Torque
Torque is defined as the tendency to
produce a change in rotational motion.
Torque is a twist
or turn that tends
to produce
rotation. * * *
Applications are
found in many
common tools
around the home
or industry where
it is necessary to
turn, tighten or
loosen devices.
What makes something rotate in
the first place?
TORQUE
How do I apply a force to make the rod rotate
about the axis? Not just anywhere!
AXIS
TORQUE
• To make an object rotate, a force must
be applied in the right place.
• the combination of force and the distance
from the axis to the point of application
(L) is called TORQUE.
lever arm, L
Axis
Force, F
Torque = force times lever arm
Torque = F  L
Torque example
F
L
What is the torque on a bolt
applied with a wrench that
has a lever arm of 30 cm
with a force of 30 N?
Torque = F  L
= 30 N  0.30 m
=9Nm
For the same force, you get more torque
with a bigger wrench  the job is easier!
Units for Torque
Torque is proportional to the magnitude of
F and to the distance L from the axis. Thus,
a tentative formula might be:
t = FL
Units: Nm or lbft
t = (40 N)(0.60 m)
= 24.0 Nm,
t = 24.0 Nm,
60 cm
40 N
Direction of Torque
Torque is a vector quantity that has
direction as well as magnitude.
Turning the handle of a
screwdriver clockwise
(negative) and then
counterclockwise (positive)
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
Line 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.
F1
F2
Line of
action
F3
The Moment Arm
The lever arm of a force is the perpendicular
distance from the line of action of a force to the
axis of rotation.
F1
F2
L
L
L
F3
Calculating Torque
• Read problem and draw a rough figure.
• Extend line of action of the force.
• Draw and label Lever arm.
• Calculate the Lever arm if necessary.
• Apply definition of torque:
t = FL
Torque = force x Lever arm
Torque
• If we know the angle  between F
and r, we can calculate torque!
t = F L
– r is the total length
– F is force
L = r sin 
t = r sin  F
Hinge (rotates)
r
Direction of
rotation
L
–  is angle between F and r
• The SI unit of torque is the
Nm.
F

Extend the
line of action
Example 1: An 80-N force acts at the end of
a 12-cm wrench as shown. Find the torque.
L
• Extend line of action, draw, calculate L
t = (80 N)(.12m sin 60o) = 8.31 N m
Net Torque
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
v
a
0
t


0
t
What mass is needed for the
levers to be in equilibrium?
Weights are attached to 8 meter long levers at rest.
Determine the unknown weights below
??
20 N
??
20 N
20 N
??
What mass is needed for the
levers to be in equilibrium?
Upward force
from the fulcrum
produces no torque
(since r = 0)
r1 = 4 m
r2 = 4 m
0 = -F2r2 + F1r1
0 = -(F2)(4)+ (20)(4)
F2 = 20 N … same as F1
F1 = 20 N
F2 =??
What mass is needed for the
levers to be in equilibrium?
r1 = 4 m
r2 = 2 m
0 = -F2r2 + F1r1
0 = -(F2)(2) + (20)(4)
F1 = 20 N
F2 = 40 N
F2 =??
(force at the fulcrum is not shown)
More interesting problems
(the pivot is not at the center of
mass)
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
CM
20 N
??
More interesting problems
(the pivot is not at the center of mass)
Trick: gravity applies a torque “equivalent to”
(the weight of the lever)(Rcm)
tcm =(mg)(rcm) = (100 N)(2 m) = 200 Nm
CM
??
20 N
Weight of lever
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
CM
R1 = 6 m
Rcm = 2 m
R2 = 2 m
0 = -F2r2 + F1r1 + FcmRcm
0 = -(F2)(2) + (20)(6)+(100)(2)
Fcm = 100 N
F2 = 160 N
F1 = 20 N
F2 = ??
Diving board
A 8 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
bolt
tcm = (392 N) 2 m = 784 Nm
F1 r1 = F2 r2 + tcm
The Pivot point is not at the
center of mass
Diving board
A 8 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
(Balance Torques)
bolt
R1 = 2
Fbolt = ? N
F1 r1 = tcm = 784
Nm
R =2
cm
Fcm = 392 N
F1 = 784Nm = 392 N
2m
Diving board
A 4 meter long diving board with a mass of 40 kg.
b. Determine the upward force applied by the fulcrum.
(Balance Forces)
F = 784 N
bolt
Fbolt = 392 N
Fcm = 392 N
The total upward
force has to
equal the total
downward for the
object to be
stable.
Remember:
An object is in “Equilibrium” when:
a. There is no net Torque
t  0
b. There is no net force acting on the object


F  0
Torque with two supports
FL
Fcm
Fr
1. Label all the forces
2. Choose a pivot point
3. Write the equation for net torques
Torque with two supports
F1
F2
Fcm
Lcm
cm
L4
L1
T net = 0
-F1 L1 + Fcm L cm + F2 (0)=0
Pivot
point
What is the force exerted at each end?
0 = F1 (0) -755 (1.40) – 167 (2.75) + F2 (5.50M)
F2 = 276 N
F2 + F1 = 755 + 167N
F1 = 646 N
Pick a
pivot pt.
F1
F2