Transcript Slide 1

Consider the situation of changing a tyre. How
difficult is it to remove the lug nuts. What do you
do to make it easier?
Talk about Leverage.
is the ability of a Force to rotate an object
about an
Units =
where d is the length of the
The
is the
distance from the axis of
rotation to the applied Force.
So F & d are at
to each other.
Note: For Work, Force & distance are
to each other.
Now, let’s look at when the Force is applied at an angle.
Note that:
d is the perpendicular distance
from “a of r” to the Force.
d  L in this case, but
d=
So,
=
Alternatively, we could leave L
alone and resolve the Force
such that we find the
tangential component.
i.e. FT =
So, again
=
Examples where torque is used.
When multiple Forces act on an object, then
FR = F =
When those multiple Forces can each cause an object to
rotate, then
R =  =
When a Force points up or to the right, then F is
“ “ “
“ down “ “ “ left, “ F is
By convention,
 is +ve if the Force causes the object to rotate
 is –ve if the Force causes the object to rotate
Conditions of Equilibrium
Recall from chapter 4 (Forces) that if an object is at rest, or
is in constant motion (i.e a = 0), then F =
This is known as
“
” where F =
If an object is rotating, the above condition is met, but it is
not in complete equilibrium.
Therefore we need to apply a 2nd condition which applies
to Torque.
This is known as
“
” where  =
 = 0 means that the object is either not
it is
at a constant rate.
, or that
If an object has the ability to rotate, but is not moving at all,
i.e.
F = 0
and
 = 0
Then the object is in
Applications:
Seesaws:
Consider a seesaw that is not moving, has 2 people on it
and is supported off centre.
The seesaw is supported at the Fulcrum which is also the
axis of rotation.
w3 is the weight of the seesaw which is usually located in
the centre.
F = 0
 = 0
Example:
Little Johnny (40kg) and Silly Sally (30kg) are playing on a
3.0m long seesaw (60kg) that is supported in the middle.
If Silly Sally is sitting on the very edge, how far from the
centre should little Johnny sit in order for the seesaw to
balance horizontally?
Example:
A 50kg sign that is 2.0m long, is supported by a pivot
against a wall and a steel wire from the edge of the
sign to the wall at an angle of 30o above the sign.
Calculate the Tension in the wire and the Force the wall is
resisting the sign with. (the Reactionary Force
components)
Let’s “Torque” about ladders!