Question Video: Finding the Maximum Distance a Man Can Climb on a Ladder Resting between a Rough Vertical Wall and a Rough Horizontal Floor Mathematics • Third Year of Secondary School

Video Transcript

A uniform ladder with length 𝐿 and weight 35 kilogram-weight is resting with its upper end against a rough vertical wall and lower end on a rough horizontal floor. The floor and wall both have the same coefficient of friction of one-half with the ladder. If the ladder is inclined to the horizontal at an angle whose tangent is three-quarters, find the maximum distance in terms of 𝐿 that a man weighing 78 kilogram-weight can ascend the ladder without it slipping.

Before we do anything, let’s begin by drawing a diagram. Here’s a ladder inclined at an angle whose tangent is three-quarters to the horizontal. If we call that angle 𝛼, then tan 𝛼 is equal to three-quarters. Now, what we’re not going to do is solve this equation for 𝛼 by finding the inverse tan of it. Instead, in a little while, we’ll construct a right-angled triangle that will help us find exact values for sin 𝛼 and cos 𝛼. For now, though, let’s continue with our diagram.

The ladder is uniform with length 𝐿 and a weight 35 kilogram-weight. The fact that it’s uniform means that its mass is evenly distributed throughout the ladder. And we can say that the force of weight must act exactly halfway along the ladder. And so we have a force of 35 kilogram-weight acting downward half 𝐿 along the ladder. If we say the base of the ladder is 𝐴 and the top of the ladder is 𝐵, we know we have reaction forces at these points.

These reaction forces are perpendicular to the surface that the ladder is resting on, as shown. We also know though that both the wall and the floor are rough. And that means we need to consider a frictional force. The frictional force will act against the direction in which the ladder is trying to move. So if someone stands on the ladder, it wants to slide downwards. And so, at point 𝐴, the frictional force will act towards the wall. At point 𝐵, the frictional force will act upwards away from the floor.

We know that friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction. And we’re told that the coefficient of friction of both the floor and the wall with the ladder is one-half. So the frictional force at point 𝐴 must be one-half times the reaction force there. So that’s one-half times 𝑅 sub 𝐴. Similarly, the frictional force at 𝐵 must be a half 𝑅 sub 𝐵. We want to find how far up the ladder the man can get before it slips. So we don’t know exactly how far up the ladder he can get, so we’ll call that 𝑥𝐿, where 𝑥 will be a fraction of the total distance. His downward force that he exerts on the ladder is 78, as shown.

So we’re going to do two things. Firstly, we’re going to begin by resolving forces horizontally and vertically. After we’ve done that, we’ll be able to calculate moments about a point on the ladder. Let’s begin by resolving forces in a vertical direction. We know that the sum of these forces is equal to zero. The ladder isn’t slipping, but it will be on the point of slipping. So it’s in equilibrium. If we take upwards to be positive, we see we have 𝑅 sub 𝐴 acting upwards. And then we have the frictional force at 𝐵 half 𝑅 sub 𝐵 also acting in that direction.

In the opposite direction, we have the weight of the ladder and the weight of the man. So we’re going to subtract both of those forces. And we know the sum of all of these forces is equal to zero. So 𝑅 sub 𝐴 plus a half 𝑅 sub 𝐵 minus 35 minus 78 is equal to zero. Negative 35 minus 78 is negative 113, so we can simplify this equation a little. Next, we’re going to see if we can create another equation linking 𝑅 sub 𝐴 and 𝑅 sub 𝐵 by resolving horizontally.

Let’s call the forces in this direction 𝐹 sub 𝑥. Then, the sum of these forces is once again equal to zero. If we take the direction right to be positive, we know we have a half 𝑅 sub 𝐴 acting in this direction and 𝑅 sub 𝐵 acting in the opposite direction. The sum of these forces is zero, so a half 𝑅 sub 𝐴 minus 𝑅 sub 𝐵 is equal to zero. And we can rearrange to say that a half 𝑅 sub 𝐴 is equal to 𝑅 sub 𝐵.

We’re going to multiply through by two to find an expression for 𝑅 sub 𝐴 in terms of 𝑅 sub 𝐵. We get 𝑅 sub 𝐴 is two times 𝑅 sub 𝐵. And this is really useful because we can replace 𝑅 sub 𝐴 with two 𝑅 sub 𝐵 in our earlier equation. Simplifying this becomes five over two 𝑅 sub 𝐵 minus 113 equals zero. And by adding 113 to both sides, we find five over two 𝑅 sub 𝐵 equals 113. We’ll solve this equation for 𝑅 𝐵 by dividing through by five over two or 2.5. And we find 𝑅 𝐵 to be equal to 45.2 or 45.2 kilogram-weight.

Let’s replace all instances of 𝑅 sub 𝐵 in our diagram with this value. The normal reaction force of the wall on the ladder then is 45.2, and then a half of this is 22.6. So the frictional force acting at the wall is 22.6 kilogram-weight. In fact, if we go back to this equation, we can also find the value of 𝑅 𝐴. It’s two times 𝑅 sub 𝐵, so two times 45.2, which is 90.4 kilogram-weight. Once again, we’ll replace 𝑅 sub 𝐴 in our diagram. The normal reaction force of the floor on the ladder then is 90.4 and the frictional force is half of this; it’s 45.2.

Now we’ve replaced all of our forces with numbers, we’re going to clear some space and take moments. Now we can take moments about any point on the ladder if we want to. However, we’ve measured our distances from point 𝐴. And also generally, there’s more going on at the foot of the ladder, so that’s where we tend to take moments about. Let’s say counterclockwise is positive. And then we use the formula moment is equal to force times distance, where the distance is the perpendicular distance from the pivot to the line of action of the force. So let’s begin by considering the moment of the weight of the ladder. Remember, we’re interested in the component of this force, which is perpendicular to the ladder. So we add a right-angle triangle as shown.

The included angle is 𝛼. And we want to work out this side. Let’s call that 𝑥. Its units will be kilogram-weight. This is the adjacent side in our triangle, and we know the hypotenuse is 35 kilogram-weight. Since cos 𝜃 is adjacent over hypotenuse, here, we can say that cos of 𝛼 must be 𝑥 over 35 or 𝑥 is equal to 35 times cos 𝛼. Now we can use the fact that tan of 𝛼 is equal to three-quarters to find an exact value for cos of 𝛼. By drawing a right-angled triangle and recognizing that tan of 𝜃 is opposite over adjacent, we can label the opposite side in our triangle as three units and the adjacent as four.

The Pythagorean theorem or the fact that we recognize this as a Pythagorean triple tells us that the hypotenuse is five units. We can therefore say that cos of 𝛼, which is adjacent over hypotenuse, must be four-fifths. Similarly, we can form an expression for sin of 𝛼. It’s opposite over hypotenuse, which is three-fifths. This means that 𝑥 can be rewritten as 35 times four-fifths, which is 28. The component of the weight of the ladder that is perpendicular to the ladder is therefore 28 kilogram-weight.

We know that the moment is force times distance. But this force is trying to push the ladder in a clockwise direction, so its moment is going to be negative. It’s negative 28 multiplied by a half 𝐿. Let’s repeat this process with the component of the weight of the man that’s perpendicular to the ladder. This time, we’ll call that 𝑦. Once again, we’re trying to find the adjacent, and we know the hypotenuse. But the hypotenuse here is 78, so we get cos 𝛼 is equal to 𝑦 over 78 or 𝑦 is 78 times cos 𝛼. cos 𝛼, of course, is four-fifths, so we get 78 times four-fifths. And the component of the weight of the man that’s perpendicular to the ladder is therefore 62.4 kilogram-weight. This moment is once again negative, so it’s negative 62.4 times 𝑥𝐿.

There are two more moments we need to consider, and those are the moments of the forces at 𝐵. Let’s consider the component of the reaction force at 𝐵 that’s perpendicular to the ladder. I’ve called that 𝑎 and drawn another right-angled triangle. The included angle is still 𝛼. We want to find the opposite. And this time, the hypotenuse is 45.2. So we can say sin 𝛼 is 𝑎 over 45.2 or 𝑎 is equal to 45.2 times sin 𝛼. Now, remember, we calculated that sin 𝛼 was equal to three-fifths. And so the component of this reaction force that acts perpendicular to the ladder is 27.12 kilogram-weight. Now this force is trying to turn the ladder in a counterclockwise direction. So its moment is positive. It’s 𝐿 times 27.12.

Now that we’ve considered that force, we clear some space to look at the component of the friction that’s perpendicular to the ladder. We once again have an included angle of 𝛼 and we’re trying to find the side labeled 𝑏. This is the adjacent side in our triangle, and we know the hypotenuse. So the cosine ratio says cos 𝛼 is 𝑏 over 22.6, and so 𝑏 is 22.6 times cos 𝛼 or 22.6 times four-fifths. That’s 18.08 kilogram-weight. And so the moment of this force is 𝐿 times 18.08. Now, the ladder is in equilibrium, so the sum of these moments must be equal to zero. And then we notice before we solve this equation for 𝑥, we can actually divide through by 𝐿.

Now the reason we’re allowed to divide through by an unknown is because we know the length of the ladder cannot be equal to zero. Negative 28 times one-half is negative 14. So our equation becomes negative 14 minus 62.4𝑥 plus 27.12 plus 18.08 equals zero. This simplifies to negative 62.4𝑥 plus 31.2 equals zero. And if we add 62.4𝑥 to both sides, we get 62.4𝑥 equals 31.2. We then divide through by 62.4. Now, 31.2 divided by 62.4 is one-half. So we find that 𝑥 is equal to one-half.

This means the man can get a halfway up the ladder before it slips, so we can get a half 𝐿 distance along the ladder without it slipping. The answer is a half 𝐿.