# Question Video: Analysis of the Equilibrium of a Ladder Resting between a Smooth Wall and Smooth Ground and Attached with a Horizontal String While a Man Is Climbing It Mathematics

A uniform ladder 𝐴𝐵 having a length 𝐿 and weighing 40 kg-wt is resting with one of its ends on a smooth floor and the other against a smooth vertical wall. The ladder makes an angle of 45° with the horizontal, and its lower end 𝐴 is attached to a string that is fixed to a point at the junction of the wall and floor. Given that the maximum tension the string can withstand is 60 kg-wt, find how far up the ladder a man of weight 140 kg-wt can go before the string breaks.

09:41

### Video Transcript

A uniform ladder 𝐴𝐵 having a length 𝐿 and weighing 40 kilogram weight is resting with one of its ends on a smooth floor and the other against a smooth vertical wall. The ladder makes an angle of 45 degrees with the horizontal, and its lower end 𝐴 is attached to a string that is fixed to a point at the junction of the wall and floor. Given that the maximum tension the string can withstand is 60 kilogram weight, find how far up the ladder a man of weight 140 kilogram weight can go before the string breaks.

There’s quite a lot of information here. So what we’re going to begin with doing is sketching out the scenario. Here is our uniform ladder, resting against a smooth vertical wall and a smooth floor. Now, the fact that it’s uniform means that its weight is evenly distributed across the ladder. We can therefore say that the downward force of the weight must act exactly halfway along the ladder. So it acts half 𝐿 away from 𝐴. We don’t have any units for 𝐿, so that’s absolutely fine.

We see that it makes an angle of 45 degrees with the horizontal but also that the lower end 𝐴 is attached to a string. Now, that string is fixed to a point at the junction of the wall and floor, here. And so there must be tension acting at the point 𝐴. Essentially, it’s the tension in the string that keeps the ladder in place. Now, the maximum tension the string can withstand is 60 kilogram weight. So we’re actually going to use tension equals 60 kilogram weight in this question. And then we have this man that weighs 140 kilogram weight going up the ladder.

Now, we don’t know exactly how far up the ladder he goes. He might make it less than halfway; he might make it more than halfway. We’ll add this to our diagram and say he makes it 𝑥𝐿 up the ladder. 𝑥 will be a fraction. We’re working out what fraction of 𝐿 he will be able to get up the ladder. What other forces have we got, though? Well, there must be a reaction force of the floor on the ladder and the wall on the ladder. These forces are perpendicular to the surface.

So we have a reaction force at 𝐴 acting directly up and a reaction force at 𝐵 acting to the left. There are no other forces. Remember, the wall and the floor are smooth, so there’s no frictional force. So what do we do next? Our next job is to resolve forces both horizontally and vertically. Once we’ve done that, we’ll be able to consider the moments acting about a point. Let’s begin by resolving our forces vertically. The string is on the point of breaking when its tension hits 60 kilogram weight. So we’re assuming that it’s still in equilibrium or a bit limiting equilibrium. It’s on the point of breaking. So in a vertical direction, we can say that the sum of forces must be equal to zero. That’s the sum of 𝐹 sub 𝑦 is equal to zero.

So what forces have we got acting in a vertical direction? Well, we’ve got the reaction force at 𝐴 acting upwards. So let’s take upwards to be positive. Then we’ve got the weight of the ladder acting in the opposite direction. So we’re going to add negative 40. And we’ve got the weight of the man acting in the opposite direction to the reaction force. So that’s a force of negative 140. So we say that the sum of the forces acting in the vertical direction is 𝑅 sub 𝐴 minus 40 minus 140. And this must be equal to zero. This simplifies to 𝑅 sub 𝐴 minus 180 equals zero. So 𝑅 sub 𝐴, if we add 180 to both sides, we see must be equal to 180 or 180 kilogram weight.

Now, don’t worry if you’re used to measuring forces in newtons kilogram weight; it’s just another way of doing so. Next, we’re going to resolve forces in a horizontal direction. Once again, the sum of these forces will be equal to zero. This time, let’s take the direction to the right to be positive. And we have tension acting in that direction. Then we have the reaction force at 𝐵 acting in the opposite direction. So tension minus 𝑅 sub 𝐵 must be equal to zero. And if we add 𝑅 sub 𝐵 to both sides, we find tension must be equal to 𝑅 sub 𝐵. But remember, we said that we’re taking the case where tension is equal to 60 kilogram weights, since that’s the maximum tension it can take before it breaks. So 𝑅 sub 𝐵 must be equal to 60 or 60 kilogram weight.

We add this force to our diagram, and we’re ready to start taking moments. Remember, the moment is the turning effect of a force. We’re going to take moments about the point 𝐴. Now, we could take moments about any point on this ladder. But generally there’s more going on at the point where the ladder hits the floor. So if we take moments about 𝐴, we’d basically have less to calculate. We’re going to assume that a counterclockwise direction is positive. And then we recall that we calculate the moment of a force by multiplying that force by the perpendicular distance of the line of action of the force from the point where we’re calculating. We’ll see what that looks like in a moment.

So let’s consider all of the forces acting on our ladder. Let’s begin by looking at the reaction force at 𝐵. We calculated that to be 60 kilogram weight. We need to work out the component of this force that is perpendicular to the ladder. This is acting in a counterclockwise direction. So its moment is going to be positive. And so let’s enlarge this triangle a little bit. We should see that it’s a right-angled triangle with a hypotenuse of 60 kilogram weight. The included angle is 45 degrees. And this is because the reaction force at 𝐵 is parallel to the floor. And we know alternate angles are equal.

We want to work out the component of this force that acts perpendicular to the ladder. So let’s call that 𝑎 or 𝑎 kilogram weight. This is the opposite side in our triangle. And we know the hypotenuse is 60. So we can link these using the sine ratio. sin 𝜃 is opposite over hypotenuse. So sin of 45 is 𝑎 over 60. By multiplying through by 60, we see that 𝑎 is equal to 60 sin 45. But in fact, we know that sin of 45 degrees is root two over two. So 𝑎 is 60 times root two over two or 30 root two kilogram weight.

So we found the component of this reaction force that acts perpendicular to the ladder. We need to calculate its moment. Now remember, it’s acting in a counterclockwise direction, so that’s positive. The moment is this force multiplied by the distance away from 𝐴. So that’s 30 root two multiplied by 𝐿.

And what about our other forces? Let’s look at the force which is the weight of the man. This time, the included angle is 45 degrees. And the hypotenuse is 140 kilogram weights. We’ll call the side that we’re looking to calculate, which is the component of this force which acts perpendicular to the ladder, 𝑏.

Now, this time, we’re looking to find the adjacent and we know the hypotenuse. And so we use the cosine ratio. cos of 45 degrees is 𝑏 over 140. So 𝑏 is 140 times cos of 45. But once again, cos of 45 is root two over two. So we find that 𝑏 is 140 times root two over two or 70 root two kilogram weights. Let’s now calculate the moment of this force. We’re going to subtract it since it’s acting in a clockwise direction. We said that this was 𝑥𝐿 away from 𝐴. So the moment is negative 70 root two times 𝑥𝐿.

There’s one more force to consider, and that’s the force of the weight of the ladder. If we add a right-angled triangle to this force, it looks a lot like our previous right-angled triangle. This time, though, the hypotenuse is 40. And we’ll call the length we’re looking to find 𝑐. cos of 45 is 𝑐 over 40. And so if we rearrange, we get 𝑐 equals 20 root two or 20 root two kilogram weights. And so we’re ready to find the moment. Once again, it’s acting in a clockwise direction. So it’s negative and its distance is a half 𝐿. So we have 20 root two times a half 𝐿. We know that this is an equilibrium. So the sum of these moments is zero.

And now, we’re going to look to solve for 𝑥. But there’s an extra variable here; there’s 𝐿. Luckily, we know that 𝐿, the length of the ladder, cannot be equal to zero. So we can divide through by 𝐿. And our equation becomes 30 root two minus 70 root two times 𝑥 minus 20 root two over two equals zero. But of course, 20 root two over two is 10 root two. So this equation simplifies further to 20 root two minus 70 root two 𝑥 equals zero.

In fact, we can also divide through by root two. And then adding 70𝑥 to both sides, our equation simplifies even further to 20 equals 70𝑥. To solve for 𝑥, we divide through by 70. So 𝑥 is 20 over 70, or 𝑥 is equal to two-sevenths. So we’ve said that the man can get two-sevenths of the way up the ladder. Since the length of the ladder is 𝐿, we say that the man can get two-sevenths 𝐿, length units, up the ladder before the string breaks.

Now, notice that we actually didn’t need to calculate the reaction force at 𝐴. This is because when we take moments about 𝐴, we’re multiplying each of the forces at this point by zero. So they’re actually zero. It’s often required, though, to resolve these forces in a vertical direction. So it’s always sensible to do this for completion.