Question Video: Finding the Tension in a String Attached to the End of a Ladder That Is in Equilibrium between the Wall and the Floor Mathematics

A ladder 𝐴𝐵 weighing 34 kg-wt and having a length of 14 m is resting in a vertical plane with its end 𝐵 on a smooth floor and its end 𝐴 against a smooth vertical wall. The end 𝐵, which is 3.3 m away from the wall, is attached by a string to a point on the floor directly below 𝐴. Given that the weight of the ladder is acting on the ladder itself at a point 5.6 m away from 𝐵, find the tension in the string when a man of weight 74 kg-wt stands on the midpoint of the ladder.

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Video Transcript

A ladder 𝐴𝐵 weighing 34 kilogram-weight and having a length of 14 meters is resting in a vertical plane with its end 𝐵 on a smooth floor and its end 𝐴 against a smooth vertical wall. The end 𝐵, which is 3.3 meters away from the wall, is attached by a string to a point on the floor directly below 𝐴. Given that the weight of the ladder is acting on the ladder itself at a point 5.6 meters away from 𝐵, find the tension in the string when a man of weight 74 kilogram-weight stands on the midpoint of the ladder.

Before we can go about answering this question, let’s sketch the scenario out. Here is our ladder resting with its ends on the smooth floor and against the wall. We’re told that the end 𝐵 is 3.3 meters away from the wall. And here it’s attached by a string. We’re going to add the force that keeps this point 𝐵 in place. The force that keeps it in place is the tension in the string. Let’s add any other forces we can find.

We’re told that the weight of the ladder acts on the ladder itself at a point 5.6 meters away from 𝐵. So we have this force of 34 kilogram-weights acting downwards at this point. We’re then told that the man of weight 74 kilogram-weight stands on the midpoint of the ladder. So that’s the point seven meters away from 𝐵. Remember, our ladder is 14 meters in length.

There are two further forces we can add to our diagram. These are the reaction forces of the ladder against the wall and the floor. These act perpendicular to the surface the ladder is resting on. Let’s call them 𝑅 sub 𝐴 for the reaction force at 𝐴 and 𝑅 sub 𝐵 for the reaction force at 𝐵. Now, we’re told that the floor and the wall are smooth. So there are no further forces such as friction, for example. So how are we going to answer this question?

Well, we’re going to begin by resolving forces both in a horizontal and vertical direction. Then we’re going to consider the moments acting on the ladder. Let’s begin by considering the forces in the vertical direction. Now, the ladder is in equilibrium; it’s not moving. And so we can say that the sum of the forces acting in the vertical direction, let’s call that 𝐹 sub 𝑦, is equal to zero. So what forces have we got acting in the vertical direction?

Well, we’ve got the reaction force acting upwards. Let’s take that to be positive. Then in the downward direction, we’ve got the force of the weight of the ladder and the weight of the man. So that’s negative 34 and negative 74 kilogram-weight. That’s of course equal to zero. That’s the sum of all of our forces in the vertical direction. Negative 34 minus 74 is negative 108. So our equation becomes 𝑅 sub 𝐵 minus 108 equals zero. We then solve for 𝑅 sub 𝐵 by adding 108 to both sides. And we see that 𝑅 sub 𝐵, that’s the reaction force of the floor on the ladder, is 108 or 108 kilogram-weights.

Now, you might be used to working in newtons rather than kilogram-weights. Don’t worry too much. The kilogram-weight is defined as the force needed to make one kilogram of mass accelerate by roughly 9.8 meters per square second. It’s just another way of measuring a force. As long as we’re consistent throughout this question, we’re absolutely fine to keep kilogram-weights as our force.

Let’s now consider what’s happening in a horizontal direction. Once again, the ladder is in equilibrium. So we can say that the sum of all the forces acting in this direction, let’s call that 𝐹 sub 𝑥, is equal to zero. So what forces have we got acting in this direction?

We got tension acting to the right. Let’s take that to be the positive direction. Then we have the reaction force at 𝐴, 𝑅 sub 𝐴, acting in the opposite direction. So the sum of the forces in a horizontal direction is 𝑇 minus 𝑅 sub 𝐴, and that’s equal to zero. We can therefore say that by adding 𝑅 sub 𝐴 to both sides, the tension must be equal to this reaction force.

Now, we don’t know what this reaction force is yet. So we’re not yet able to calculate the tension. We’ll add this information next to our diagram. And next, we’re going to clear some space and consider moments. Before we do though, let’s just work out the size of the included angle. That’s the angle that the ladder makes with the floor. I’ve called that 𝜃.

Since the wall is vertical, we can say that the ladder, the wall, and the floor make a right-angled triangle. We see that the included angle is 𝜃. The distance between the ladder at 𝐵 and the wall is 3.3 meters. And the length of the ladder is 14 meters. In fact, there are two things that we could do now. We could work out the size of this length here. Let’s call that 𝑥 or 𝑥 meters. That will allow us to then create expressions for cos of 𝜃, sin of 𝜃, and so, should we need it, tan of 𝜃. But we’re actually going to work out the value of 𝜃.

We know the length of the adjacent and the hypotenuse, so we’re going to use the cosine ratio. cos of 𝜃 is adjacent over hypotenuse. So in this case, cos 𝜃 is 3.3 divided by 14. By taking the inverse cos of both sides, we see that 𝜃 is inverse cos of 3.3 over 14, which is 76.366 and so on degrees. Let’s add that to our diagram. And we’re now ready to take moments. We’re going to take moments about 𝐵.

Now, we could actually take moments about any point on the ladder. But it can be easier to take moments about the point where the ladder hits the floor. We’re going to take the anticlockwise direction to be positive. But what do we mean when we talk about moments?

A moment is the turning effect of the force. And we say that the moment of a force is the force multiplied by distance 𝑑. It’s important though that we remember that the distance is the perpendicular distance from the pivot to the line of action of the force. So let’s consider the moments about point 𝐵. Let’s look at the weight of the ladder. The weight of the ladder is acting in a downwards direction. We need to therefore work out the component of this force that acts perpendicular to the ladder.

And so we can add a little right-angled triangle. But we’re going to actually increase the size of this triangle so we can see what’s going on. So we have the downward force, let’s just call that 34 that we know it’s kilogram-weights. And we’re trying to find the component of this force which is perpendicular to the ladder. Let’s call that 𝑥, although we know it’s going to be 𝑥 kilogram-weight. This angle here is also 𝜃. It’s 76.3 and so on.

And so we see that we can link the adjacent side and the hypotenuse by using the cosine ratio, giving us cos 𝜃 is 𝑥 over 34 or 𝑥 equals 34 cos 𝜃. This moment is acting against the direction that we’ve defined to be positive. So force multiplied by distance will be negative 34 cos 𝜃 multiplied by 5.6. And this is because we said that the weight of the man acts at a point 5.6 meters away from 𝐵. cos of 𝜃 is cos of 76.3 and so on. And if we use that exact value, we get 33 over 140. This makes a lot of sense since we defined originally cos of 𝜃 to be equal to 3.3 over 14. These are equivalent.

We’re now going to repeat this process for the force of the man. Once again, we can add a right-angled triangle. This triangle is almost identical to the triangle we looked at before. However, the hypotenuse is 74. Let’s call the length we’re looking for, which is the component of this force that acts perpendicular to the ladder, 𝑦. This time, we get cos 𝜃 is 𝑦 over 74. So 𝑦 is 74 cos 𝜃. Its moment is, therefore, negative 74 cos 𝜃 times seven. It’s negative because it’s acting in a clockwise direction. And we’re multiplying that force by seven because the man is standing at the midpoint of the ladder. He’s seven meters away from 𝐵. Once again, we can replace cos 𝜃 with 33 over 140. That’s cos of 76.3 and so on. So what other forces have we got?

We disregard the forces acting at the point 𝐵. And that’s because they are zero meters away from 𝐵. So if we were to calculate their moment, we’d be multiplying by zero. We have, however, got this reaction force at 𝐴. We need to therefore work out the component of this force, which acts perpendicular to the ladder. So once again, we add a right-angled triangle.

Once again, the angle included in this triangle is 𝜃. And that’s because the reaction force 𝐴 is parallel to the floor. And we know that alternate angles are equal. We’re looking to find a way to link 𝑍, which is the component of this force, which acts perpendicular to the ladder, with the reaction force at 𝐴. This time, we’re dealing with the opposite and hypotenuse. So we’re going to use the sine ratio. So sin 𝜃 is 𝑍 over 𝑅𝐴. And multiplying both sides of this equation by 𝑅 sub 𝐴, we see 𝑍 is equal to 𝑅 sub 𝐴 times sin 𝜃.

We can find the moment of this force by multiplying this perpendicular component by the distance that 𝐴 is away from 𝐵. So that’s 𝑅 sub 𝐴 times sin 𝜃 times 14. We know that the ladder is in equilibrium. So the sum of all of its moments is equal to zero. But what else can we do here?

Well, we can replace 𝑅 sub 𝐴 with 𝑇. We calculated earlier that 𝑇 would be equal to 𝑅 sub 𝐴. We can also calculate the value of sin 𝜃, but we’ll do that in a moment. Let’s calculate the value of negative 34 times 33 over 140 times 5.6 and negative 74 times 33 over 140 times seven. Then we find their sum is negative 8349 over 50. By adding this value to both sides of our equation, we get 14 sin 𝜃 times 𝑇 equals 8349 over 50.

Of course, 14 sin 𝜃 we can now calculate the value of. We have 𝜃 as 76.3 and so on. Ideally, we use the exact value of 𝜃. And when we do, we get 13.6 and so on. To solve for 𝑇, we need to divide both sides of our equation by this value. That’s 𝑇 equals 8349 over 50 divided by 13.6, which gives us 12.272 and so on. Rounding this answer to two decimal places and continuing with the force units we defined at the start, we find tension is approximately equal to 12.27 kilogram-weights.

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