Video Transcript
A uniform ladder is resting in a
vertical plane with its upper end against a smooth vertical wall and its lower one
on a rough horizontal floor, where the coefficient of friction between the ladder
and the floor is two-thirds. The ladder is inclined to the
horizontal at an angle measuring 48 degrees. Given that the ladder weighs 295
newtons and has a length 𝐿, find, in terms of 𝐿, the maximum distance a man
weighing 610 newtons can climb up the ladder without it slipping, rounding your
answer to two decimal places.
To answer this question, we’re
simply going to begin by drawing a diagram. The angle that the ladder makes
with the horizontal is 48 degrees. We’re told that the ladder is
uniform and that it weighs 295 newtons. The fact that it is uniform means
that its weight is evenly distributed along its length. And so when we model this, we can
model its weight as acting exactly halfway along the ladder. Well, since the ladder is 𝐿 long
and we’re not told the units, we know that this acts at a distance of a half 𝐿 from
the base of the ladder. We’re told that the floor is rough,
and so this means there’s a frictional force of the floor on the ladder. This acts against the direction in
which the ladder is looking to slide. And so, in our diagram, it acts to
the right.
There are two further forces we’re
interested in. Let’s label the base of our ladder
𝐴 and the top of the ladder, the point where it meets the wall, as 𝐵. There are two reaction forces at
these points. 𝑅 sub 𝐴 is the reaction force of
the floor on the ladder, and it acts perpendicular to the floor. And 𝑅 sub 𝐵 is the reaction force
of the wall on the ladder, and it acts perpendicular to the wall. The question is asking us how far
up the ladder a man weighing 610 newtons can climb before the ladder slips. Now what makes this a little bit
tricky is we don’t know whether he can get halfway or further than halfway. On our diagram, we’ll model him as
getting further than halfway. But in fact, the maths will answer
this for us.
Let’s say he makes it 𝑥 of the way
up the ladder, where 𝑥 is some number between zero and one. We can say that the distance he
will travel from 𝐴 is 𝑥𝐿 units. Now that we’ve got our diagram,
we’re ready for our next steps. This coefficient of friction value
we call 𝜇, and this relates to the frictional force of the floor on the ladder. To answer this question, let’s
begin by resolving our forces in a vertical direction. We’ll then resolve horizontally
before finally considering moments acting on the ladder.
We’re going to say that the ladder
is on the point of slipping, and so it’s in limiting equilibrium. For this to be the case, the sum of
all its vertical forces, let’s call that 𝐹 sub 𝑦, must be equal to zero. If we take upwards to be the
positive direction, we know we have 𝑅 sub 𝐴 acting upwards, and acting in the
opposite direction, we have the weight of the ladder and the weight of the man. So the sum of the forces in a
vertical direction is 𝑅 sub 𝐴 minus 295 minus 610. We know that the ladder is in
equilibrium, so this is equal to zero. Negative 295 minus 610 is negative
905. And so adding 905 to both sides of
this equation tells us that 𝑅 sub 𝐴 is 905 newtons.
Next, we resolve in a horizontal
direction. Once again, we know that the sum of
these forces will be equal to zero. Let’s take the direction to the
right to be positive. Then we see our frictional force is
positive. The reaction force at 𝐵 acts in
the opposite direction. So the sum of our forces in a
horizontal direction is friction minus 𝑅 sub 𝐵. And of course, this is equal to
zero. But in fact, we know that the
friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the
reaction force at that point. The coefficient of friction in this
question is two-thirds, and the reaction force 𝐴, which is where the friction is
acting, is 905. So we get two-thirds times 905
minus 𝑅 𝐵 equals zero. Two-thirds times 905 is 1810 over
three. So we add 𝑅 sub 𝐵 to both sides
of our equation, and we see that 𝑅 sub 𝐵 is 1810 over three newtons.
Now that we’ve resolved
horizontally and vertically and worked out the value of our reaction forces, we’re
actually able to take moments. Now, we can take moments at any
point on the ladder. In reality, though, it’s often
sensible to take moments about the point where the ladder meets the floor. Usually, there are several forces
acting at this point, so it minimizes the calculations we need to do. We’re going to take moments then
about 𝐴, and we’re going to define a positive direction. Let’s say counterclockwise is
positive. We recall that a moment is equal to
force times distance, where the distance 𝑑 is the perpendicular distance from the
pivot to the line of action of the force.
Once again, since the ladder is in
limiting equilibrium, we know that the sum of our moments is going to be equal to
zero. So, let’s begin by looking at the
weight of the ladder and the moment of this force. We want the component of this force
that’s perpendicular to the ladder. And so we add a right-angled
triangle with an included angle of 48 degrees. We know the hypotenuse of this
force triangle is 295 newtons. And we want to find the adjacent
side. If we define this component of the
force to be equal to 𝑎 or 𝑎 newtons, we see we can use the cosine ratio. cos of 48 is 𝑎 over 295. We then multiply both sides by 295,
and we found the value of 𝑎; it’s 295 times cos of 48 degrees. And of course, this is in
newtons.
This force is trying to move the
ladder in a clockwise direction. So its moment is going to be
negative. The moment is force times
distance. So, this moment is negative 295 cos
of 48 times a half 𝐿. And we’ve chosen a half 𝐿 because
we said that it’s the distance along the ladder where this force acts. We’ll now consider the weight of
the man. Once again, we add a right-angle
triangle. Now this right-angle triangle is
similar to our previous one. The only difference is the
hypotenuse; this time it’s 610. And we denote the side we’re
looking for as 𝑏. Remember, we want the component
that is perpendicular to the ladder. So we get cos 48 equals 𝑏 over
610, meaning 𝑏 is 610 times cos of 48.
Once again, the moment is negative,
so we’re going to subtract 610 cos 48 multiplied by the distance, which we said is
𝑥𝐿. There’s one more force we’re
looking at. This time, we’re interested in the
component of the reaction force of 𝐵, this perpendicular to the ladder. So we add another right-angled
triangle in. Once again, we know the
hypotenuse. It’s 𝑅 sub 𝐵, which we’ve already
calculated. But this time we want the opposite
side. Let’s call that 𝑐. And so to link the opposite and
hypotenuse, we use the sine ratio, sin 48 equals 𝑐 over 𝑅 sub 𝐵. So, we multiply by 𝑅 sub 𝐵, and
we find 𝑐 is 𝑅 sub 𝐵 sin 48.
Now, this force is trying to move
the ladder in a counterclockwise direction. So its moment is positive. We multiply that force by the
distance away from 𝐴. So that’s 𝑅 sub 𝐵 sin 48 times
𝐿. We know that the sum of all of
these is equal to zero. Now, notice that had we taken a
moment about a different point, say, 𝐵 for example, we would’ve had two forces to
consider at 𝐴, whereas at 𝐵 we had just one force to consider. So, there’s a little bit less in
the way of calculations. Now, before we do anything, we see
that we can divide through by 𝐿. Remember, we can only do this if 𝐿
is not equal to zero. Well, 𝐿 is the length of the
ladder, so we know that that’s true.
Next, we simplify. And at the same time, we’re going
to replace 𝑅 sub 𝐵 with 1810 over three. And so our equation is negative 295
over two times cos 48 minus 610𝑥 cos 48 plus 1810 over three sin 48 equals
zero. To solve for 𝑥, we’ll add 610
times 𝑥 cos 48 to both sides. And then we’ll evaluate what’s left
on the left. Typing this into our calculator
gives us 349.667 and so on. To solve for 𝑥, we’ll divide
through by 610 cos 48. 349.667 divided by 610 cos 48 gives
us a value of 0.85667 and so on. Correct to two decimal places, it’s
0.86. And so we see that the man can get
0.86 of the way up the ladder before it slips. In terms of 𝐿, the distance he
travels then is 0.86𝐿.
