must be zero
S F=0
S Fx = 0 and S Fy = 0
§ 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
► This
t=0
is a statement of rotational equilibrium
Calculate using equilibrium
#6.
§ 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?
Calculate using equilibrium
#6.
§ 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)
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 = 0
Solve for FB
Plug value of FB into first equation to
solve for FA:
Rotational Equilibrium