Download Slide 1

Torque
We know that Newton’s second law (  F  ma )
explains that the net force is the source of an
object’s acceleration.
What is the source of a rotating object’s angular
acceleration? It can’t be just a force, because it
matters where on the object that force is applied.
The answer lies in the quantity called torque.
Torque…

Torque, t, is the tendency of a force to rotate
an object about some axis


Torque is a vector
t = r F sin f = F d



F is the force
f is the angle the force makes with the horizontal
d is the moment arm (or lever arm)
…Torque…

The moment arm, d, is
the perpendicular
distance from the axis
of rotation to a line
drawn along the
direction of the force

d = r sin Φ
…Torque


The horizontal component of F (F cos f) has
no tendency to produce a rotation
Torque will have direction


If the turning tendency of the force is
counterclockwise, the torque will be positive
If the turning tendency is clockwise, the torque will
be negative
Conceptest…
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?
A.yes
B.no
…Conceptest
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?
A.yes
B.no
Conceptest…
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:
…Conceptest
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:
Conceptest…
A plumber pushes
straight down on the
end of a long wrench
as shown. What is the
magnitude of the
torque he applies about
the pipe at lower right?
A. (0.80 m)(900 N)sin 19°
B. (0.80 m)(900 N)cos 19°
C. (0.80 m)(900 N)tan 19°
D. none of the above
…Conceptest
A plumber pushes
straight down on the
end of a long wrench
as shown. What is the
magnitude of the
torque he applies about
the pipe at lower right?
A. (0.80 m)(900 N)sin 19°
B. (0.80 m)(900 N)cos 19°
C. (0.80 m)(900 N)tan 19°
D. none of the above
Net Torque



The force F1 will tend to
cause a
counterclockwise
rotation about O
The force F2 will tend to
cause a clockwise
rotation about O
St  t1 + t2  F1d1 –
F2d2
Torque vs. Force

Forces can cause a change in linear
motion


Described by Newton’s Second Law
Forces can cause a change in rotational
motion


The effectiveness of this change depends on
the force and the moment arm
The change in rotational motion depends on
the torque
Torque Units

The SI units of torque are N.m


Although torque is a force multiplied by a
distance, it is very different from work and energy
The units for torque are reported in N.m and not
changed to Joules
Torque and Angular Acceleration,
Wheel Example

The wheel is rotating
and so we apply St  Ia


The tension supplies the
tangential force
The mass is moving in
a straight line, so apply
Newton’s Second Law

SFy = may = mg - T
Torque and Angular Acceleration,
Multi-body Ex., 1

Both masses move in
linear directions, so
apply Newton’s Second
Law

Both pulleys rotate, so
apply the torque
equation
Torque and Angular Acceleration,
Multi-body Ex., 2


The mg and n forces on each pulley act at the
axis of rotation and so supply no torque
Apply the appropriate signs for clockwise and
counterclockwise rotations in the torque
equations
Problem
A model airplane with mass 0.750 kg is tethered by a
wire so that it flies in a circle 30.0 m in radius. The
airplane engine provides a net thrust of 0.800 N
perpendicular to the tethering wire.
(a) Find the torque the net thrust produces about the
center of the circle.
(b) Find the angular acceleration of the airplane when
it is in level flight.
(c) Find the linear acceleration of the airplane tangent
to its flight path.
Answers
a)
b)
c)
24.0 N-m
0.0356 rad/s2
1.07 m/s2
Review: The Vector Product

Given two vectors, A and B

The vector (“cross”) product of A and B is
defined as a third vector, C

C is read as “A cross B”

The magnitude of C is AB sin q

q is the angle between A and B
More About the Vector Product



The quantity AB sin q is
equal to the area of the
parallelogram formed by A
and B
The direction of C is
perpendicular to the plane
formed by A and B
The best way to determine
this direction is to use the
right-hand rule
Properties of the Vector Product



The vector product is not commutative. The
order in which the vectors are multiplied is
important
To account for order, remember
AxB=-BxA
If A is parallel to B (q = 0o or 180o), then A x
B=0

Therefore A x A = 0
Vector Products of Unit Vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  ˆj  ˆi  kˆ
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  ˆi  ˆi  kˆ  ˆj
Signs are interchangeable in cross products
A x (-B) = - A x B
 
 ˆi  ˆj  ˆi  ˆj
The Vector Product and Torque



The torque vector lies in a
direction perpendicular to
the plane formed by the
position vector and the
force vector
t=rxF
The torque is the vector (or
cross) product of the
position vector and the
force vector
Torque Vector Example

Given the force
F  (2.00 ˆi + 3.00 ˆj) N
r  (4.00 ˆi + 5.00 ˆj) m

t=?
t  r  F  [(4.00ˆi + 5.00ˆj)N]  [(2.00ˆi + 3.00ˆj)m]
 [(4.00)(2.00)ˆi  ˆi + (4.00)(3.00)ˆi  ˆj
+(5.00)(2.00)ˆj  ˆi + (5.00)(3.00)ˆi  ˆj
 2.0 kˆ N  m