Chapter 9
Torque
9.1 The Action of Forces and Torques on Rigid Objects
A net force causes an object to accelerate. This is a
translational acceleration.
What causes a rotational, or angular, acceleration?
Opening a door
Consider the common experience of pushing open a
door. Shown is a top view of a door hinged on the left.
Four pushing forces are shown, all of equal strength.
Which of these will be most effective at opening the
door?
The ability of a force to cause a rotation depends on
three factors:
1. the magnitude F of the force.
2. the radial distance r from the point of application of
the force to the hinge, or pivot.
3. the angle at which the force is applied.
Some vocabulary
• Line of action, line of force (F) (blue dashed
line) – the line along which the force acts.
• Axis of rotation (hinge, pivot) – in the examples
above, the axis of rotation is out of the page (z
axis).
More Vocabulary
•Radial axis (r) top of the door - in all three the example above, the radial
axis is the radius of the circle the door makes as it swings around the
hinge.
•Lever arm (ℓ) (moment arm) – red dashed line – the line that makes a
900 between the axis of rotation and the line of action. In the first
diagram, the lever arm and the radial axis coincide. In the third diagram,
there is no lever arm.
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
  F
ℓ = r sin θ
Where θ is the angle between the line of action and the radial axis
Direction of Torque: The torque is positive when the force tends to
produce a counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
9.1 The Action of Forces and Torques on Rigid Objects
The amount of torque depends on where and in what direction the
force is applied, as well as the location of the axis of rotation.
Rank in order, from largest to smallest, the
five torques τa − τe. The rods all have the
same length and are pivoted at the dot.
(a)
(b)
(c)
(d)
(e)
Rank in order, from largest to smallest, the
five torques τa − τe. The rods all have the
same length and are pivoted at the dot.
(a)

= r sin θ
(b)
(c)
(d)
(e)
EOC #9
The left end of the meter stick is pinned to the table so the
stick can rotate freely parallel to the table top. Two
forces, both parallel to the table are applied to the stick.
The net torque is zero. The forces and angles are
shown. How far from the pivot point (or axis of rotation)
is the 6.00 N force applied?
EOC #9 - Analysis
The drawing below acts as an extended free body diagram. The weight
force and the normal force from the table are parallel to the axis of
rotation (pivot point), and provide no torque. The 6.00 N force tends
to cause cw rotation, so the torque about the pivot from that force is
negative. The 4.00 N force tends to cause ccw rotation, so the
torque about the pivot from that force is positive. Since the problem
states the net torque is zero, these two torques must have an equal
magnitude.
knowns
Find r2
F1 = 4.00N
θ1 = 90°
F2 = 6.00 N
θ2 = 60°
r1 = 1 m
‫ =ז‬Fr(sin θ)
EOC #9 - Answer
Σ‫ = ז‬0
‫ז‬1 – ‫ז‬2 = 0
F1r1(sin θ1) – F2 r2 (sin θ2) = 0
4 N(1m) – 6N (r sin 60°) = 0
r = 0.77 m
knowns
Find r2
F1 = 4.00N
θ1 = 90°
F2 = 6.00 N
θ2 = 60°
r1 = 1 m
‫ =ז‬Fr(sin θ)
EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium if:
F
x
0
F
y
0
  0
This analysis will most likely involve multiple equations
with more than one unknown.
Reasoning Strategy
1. Select the object to which the equations for equilibrium are to be applied.
2. This is the most important step. Draw an extended free-body diagram that
shows all of the external forces acting on the object. You will no longer be
able to use a particle to represent the object.
3. Choose a convenient set of x, y axes and resolve all forces into components
along these axes. The weight force acts at the center of gravity. Assume the
center of gravity is at the mid point of the object, unless told otherwise.
4. Apply Newton’s 1st Law (since ma=0 for equilibrium) in component form.
5. Select a pivot point where one or more of the unknown forces will have a
torque of zero. Set the sum of the torques about this axis equal to zero.
The pivot you pick does not have to be an actual axis of rotation for this
object!
6. Solve the equations for the desired unknown quantities.
Example 3 A Diving Board
A woman whose weight is 530 N is
poised at the right end of a diving board with
length 3.90 m. The board has negligible
weight and is supported by a fulcrum 1.40 m
away from the left end. Find the forces that the
bolt and the fulcrum exert on the board.
known
W - normal force of woman on board,
numerically equal to her weight,
Lw = 3.90m
L2 = 1.4 m
Find
F1 force of bolt on board
F2 force of fulcrum on board
A diving board
1. Select the object (the board).
2. Draw a free-body diagram
(diagram b). Note the
direction of forces. The
woman pushes down on the
board. The fulcrum pushes
up (a normal force) With the
woman on the board, the
diving board would rotate in a
clockwise direction about the
fulcrum. What prevents this?
The downward force of the
bolt.
A diving board
3. Choose a convenient
set of x, y axes and
resolve all forces
into components
along these axes (All
forces are in the y
direction –object
weight negligible –
so true horizonal/
vertical axes are
appropriate).
y
x
A diving board
4. Apply Newton’s Law along
the y axis:
 Fy  0
F2 - F1 – W = 0
F2 = F1 + W
Both F1 and F2 are unknowns
Uh-oh
known
W - normal force of woman on board,
numerically equal to her weight,
Lw = 3.90m
L2 = 1.4 m
Find
F1 force of bolt on board
F2 force of fulcrum on board
y
x
A diving board
5. Select a pivot point where
one or more of the
unknown forces will
have a torque of zero.
The pivot point is not
necessarily the point
about which the object
is most likely to pivot. In
this case, we can
choose as the pivot
point the position of the
fulcrum, or the position
of the bolt, since they
both have one unknown
force with a torque of
zero about that point (no
lever arm).
y
x
A diving board
5 (cont’d) In this problem, we
choose the bolt to be the pivot
point. Set the sum of the
torques about this axis equal
to zero. Negative sign is if the
force causes the board to
rotate cw
  0
  F 
2 2
F2
about the bolt
y
 W W  0

530 N 3.90 m 

 1480 N
1.40 m
x
Diving Board
F
y
 F1  F2  W  0
obvious that F1 must point
down for board to be in
equilibrium, so it gets a
negative sign in Newton’s 2nd
Law
 F1  1480 N  530 N  0
F1  950 N
Walking the Plank
Adrienne (50 kg) and Bo (90 kg) are playing on a
100 kg rigid plank resting on the supports
shown. How far can Bo walk before the plank
tips? (hint, the plank “tips” when it begins to rise
up from the left support i.e. when nL = 0).
Walking the Plank
Adrienne (50 kg) and Bo (90 kg)
are playing on a 100 kg rigid
plank resting on the supports
shown. How far can Bo walk
before the plank tips?
Using the left support as the pivot
point we calculate the sum of the
forces and the sum of the torques
with n1 = 0. This results in db = 6.3
m, which means Bo can walk 8.3
meters away from Adrienne before
the plank begins to tip.
Will the ladder slip?
A 3.0-m long ladder
leans against a
frictionless wall at an
angle of 60°. What is
the minimum value of
us that prevents the
ladder from slipping?
Will the ladder slip?
A 3.0-m long ladder
leans against a
frictionless wall at an
angle of 60°. What is
the minimum value of
us that prevents the
ladder from slipping?
Answer:
us-min = 1/(2 tan 60°) =
0.29
The uniform beam has a weight of
1220 N. It is attached to a
vertical wall at one end and is
supported by a cable at the
other end. A 1960-N block
hangs from the far end of the
beam. Find:
a. The magnitude of the tension in
the cable.
b. The magnitude of the horizontal
c. and vertical components of the
force that the wall exerts on the
left end of the beam.
The uniform beam has a weight of
1220 N. It is attached to a
vertical wall at one end and is
supported by a cable at the
other end. A 1960-N block
hangs from the far end of the
beam. Find:
a. The magnitude of the tension in
the cable. 2260N
b. The magnitude of the horizontal
c. and vertical components of the
force that the wall exerts on the
left end of the beam. Both =
1450N.