Moment of a Force about a Point Video Lecture | Engineering Mechanics - Civil Engineering (CE)

hey guys welcome back just one more
video on moments and how to find a
moment of a force about a point on an
object so in this case we have this sort
of bracket or shelf or something that's
attached to a wall we're applying a
force over here and we want to figure
out what is the moment that that force
is creating about point a so there's two
ways that we can do that we can we can
use the perpendicular distance from the
point to the line of action of the force
and then multiply it by the magnitude of
the force or we can use the vector cross
product so we'll do both examples in
this video but I guess we'll start with
a perpendicular distance method first so
first of all we have the moment about
point a what we have to do is have to
define a positive sense for the moment
so basically if we have a moment that's
attending to want to create a
counterclockwise rotation about this
point a then that would be a positive
moment otherwise it'd be a negative
moment and so then all we have to do is
we have our perpendicular distance times
the magnitude of the force all right now
if we wanted to find the perpendicular
distance what we need to do is first we
have the line of action of the force
which would fall on this sort of incline
here and then the perpendicular distance
from the point to that line of action
would be this red line so all we would
need to know is the length of this red
line and we multiply it by 500 Newtons
and then we would get our answer which
is the moment in newton meters but it's
actually kind of hard to find the length
of this line and so I think the easier
way to do it is to not worry about that
will break the specter or will break
this force into its vector components
and then we'll solve we'll add them up
to get the to get the equivalent answer
all right
so to do that basically what we're doing
is instead of just finding the
perpendicular distance to the line of
action of the force we're finding the
perpendicular distance from the point to
the line of action of the forces of the
x and y components and then multiplying
those those magnitudes throughout so we
would have the perpendicular distance
from the point to the line of action of
the X component of the force times the
magnitude of the X component of the
force and then we would do the same
thing with the Y component of the force
so what we should do is you should draw
on our our components here so you'd have
F X and then we would have down here F Y
would look something like that all right
so let's just
or draw our coordinate axis over here
just so we don't get lost and then we
know that FX is just going to be equal
to 500 times cos 30 and that actually
works out to 433 Newtons and then FY
down here is 500 sine 30 and that
actually works out to 250 Newton's 250
Newtons those are the magnitudes the
directions would be like that and like
that
okay so then we can plug that
information basically into this formula
and we'll get the perpendicular distance
that the line of action of the X
component of the force well the X
component of the force line of action is
right here this horizontal line so the
perpendicular distance from that point a
to the line of action of the force is 2
meters so we get 2 meters times the
magnitude of the X component of the
force which is 433 Newtons
now what we have to do is we have to
consider if this will be causing a
positive or a negative moment so if we
imagine that this sort of bracket or
l-shaped object here can rotate could
rotate about point a well if we just
plot if we just apply this FX it would
want to kind of give it the tendency to
rotate this way about point a if point a
could rotate so we say that that has a
negative moment because that's opposite
the positive sense that we've defined a
beer so what we have to do is we have to
give that a negative sign now for FY
same thing this X this Y component of
the force if the object could rotate
about point a if we just apply it FY it
would kind of make the object want to
rotate this way spin this way around
point a so again we're seeing a negative
moment because of the Y component of the
force okay so knowing that the distance
the perpendicular distance from point a
to the line of action of the force which
is this vertical line here is 6 meters
six meters and the magnitude of the Y
component of the force is 250 Newtons
all right now if we just add these all
together or simplify this we actually
get negative 800
66 and this becomes minus 1500 and that
all works out to be we can write it down
here that was actually Newton meters and
so we end up with a final answer here of
negative two thousand three hundred and
sixty-six Newton meters and it's
negative again because it's creating
this if it was able to rotate about
point a it would be creating this sort
of clockwise rotation about a so the
other thing we could say if we want to
do we could drop the negative sign and
we could just say that it's two thousand
three hundred and sixty-six Newton
meters in the clockwise direction at a
clockwise sense all right so that's
that's how we use the perpendicular
distance and the magnitude of the force
now let's use the vector cross product
so the other thing we can do is we can
write it like this at the moment about
point a we will define the same sense to
be positive and then we just say we have
it's R cross F now R is our position
vector from the point to anywhere on the
line of action of the force so it could
be here it could be here it could be
here it doesn't matter as long as it's
on the line of action of the force but
what we want to do is we'll pick this
point because we actually know the
coordinates at that point so that's
going to help us out a little bit make
our lives a little bit easier
so the Rigo we have our okay so when we
want to do this R cross F I guess we'll
have the X component of R is six meters
the Y component of R is two meters in
the Z component of R is zero because
this vector is in the XY plane there's
no Z component and then for our for our
vector form of our force the X component
of that is 433 the Y component is
negative 250 we didn't include the
negative sign before because we're
strictly talking magnitudes but here
it's really important for those sort of
orientation of it and then Z again this
is an XY plane so we have zero component
in the Z direction alright so now we can
go ahead and cross these like we've been
doing in previous videos so for the
first component the or for the yeah for
the X component
would be two times zero minus zero times
negative 250 so two times zero minus
zero times negative 250 just like that
if we can make a zero for the last
component it would be the same method
but just switched so six times negative
250 six times negative 250 minus 2 times
433 two times four hundred and thirty
three and the middle component we do it
the reverse direction so we start at the
bottom here go zero times four hundred
and thirty-three minus six times zero
minus six times zero all right so when
we go and simplify this we have two
times zero minus zero times 250 so that
becomes 0 we have 0 times something
minus something times zero so the Y
component becomes zero and down here we
have six times negative 250 well that's
actually negative fifteen hundred minus
two times 433 so that's a 66 and we can
see that this would actually reduce to
zero zero negative two thousand three
hundred and sixty-six alright so now
this is actually looking suspiciously
similar to what we have up here what we
can do is we can just go one step
further or a few steps further I guess
and we'll just pull out this because of
scalar multiplication we can do that so
we'll have negative two thousand three
hundred and sixty six times zero zero 1
and this is 0 0 1 vector is actually our
unit vector we call k hat because as a
length of 1 and a direction of and it's
in the positive K direction which is the
positive z direction so we can also
write this as negative 2366 k hat and if
you recall this whole time our units of
our was meters and our units of F was
Newton's so this is actually Newton
meters just like that now this is
actually the answer this is the complete
answer and this is equivalent to saying
all of any of these so negative two
thousand three hundred sixty-six K hat
is equal to
basically or is equivalent to this
clockwise rotation in the XY plane for
describing our moment and the reason why
that is is because if you remember from
the the cross-product video we did we
have our first vector so this will be R
and then we're crossing it with F which
looked something like that what you do
is you take your right hand and you
point it in the line you point it to the
tip of the first arrow so you pointed to
the direction that R is pointing in and
then you sweep with your palm towards
the next vector so when you sweep down
your fingers will start pointing towards
F as you sweep and when you do that
you'll notice your thumb is actually
pointing into the screen and that is the
direction sort of that we say the moment
has or that's the direction that the
cross-product is basically going in and
so mmm that's just the equivalent thing
of you know we can we can talk about we
can talk about a moment having a sort of
a sense like a rotational sense but also
because of the definition of the cross
product when you do rotate with your
hand like that the way that you're the
way that your thumb points in is also
another way to describe that rotation so
sometimes you'll be you know especially
when you're working with moments in 3d
instead of 2d this starts to become
really important about sort of which
where the components are but basically
what this is just saying is that you
know if we had if this was actually a
positive value well then it would be
going in the positive z direction and it
would be coming out of the page but
that's not how we draw our axes well
that is how we draw axes but if we have
a negative value it just basically flips
the direction so that's what you're
seeing here all right well I hope that
video is helpful and I will see you in
the next one