Physics 2210 (Fall 2010)
Torque Supplement
Mike McLaughlin
Quick Note: We ran out of time today in section between trying to cover moment of inertia
and torque. These supplemental notes should provide you with a reasonable foundation on
the topic of torque.
1
Introduction
Torque, denoted by the Greek letter tau (τ ), is the rotational analog of force. Think of
torque as a measure of a force’s ability to cause rotation about a specified axis. All torques
are the result of a force, but not all forces generate torques.
1.1
Direction of Rotation
Torque is a vector (technically, it’s a pseudo-vector but we’ll ignore that issue in 2210) and
therefore it has a magnitude and direction. We’ll learn how to calculate the magnitude
later in these notes, but an important point needs to be made about the direction first.
Torques generate rotation of an object about an axis. These rotations are either clockwise
or counter-clockwise. The convention in Physics is to assign a positive sign (+) to any
torques that generate counter-clockwise rotations and a negative sign (-) to any torques
that generate clockwise rotations. (Note that we don’t really have the freedom of choice
here that we had with up/down and left/right of forces. The lack of choice is not because
it wouldn’t work, it’s simply because Physics has chosen a convention and you’ll get the
answer wrong if you have the “wrong” sign. I apologize on behalf of Physics for the
arbitrariness of this convention.)
2
Calculating Torque
We’ll consider the case of a very thin rod of length L = 3.1 m attached to a wall. The
rod is free to rotate in the vertical plane about the pivot point (noted by a red circle on
the diagram). We’ll ignore any influence of gravity in this problem and focus instead on
three forces (F~1 = 10 N, F~2 = 15 N, and F~3 = 20 N). The measure of θ is 28◦ . Our
goal: determine the torque generated by each of the three forces. We’ll do this using three
different methods.
F1
F3
θ
F2
2.1
Using the Cross Product - Part I
One definition of torque uses the vector cross product:
~τ = ~r × F~ = |~r||F~ | sin θ
(1)
Here, ~r is the position vector for the object as measured from the rotational axis point to
the point of contact with the force. As seen in the figure above, ~r would be a vector of
length L pointing in the x̂ direction. θ is defined as the angle measured counter-clockwise
from the direction of ~r to the direction of F~ . With these definitions in mind, let’s compute
the torque for each of the three forces in the diagram.
2.1.1
Torque Due to Force 1
From the diagram, you should see that the angle between ~r and F~1 is 270◦ . Therefore, our
torque calculation goes as follows:
τ~1 = ~r × F~1 = |~r||F~1 | sin θ1
(2)
τ1 = LF1 sin 270◦
(3)
τ1 = LF1 = −31 N · m
(4)
The torque is negative and therefore causes a clockwise rotation of the rod about the pivot
point (but you can also figure that out with common sense). The actual direction of the
torque vector is found by applying the right-hand-rule. In this case, as your fingers run
along the direction of ~r, you let them curl naturally in the direction of F~1 , and your thumb
indicates the direction of τ~1 which happens to be into the page.
2.1.2
Torque Due to Force 2
From the diagram, you should see that the angle between ~r and F~2 is 180◦ . Therefore, our
torque calculation goes as follows:
τ~2 = ~r × F~2 = |~r||F~2 | sin θ2
(5)
τ2 = LF2 sin 180◦
(6)
τ2 = LF2 = 0 N · m
(7)
The torque is zero and therefore causes no rotation of the rod about the pivot point.
2.1.3
Torque Due to Force 3
From the diagram, you should see that the angle between ~r and F~3 is 208◦ . Therefore, our
torque calculation goes as follows:
τ~3 = ~r × F~3 = |~r||F~3 | sin θ3
(8)
τ3 = LF3 sin 208◦
(9)
τ3 = LF3 = −29 N · m
(10)
The torque is negative and therefore causes a clockwise rotation of the rod about the pivot
point (but you can also figure that out with common sense). The actual direction of the
torque vector is found by applying the right-hand-rule. In this case, as your fingers run
along the direction of ~r, you let them curl naturally in the direction of F~3 , and your thumb
indicates the direction of τ~3 which happens to be into the page.
2.2
Using the Cross Product - Part II
~ = (−2, 1, 0)). You
Sometimes you are given vectors that are in coordinate form (e.g. A
can still perform the cross product in this instance. Imagine that our force vectors and
position vector from the example above were given in terms of coordinate notation:
F~1 = (0, −10, 0)
F~2 = (−15, 0, 0)
F~3 = (20 cos 208◦ , 20 sin 208◦ , 0)
(11)
and
~r = (3.1, 0, 0)
(12)
Let’s apply the cross product to these vectors to determine the torques that the forces
generate. If you’re unfamiliar with the cross product, be sure to check out the vector cross
product supplement at the same location where you found this document.
2.2.1
Torque Due to Force 1
τ~1 = ~r × F~1 = (3.1, 0, 0) × (0, −10, 0)
(13)
The only non-zero components are rx and Fy , therefore only the z-component of the torque
vector will be non-zero (see vector cross product supplement for details).
τ~1 = (0, 0, −31)
(14)
This agrees, of course, with our earlier method. The torque is negative, indicating a
clockwise rotation about the axis. The torque points in the - ẑ direction which would be
into the page.
2.2.2
Torque Due to Force 2
τ~2 = ~r × F~2 = (3.1, 0, 0) × (−15, 0, 0)
(15)
The only non-zero components are rx and Fx , therefore all of the torque components will
be zero (see vector cross product supplement for details).
τ~2 = (0, 0, 0)
(16)
This agrees, of course, with our earlier method.
2.2.3
Torque Due to Force 3
τ~3 = ~r × F~3 = (3.1, 0, 0) × (20 cos 208◦ , 20 sin 208◦ , 0)
(17)
The only non-zero components are rx , Fx and Fy , therefore only the z-component of the
torque vector will be non-zero (see vector cross product supplement for details).
τ~1 = (0, 0, −29)
(18)
This agrees, of course, with our earlier method. The torque is negative, indicating a
clockwise rotation about the axis. The torque points in the - ẑ direction which would be
into the page.
2.3
Using the Lever Arm
An alternative to using the cross product for determining the torque employs the concept
of the lever arm. The lever arm is defined as the perpendicular distance between the line
of force and the pivot point of rotation. The formula for the magnitude of the torque in
this instance is:
τ = F rla
(19)
where rla is the lever arm. This method requires only a few steps of effort. First, for a
particular force, extend the force vector in your diagram a bit in both directions (make
it longer). Next, draw a line that connects the pivot point with the line of force in such
a way that the connecting line you’ve drawn is perpendicular to the line of force (this
connecting line is your lever arm). The computation of the torque will simply be the
product of the force with the lever arm.
2.3.1
Torque Due to Force 1
Going back to our original diagram, it should be clear to you that a line that connects the
pivot point to F~1 in a perpendicular manner is simply the length of the rod, L. Thus,
τ1 = F1 rla1 = F1 L = 31 N · m
(20)
Note that this method only gets you the magnitude of the torque (the direction would be
found using the right-hand-rule).
2.3.2
Torque Due to Force 2
If you extend your line of force for F~2 in both directions, you’ll notice that the line intersects
the pivot point. If you were to draw a perpendicular connecting line (the lever arm) from
the pivot point to this line of force, you’ll see that it has a length of zero. The lever arm
is zero since the line of force intersects the pivot point. If the lever arm is zero, then the
torque is zero as well.
τ2 = F2 rla2 = 0 N · m
(21)
2.3.3
Torque Due to Force 3
The determination of the lever arm for τ3 is shown below in a figure. The lever arm is the
red line which is part of a right triangle with the rod length serving as the hypotenuse.
F3
L
θ
θ
The lever arm is clearly L sin θ and therefore:
τ3 = F3 rla3 = F3 L sin 28◦ = 29 N · m
The direction would be determined in the usual way using the right-hand-rule.
(22)