# Question Video: A Ladder Resting in Equilibrium Between a Smooth Floor and a Smooth Vertical Wall with a String Attached to Its Lower End

A ladder 𝐴𝐵 weighing 40√3 kg-wt and having a length of 5 m is resting in a vertical plane with end 𝐵 on a smooth floor and end 𝐴 against a smooth vertical wall. End 𝐵 is attached by a string to a point on the floor vertically below 𝐴. Given that 𝐵 is 2.5 m away from the wall, and the weight of the ladder is acting at a point on the ladder 2 m away from 𝐵, find the tension in the string.

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### Video Transcript

A ladder 𝐴𝐵 weighing 40 root three kilogram-weight and having a length of five meters is resting in a vertical plane with end 𝐵 on a smooth floor and end 𝐴 against a smooth vertical wall. End 𝐵 is attached by a string to a point on the floor vertically below 𝐴. Given that 𝐵 is 2.5 meters away from the wall and the weight of the ladder is acting at a point on the ladder two meters away from 𝐵, find the tension in the string.

Before we perform any calculations, we’re going to begin by drawing a free body diagram. This is a very simple sketch of the scenario that shows all of the forces we’re interested in. Here is a ladder with a length of five meters resting against the smooth vertical wall. We don’t know the angle that the ladder makes with the horizontal, but we are going to need to calculate it at some point. So let’s call that 𝜃. We’re told that end 𝐵 is attached by a string to a point on the floor vertically below 𝐴. This means we need to add a tensional force 𝑇 to the diagram as shown. Then we’re told that 𝐵 is 2.5 meters away from the wall. Now this will be useful in a moment cause we’re going to be able to use right angle trigonometry to calculate the value of 𝜃.

We’re then told that the weight of the ladder acts at a point on the ladder two meters away from 𝐵. So that’s a downwards force of 40 root three kilogram-weight. There are two further forces that we’re interested in. We know by Newton’s third law of motion that since the ladder will be exerting a force on the floor and the wall, there will be normal reaction forces of the floor and the wall on the ladder, respectively. Let’s call them 𝑅 sub 𝐵 and 𝑅 sub 𝐴. We now have all the key information and we’re trying to find the tension in the string. We’ve called that 𝑇. So let’s clear some space.

So what do we do next? Well, we know that the ladder is resting in its position. And so we can assume that it’s in equilibrium. For a rigid body such as this ladder to be in equilibrium, there are two conditions that need to be met. Firstly, the sum of all the forces acting on the body must be equal to zero. We often break these down into horizontal and vertical forces such that the sum of the horizontal forces is zero and the sum of the vertical forces is zero. We’re also able to say that the sum of all the moments acting on the body must be equal to zero, where moment is calculated by finding the product of the force 𝐹 and the distance 𝑑, where 𝑑 is the perpendicular distance of the line of action of the force from the point about which the object is trying to rotate.

As with forces, we also give a direction to our moments. This time those are clockwise and counterclockwise. So let’s begin by finding the sum of the forces on our diagram. We’re trying to calculate the tension. And we really aren’t interested in 𝑅 sub 𝐵 at all. So let’s begin by calculating the sum of the horizontal forces. Let’s take the direction in which the tension force is acting to be positive. Then we have 𝑅 sub 𝐴 acting in the other direction. So the sum of the forces in a horizontal direction is 𝑇 minus 𝑅 sub 𝐴. And that, of course, is going to be equal to zero.

Next, we’re going to take moments about 𝐵. Now we can take moments about any point on the ladder, although we often choose the foot of the ladder since there are more forces acting there. Now, in fact, there are two forces that we’re going to be looking at, and that is the downwards force of the weight and the reaction force at 𝐴. But remember, we said the moment was calculated by multiplying the force with its perpendicular distance from the pivot. And so we’re going to need to resolve these forces into their components which act perpendicular to the ladder. We can add right-angled triangles as shown with an included angle of 𝜃. So we actually need to calculate the value of 𝜃 next.

By enlarging just the outline of the floor, the wall, and the ladder, we see we can use right angle trigonometry. We have an adjacent side of 2.5 meters and a hypotenuse of five. And so we use the cosine ratio. cos of 𝜃 is adjacent over hypotenuse. In our triangle, that’s cos of 𝜃 is 2.5 divided by five. But 2.5 divided by five is equal to one-half. And since we know cos of 60 is equal to one-half 𝜃 must be equal to 60 degrees. We can therefore add 60 degrees in place of 𝜃. And we’re ready to start calculating the components of our weight and reaction force that are perpendicular to the ladder.

Let’s begin with the component of the weight. We enlarge this triangle a little bit. Let’s call the component we’re trying to find 𝑥. 𝑥 is the adjacent side in our triangle, whereas the hypotenuse is this 40 root three kilogram-weight. We can therefore use the cosine ratio once again to find the value of 𝑥. We get cos of 60 is 𝑥 over 40 root three. But remember, cos of 60 degrees was one-half. So let’s solve this equation by multiplying both sides by 40 root 3. One-half times 40 root three is 20 root three. So the component of the weight force that acts perpendicular to the ladder is 20 root three kilogram-weight.

So what is the moment of this force? We’re going to say it’s negative, since we chose counterclockwise to be positive. It’s negative 20 root three times the distance from that pivot; that’s two. We’re now going to repeat this process for 𝑅 sub 𝐴. Let’s call the component of this force that we’re trying to find 𝑦. This time, 𝑦 is the opposite side in our triangle. And we have an expression for the hypotenuse. The trigonometric ratio that links these is the sine ratio. So let’s substitute what we know about our triangle into this formula.

When we do, we get sin of 60 is equal to 𝑦 over 𝑅 sub 𝐴. In fact, though, sin of 60 is root three over two. So let’s multiply both sides of this equation by 𝑅 sub 𝐴 to find an expression for 𝑦. When we do, we get 𝑦 is equal to root three over two times 𝑅 sub 𝐴. Now, before we calculate the moment, let’s go back to that first equation we wrote. If we add 𝑅 sub 𝐴 to both sides, we find that 𝑇 is equal to 𝑅 sub 𝐴. Now this is really useful because if we replace 𝑅 sub 𝐴 with 𝑇, we find 𝑦 is equal to root three over two 𝑇. And when we complete the equation that shows the sum of our moments is equal to zero, we’re going to have an equation purely in terms of 𝑇.

The moment of this force, then, is root three over two 𝑇 times the distance from the pivot, so that’s five. And of course, we know that sum of these moments is zero. We need to solve this equation for 𝑇, so the first thing we’re going to do is just divide through by root three. When we do, our equation becomes negative 40 plus five over two 𝑇 equals zero. We can add 40 to both sides, and our final step will be to divide through by five over two. This is the same as dividing by five and multiplying by two. 40 divided by five over two then is 16. And so we’ve calculated the tension in the string. We know our units for force in this question are kilogram-weight. So the tension is 16 kilogram-weight.