Transcript Slide 1

TORQUE AND LEVERS
Lesson 5
FEELING TORQUE: ACTIVITY


When a force or set of forces causes a rigid body
to rotate, we say a torque has been applied.
Torque – the turning effect caused by a force on
a rigid object around a axis or fulcrum, symbol T;
it is measured in Newton-meters, or Nm; it can
be called a
“moment force”
EXAMPLE:
Every time you open a door, you are producing a
torque on the door.
 A small force applied far from the hinges can
produce the same amount of torque as a large
force applied closer to the hinges.

Distance
Distance
In order to create the largest amount of torque
possible when pushing on the door, the force
generated must be at a 90 degree angle to the
door.
 Example 2

 The
amount of torque produced depends
on two factors.
1. The magnitude of the force (F)
applied to the rigid object.
2.
The distance (d) between the force and
the axis or fulcrum.
Using the symbol T for the magnitude of torque,
the following statements hold true:
 T increases as F increases ( T  F)
 T increases as d increases ( T  d)


Torque = force x distance or T = Fd (where F
is perpendicular to the ridge object)
EXAMPLE PROBLEM 1:

Calculate the magnitude of a wrench
experiencing a force of 84 N 0.25 m away from
the bolt.
F
= 84 N
 D = 0.25 m
T = ?
T
= Fd
 = (84 N) ( 0.25 m)
= 21 Nm
Therefore,
the magnitude of the
torque on the wrench is 21 Nm.
TORQUE ON LEVERS


Two torques can be calculated for a lever: the
effort torque (TE) and the load torque (TL).
The associated distances are the effort distance,
or effort arm (dE), and the load distance, or load
arm (dL).
dE
TE
dL
TL

Effort torque = effort force x effort arm
 TE = FEdE



Or
Load Torque = load force x load arm
 TL = FLdL
In each case, the force is perpendicular to the
lever, which allows us to deal with magnitudes
only, thus avoiding vector signs.
Example problem 2:
 A camper is using a large plank as a first class
lever to move a rock. The effort force has a
magnitude of 4.5 x 102 N, and the distance from
the fulcrum to the effort force is 2.2 m. What is
the magnitude of the effort torque produced?
(ignore the mass of the plank)

 FE
= 4.5 x 102 N
 dE = 2.2 m
 TE = ?
 TE
= FEdE
 = (4.5 x 102 N) (2.2 m)
 = 9.9 x 102 Nm

Therefore, the magnitude of the effort torque
produced is 9.9 x 102 Nm.
STATIC EQUILIBRIUM OF LEVERS
 The
word static means at rest. A rigid
object that is in static equilibrium is at
rest in two ways.
1. It is not moving in any direction
2. It is not rotating
LAW OF THE LEVER
When
a lever is in static
equilibrium, the magnitude
of the effort torque equals the
magnitude of the load torque.
This law can be written in the equation form
 Effort torque = load torque

Effort force x effort arm = load force x load arm
 FEdE = FLdL
 For this equation, only the magnitudes of the
quantities are considered. This eliminates the
need for positive or negative signs.


Any of the 4 variables can be found by
rearranged the equation.
EXAMPLE PROBLEM 3:

A camper wants to mount a trailer on blocks for
the winter. One corner of the trailer is lifted by
applying an effort force using a 3.00 m steel bar.
The trailer is applying a load force of 1.8 x 103 N
0.45 m away from the fulcrum. Determine the
magnitude of the effort force required (ignore the
mass of the bar)
 FL
= 1.8 x 103 N
 dL = 0.45 m
 dE = 3.00 m – 0.45 m = 2.55 m
 FE = ?
FE = 3.2 x 102 N
 Therefore the force needed to lift the trailer is 3.2
x 102 N.

For any ridged object, the law of the lever can be
stated in more general terms based on which way
it is turned.
 The clockwise torque is balanced by the counter
clockwise torque.
 TCW = TCCW
 Where TCW = magnitude of the clockwise torque
on an object around the fulcrum.
 Where TCCW = magnitude of the counter clockwise
torque on an object around the fulcrum.

QUESTIONS: HAND IN
1.
2.
3.
A mechanic applies a force of 540 N
perpendicular to a wrench to loosen a nut.
Calculate the magnitude of the torque if the
distance from the applied force to the nut is (a)
0.30 m and (b) 0.50 m T (2)
A person applies a force of 150 N at the hinge
side of a door, which means that d is zero. What
is the magnitude of the torque produced on the
door? C (1)
Estimate the magnitude of the maximum effort
torque you could produce on a wheel nut using a
tire wrench that is 50 cm long. (Hint, Do you
think your maximum effort force could be
greater than your own weight?) T (1) C (1)

For the following questions, solve for the
unknown.
 Find dE given FE = 15 N, FL = 75 N, and dL =
0.15 m T (1)
 Find FL given FE = 64 N, dE = 3.5 m, and dL =
0.70 m T (1)
 Find dL given FE = 32 N, FL = 640 N, and dE =
2.0 m T (1)

Calculate the effort force needed for the situation
shown below. T (2)
1.6 m
0.8 m
Effort force
640 N