Resource Lesson
Torque: An Introduction
Printer Friendly Version
Basic Definition and Rotating Beams
Whenever a force is applied to a rigid body (a bar, a beam, a pole) it usually results in the rigid body rotating about an axis or pivot - that is, a
torque
has been applied.
In the following diagrams, we first see a uniform beam balanced on a knife-edge. This is accomplished by placing the knife-edge in the exact center directly under the beam's center of gravity.
If the beam is shifted to either the right or left side of the knife-edge, it will no longer remain in equilibrium since the weight of the beam will be producing a torque. The beam will now be tilted.
If a mass having the same weight of the beam is placed equidistant from the pivot on the opposite side, the beam will once again be brought back to equilibrium and return to its original level orientation.
To calculate the magnitude of the torque produced by a force we use the formula
or
where
F
represents the magnitude of applied force
r
represents its moment arm
represents the angle formed between
F
and
r
If
F
and
r
are mutually perpendicular, sin 90º = 1 and
. The magnitude of the moment arm,
r
, is often written as
since it, more often than not, represents a length along a beam. The
moment arm
is defined as the perpendicular distance from the line of action of the force to the pivot point. Since
F
is measured in newtons and
is in meters, torque is measured in m nt.
Alert! torque is not measured in joules, even though a joule equals a newton-meter.
Refer to the following information for the next three questions.
Now let's practice identifying the magnitude of some moment arms and the direction of the resulting torque on a horizontal beam. In each diagram, the green triangle represents the pivot point.
A force F is applied to a bar midway between the pivot and the end of the bar.
A force F now pulls downward on the end of the bar at an angle θ.
A force F now pulls upward on the end of the bar at an angle θ.
If you would like to practice more on calculating moment arms, torques, and the direction of rotation on horizontal beams, then use
this accompanying worksheet
.
Torque as a Vector Cross Product and the Right Hand Rule
Newton's Second Law, net F = ma, has an analogous rotational expression,
net τ = Iα
. It is this analogous expression that we are going to study in this lesson. Remember that we previously calculated
torque
as the product of a force and its moment arm.
Torque is an example of a vector quantity formed by a
cross product
of two vectors
r
and
F
. One mathematical equation used to calculate this vector cross product is
where
θ represents the angle between vectors
r
and
F
and the double bars
represent the magnitude, or norm, of vectors A and B
θ's location is determined by following this procedure: with
r
and
F
drawn head-to-tail, θ begins on
r
and terminates on
F
. The direction of the cross product is determined by a right hand rule (RHR). With your right hand, curl your fingers in the direction of θ. The extended thumb points in the direction of
r x F
.
Refer to the following information for the next question.
For the diagram shown above, use the RHR to determine the direction of
A x B
?
Refer to the following information for the next three questions.
First, let's practice identifying the magnitude of some moment arms and the direction of the resulting torque on a circular wheel. In each diagram, the green dot represents the axis of revolution.
Each force has a magnitude of F and is being applied tangentially to the wheel. What is the moment arm of each force and in what direction would the wheel rotate?
Each force has a magnitude of F and is being applied tangentially to the wheel. What is the moment arm of each force and in what direction would the wheel rotate?
Each force has a magnitude of F and is being applied radially to the wheel. What is the moment arm of each force and in what direction would the wheel rotate?
Refer to the following information for the next three questions.
If a spool of thread were sitting on a horizontal table, which way would it move if it were to be pulled by each of the three strings shown below?
You might want to take a moment and practice the CP workbook page on
cams and spools
.
F
_{1}
?
F
_{2}
?
F
_{3}
?
Determinants
Numerically, a
cross product
can also be calculated as a determinant. For torque,
where
r
_{x}
,
r
_{y}
, and
r
_{z}
are the components of the radius vector
r
and
F
_{x}
,
F
_{y}
, and
F
_{z}
are the components of the force vector
F
. The letters
i
,
j
,
k
represent the unit vectors along the x, y, and z axis.
i
= (1, 0, 0)
j
= (0, 1, 0)
k
= (0, 0, 1)
Let's look at some numerical examples.
Force F equals 100 N and is applied 25 cm from the pivot. What is the magnitude and direction of the torque it creates?
Force F equals (100 N, 37º) and is applied at the end of the beam, 50 cm from the pivot. What is the magnitude and direction of the torque it creates?
Force F equals (100 N, -37º) and is applied at the end of the beam, 50 cm from the pivot. What is the magnitude and direction of the torque it creates?
What torque is produced by the weight of a 60-kg girl who has climbed 1 meter along a ladder that is leaning against a wall at an angle of 53º?
Notice in the last example that using a determinant allows you to not need to calculate the angle between
r
and
F
- an area where some students might make a mistake.
Related Documents
Lab:
Labs -
A Physical Pendulum, The Parallel Axis Theorem and A Bit of Calculus
Labs -
Conservation of Momentum in Two-Dimensions
Labs -
Density of an Unknown Fluid
Labs -
Mass of a Paper Clip
Labs -
Moment of Inertia of a Bicycle Wheel
Labs -
Rotational Inertia
Resource Lesson:
RL -
A Chart of Common Moments of Inertia
RL -
A Further Look at Angular Momentum
RL -
Center of Mass
RL -
Centripetal Acceleration and Angular Motion
RL -
Discrete Masses: Center of Mass and Moment of Inertia
RL -
Hinged Board
RL -
Introduction to Angular Momentum
RL -
Rolling and Slipping
RL -
Rotary Motion
RL -
Rotational Dynamics: Pivoting Rods
RL -
Rotational Dynamics: Pulleys
RL -
Rotational Dynamics: Rolling Spheres/Cylinders
RL -
Rotational Equilibrium
RL -
Rotational Kinematics
RL -
Rotational Kinetic Energy
RL -
Thin Rods: Center of Mass
RL -
Thin Rods: Moment of Inertia
Worksheet:
APP -
The Baton Twirler
APP -
The See-Saw Scene
CP -
Center of Gravity
CP -
Torque Beams
CP -
Torque: Cams and Spools
NT -
Center of Gravity
NT -
Center of Gravity vs Torque
NT -
Falling Sticks
NT -
Rolling Cans
NT -
Rolling Spool
WS -
Moment Arms
WS -
Moments of Inertia and Angular Momentum
WS -
Practice: Uniform Circular Motion
WS -
Rotational Kinetic Energy
WS -
Torque: Rotational Equilibrium Problems
TB -
Basic Torque Problems
TB -
Center of Mass (Discrete Collections)
TB -
Moment of Inertia (Discrete Collections)
TB -
Rotational Kinematics
TB -
Rotational Kinematics #2
PhysicsLAB
Copyright © 1997-2018
Catharine H. Colwell
All rights reserved.
Application Programmer
Mark Acton