Question Video: Finding the Maximum Distance a Man Can Climb on a Ladder Resting between a Smooth Vertical Wall and a Rough Horizontal Floor Mathematics

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 2/3. The ladder is inclined to the horizontal at an angle measuring 48°. Given that the ladder weighs 295 N and has a length 𝐿, find, in terms of 𝐿, the maximum distance a man weighing 610 N can climb up the ladder without it slipping, rounding your answer to two decimal places.

08:54

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.