Video Transcript
A uniform rod rests with one end
against a rough vertical wall, and the coefficient of friction between the rod and
wall is three-tenths. The other end of the rod sits on a
rough horizontal floor, and the coefficient of friction between the rod and the
floor is one-third. Calculate the angle of inclination
𝜃 between the rod and the floor when it is in limiting equilibrium, giving our
answer to the nearest minute.
Before we perform any calculations
here, we’re going to simply begin by drawing a free-body diagram. Remember, this is a really simple
sketch that just shows all of the key forces we’re interested in. Here is a rigid rod inclined at an
angle of 𝜃 degrees to the horizontal. We’re going to call end 𝐵 the end
that sits against the wall and end 𝐴 the end that sits on the floor. We’re also told the rod is
uniform. This means the downwards force of
its weight will act at the point exactly halfway along the ladder.
For ease, let’s define then the
length of the ladder to be two 𝑥 meters. And so, the weight force acts at a
point 𝑥 meters from point 𝐴. Then, we consider Newton’s third
law of motion. This tells us that if the rod
exerts a force on the wall and the floor, the wall and the floor must themselves
exert a normal reaction force on the rod. We’re going to call the normal
reaction force of the floor on the rod 𝑅 sub 𝐴 and the normal reaction force of
the wall on the rod 𝑅 sub 𝐵.
Then, there are two further forces
that we’re interested in. We’re told that the wall and the
floor are rough. This means there must be frictional
forces at play. We’re also told that the rod is in
limiting equilibrium. That means it’s on the point of
sliding. Remember, the frictional force will
act against the direction in which the body is trying to travel. So, the frictional force at 𝐴 will
act towards the wall as shown. And the frictional force at 𝐵,
let’s call that Fr sub 𝐵, must be acting upwards.
And at this point, we can recall a
formula to help us calculate the friction. Friction is equal to 𝜇𝑅, where 𝜇
is the coefficient of friction and 𝑅 is the normal reaction force at that
point. We’re told the coefficient of
friction between the rod and the floor, so at point 𝐴, is one-third. So, 𝜇𝑅 here would be a third 𝑅
sub 𝐴. Similarly, the coefficient of
friction between the rod and the wall is three-tenths. So, 𝜇𝑅 at point 𝐵 must be
three-tenths times 𝑅 sub 𝐵.
We’ve now extracted all the
relevant information we need from this question. So, let’s clear some space. So, once we have all the forces
labeled, what do we do next? Well, we recall that the body was
in limiting equilibrium. We discussed what the limiting part
meant. That meant it’s on the point of
sliding. But for a body to be in
equilibrium, there are two conditions we need to consider. Firstly, the sum of all the forces
acting on the body must be equal to zero. We tend to break these down
directionally and say that the sum of the horizontal forces is zero and the sum of
the vertical forces is zero.
We also know that the sum of all
the moments acting on the body must also be equal to zero, where moment is 𝐹𝑑. 𝐹 is the force, and 𝑑 is the
perpendicular distance of the line of action of this force from the point about
which the body is trying to pivot. And so, what we’re going to do here
is begin by resolving the forces on our diagram. Then, we’ll consider moments.
We’re going to resolve both
horizontally and vertically. Let’s begin by resolving
vertically, and we’re going to take upwards to be positive. Acting upwards, we have the
reaction force at 𝐴 and the frictional force at 𝐵. And then acting downwards, we have
the weight of the body. The sum of these forces is, of
course, equal to zero. Now, we’re actually going to
replace the frictional force at 𝐵 with the expression we developed earlier,
three-tenths 𝑅 sub 𝐵.
So, we get 𝑅 sub 𝐴 plus
three-tenths 𝑅 sub 𝐵 minus 𝑤 equals zero. Next, we add 𝑤 to both sides. Eventually, we want to be getting
rid of all 𝑤’s in our equation. So, we get 𝑅 sub 𝐴 plus
three-tenths 𝑅 sub 𝐵 equals 𝑤.
Next, we resolve horizontally. And we’re going to take the
direction in which the frictional force at 𝐴 is acting to be positive. In this direction, we have
frictional force of 𝐴. And in the other direction, we have
the reaction force at 𝐵. So, their sum is Fr sub 𝐴 minus 𝑅
sub 𝐵 and that, of course, is equal to zero. Now, we’re going to replace the
frictional force at 𝐴 with the expression we developed earlier, a third 𝑅 sub
𝐴.
So we get one-third 𝑅 sub 𝐴 minus
𝑅 sub 𝐵 equals zero. We’re going to make 𝑅 sub 𝐴 the
subject here. And eventually, what we’re going to
do is try and find an expression purely in terms of 𝑅 sub 𝐵. If we add 𝑅 sub 𝐵 to both sides
of our equation and then multiply by three, we find that 𝑅 sub 𝐴 is three times 𝑅
sub 𝐵. Now, this is really useful because
we can now replace this 𝑅 sub 𝐴 in our earlier equation. And so, notice we have an equation
just in terms of two variables now. We have 𝑤 equals three times 𝑅
sub 𝐵 plus three-tenths 𝑅 sub 𝐵 or 𝑤 equals 33 over 10 𝑅 sub 𝐵.
We’re now going to calculate
moments. We begin by finding the point about
which we want to calculate those moments. It doesn’t really matter which
point we choose. We should get the same answer
either way, but we tend to choose the foot of the ladder, since there’s usually more
forces acting here. We’re also going to define the
counterclockwise direction to be positive. And so, we’ll begin with the weight
force.
Now, remember, when we calculate a
moment, we want to find the component of each force that acts perpendicular to the
body. We add a right-angled triangle in,
and we need to find the value of 𝑥. Now, 𝑥 is the component of the
weight force we’re interested in. It’s not the same as the length of
the ladder. The included angle here is 𝜃, 𝑤
is the hypotenuse, and 𝑥 is the adjacent. So, we can use the cosine ratio to
find some sort of expression for 𝑥.
When we do, we get cos of 𝜃 is 𝑥
over 𝑤. And then when we multiply by 𝑤, we
get 𝑥 equals 𝑤 cos 𝜃. Now, of course, we said 𝑤 is
thirty-three tenths 𝑅 sub 𝐵 And so, we have an expression for 𝑥 in terms of 𝜃
and 𝑅 sub 𝐵. The moment is going to be negative
since this is trying to turn the rod in a clockwise direction. Force times distance is
thirty-three tenths 𝑅 sub 𝐵 cos 𝜃 times 𝑥.
There are two further moments we
need to consider; these are the moments to do with 𝑅 sub 𝐵 and the frictional
force at 𝐵. Once again, we’re going to resolve
𝑅 sub 𝐵 to find the component of this force that acts perpendicular to the
rod. This time, 𝑅 sub 𝐵 is the
hypotenuse, but we’re trying to find the value of 𝑦, which is the opposite. So, we’re going to use the sine
ratio. When we do, we find that sin 𝜃 is
𝑦 over 𝑅 sub 𝐵. And we can multiply both sides of
this equation by 𝑅 sub 𝐵. So, 𝑦 is 𝑅 sub 𝐵 sin 𝜃. This time, this moment is going to
be positive since it’s acting in the counterclockwise direction. And force times distance is 𝑅 sub
𝐵 sin 𝜃 times two 𝑥.
There’s one more force we’re
interested in, and that’s the frictional forces at 𝐵. Once again, we can add a
right-angled triangle here. And we’re trying to find the value
of 𝑧; this is the component of the frictional force that’s perpendicular to the
rod. It’s a little bit trickier to find
the angle we’re interested in here. In fact, though, this angle, the
angle at 𝐵, is 90 minus 𝜃. We can convince ourselves this is
true because the angle between 𝑅 sub 𝐴 and the frictional force at 𝐴 must be
90. Then, we have a pair of
corresponding angles. And so, this angle up here must
itself be 𝜃, since angles in a triangle add up to 180.
We know the hypotenuse and the
adjacent in this triangle then. So, we go back to the formula cos
of 𝜃 is adjacent over hypotenuse. So, cos of 𝜃 is 𝑧 over the
frictional force at 𝐵. We multiply both sides of this
equation by the frictional force at 𝐵. But remember, we have an expression
for that; it’s three-tenths 𝑅 sub 𝐵. And we found the component of the
frictional force at 𝐵 that’s perpendicular to the rod; it’s three-tenths 𝑅 sub 𝐵
times cos 𝜃. And so, we can find the moment of
this force. Once again, it’s positive. It’s three tenths 𝑅 sub 𝐵 cos 𝜃
times two 𝑥.
The body is in equilibrium. So, we set this equal to zero. Now, we want to solve for 𝜃, but
we actually have three unknowns here. We have 𝑅 sub 𝐵, 𝜃, and 𝑥. However, we defined two 𝑥 to be
the length of the ladder, so 𝑥 was half that length. And there’s no way 𝑥 can be equal
to zero. We’re therefore able to divide
through by 𝑥. Similarly, we know the reaction
force at 𝐵 cannot be equal to zero either. So, we divide through by 𝑅 sub
𝐵. And so, we’re left with negative 33
over 10 cos 𝜃 plus two sin 𝜃 plus three-fifths cos 𝜃 equals zero.
Notice now we’re working with just
one variable. We’re going to gather together the
cos 𝜃s first. So, two sin 𝜃 minus 27 over 10 cos
𝜃 equals zero. And then, we’re going to recall one
of our key trigonometric identities; that is, tan 𝜃 is equal to sin 𝜃 over cos
𝜃. We want our equation to look a bit
like this. So, we’re going to add 27 over 10
cos 𝜃 to both sides. Next, we’re going to divide through
by cos 𝜃. So, two sin 𝜃 over cos 𝜃 equals
twenty-seven tenths. But of course, we can replace sin
𝜃 over cos 𝜃 with tan 𝜃, and now we’re going to divide through by two.
We’re nearly done. We know that tan 𝜃 is equal to
twenty-seven twentieths. So, 𝜃 must be equal to the inverse
or arc tan of this value. That gives us 53.471 degrees. Now, of course, we were told to
give our answer to the nearest minute. And so, we use the relevant button
on our calculator. And we find that 𝜃, which is the
angle of inclination of the rod to the floor, is 53 degrees and 28 minutes.
