Question Video: Analysis of the Equilibrium of a Ladder Resting between a Smooth Wall and a Rough Ground Mathematics

𝐴𝐵 is a uniform ladder with weight 177 N. End 𝐴 rests on rough horizontal ground and end 𝐵 rests against a smooth vertical wall. The ladder is inclined to the horizontal at an angle of 60°, and the coefficient of friction between the ground and the ladder is √(3/4). Find the maximum weight that can be suspended from 𝐵 without the ladder slipping.

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Video Transcript

𝐴𝐵 is a uniform ladder with weight 177 newtons. End 𝐴 rests on rough horizontal ground and end 𝐵 rests against a smooth vertical wall. The ladder is inclined to the horizontal at an angle of 60 degrees, and the coefficient of friction between the ground and the ladder is root three over four. Find the maximum weight that can be suspended from 𝐵 without the ladder slipping.

Let’s begin by sketching this scenario out. We have a ladder that is inclined to the horizontal at an angle of 60 degrees as shown. We’re told that it’s uniform, so we can assume that its weight is evenly distributed across the ladder. We can, therefore, say that the downward force of the weight, which is 177 newtons, must act exactly halfway along the ladder. We are, however, told that end 𝐴 of the ladder rests on rough horizontal ground. This means the ground exerts a frictional force on the ladder.

We’re interested in the scenario where the ground is stopping the ladder from slipping. In other words, the frictional force must be acting in the direction shown. Now, since the ladder is resting on the ground, we know that there must be a reaction force of the ground on the ladder. And that reaction force is perpendicular to the ground as shown. Similarly, we’re told end 𝐵 of the ladder rests against a smooth vertical wall. Since the wall is smooth, there are no forces aside from the reaction force of the wall on the ladder. Once again, this force acts perpendicular to the wall as shown.

We’re told that the coefficient of friction between the ground and the ladder is root three over four. Let’s define that to be equal to 𝜇. Now, our job is to find the maximum weight that can be suspended from 𝐵. So let’s add this downward force at 𝐵 and let that be equal to 𝑤 newtons. Ultimately, that’s what we’re going to be looking to calculate.

Now, this is really a moments question. To answer the question, we’ll begin by calculating the horizontal and vertical forces that act on the ladder. And then we’re going to resolve moments about one end of our ladder.

Let’s begin by resolving our forces in a vertical direction. We know that the ladder is in equilibrium, it’s on the point of slipping, but it’s not slipping. So the sum of all the forces, let’s call that 𝐅 sub 𝑦, in the vertical direction must be equal to zero. Now, acting upwards on our diagram, we have the reaction force at 𝐴. And then we have two downwards forces. We have the weight of the ladder, which is 177 newtons, and the weight suspended at 𝐵, which we’ve let be equal to 𝑤. These two forces are acting in the opposite direction to our reaction force at 𝐴. So the sum of these forces must be 𝑅 𝐴 minus 177 minus 𝑤, which we of course know is equal to zero. We can rearrange to make 𝑅 sub 𝐴 the subject, and we get 𝑅 sub 𝐴 equals 177 plus 𝑤.

Let’s now look to resolve forces horizontally. Once again, the ladder hasn’t actually slipped, so it’s in equilibrium. And we can say that the sum of the horizontal forces must be equal to zero. That’s the sum of 𝐅 sub 𝑥 must be equal to zero. Well, we have frictional force acting, let’s say, in the positive direction. And then in the opposite direction, we have 𝑅 sub 𝐵, the reaction force at 𝐵. Their sum then must be friction minus 𝑅 sub 𝐵. And their sum is equal to zero.

But actually, we have a formula that helps us to link the frictional force and the coefficient of friction. Its friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the reaction force at that point. The reaction at that point is 𝑅 sub 𝐴, so friction is root three over four times 𝑅 sub 𝐴. But we can replace 𝑅 sub 𝐴 with 177 plus 𝑤. So we found that friction must be equal to root three over four times 177 plus 𝑤. And if we were to add 𝑅 sub 𝐵 to both sides of our earlier equation, we see that the friction must be equal to 𝑅 sub 𝐵. So we can also say that the reaction force at 𝐵 is root three over four times 177 plus 𝑤.

Now that we’ve resolved forces horizontally and vertically, we’re going to take moments. Now, we can actually take moments about any point on our ladder. It makes sense to take moments about one end. And generally, it can be easier to take moments about the foot of the ladder, in other words, the point where that ladder meets the floor. So we’re going to take moments about 𝐴, taking the anticlockwise direction to be positive.

We recall, though, a moment is essentially a turning force. The formula to calculate the moment of a force is moment is equal to force times distance. Here, distance is the perpendicular distance from the pivot to the line of action of the force. Once again, since our ladder’s in equilibrium, we know that the sum of all of the moments must be equal to zero. So since we’re starting by taking moments about 𝐴, let’s look at the force that’s furthest away from 𝐴. It’s the forces at 𝐵.

There are two forces at 𝐵. There’s the downwards force of the weight of the object suspended from 𝐵. And then there’s the force perpendicular to the wall, the reaction force. But we want the components of these forces that are perpendicular to the ladder. So we begin by adding a right-angled triangle here. Our job is to calculate the component of this reaction force that is perpendicular to the ladder.

Now, if we make this triangle a little bit bigger, we can see that it’s a right-angled triangle and the hypotenuse represents the reaction force at 𝐵. The included angle must be 60 degrees since we know that alternate angles are equal. In other words, it’s the same as this angle down here. By using trigonometric conventions for labeling our triangle, we see we want to find the length of the opposite side in this triangle. And we know the hypotenuse. Remember, we want the opposite side because we’re interested in the component of the force that’s perpendicular to the ladder.

If we let that measurement be equal to 𝑥 or 𝑥 newtons, we can use the sine ratio, where sin 𝜃 is opposite over hypotenuse. sin 60 is 𝑥 over 𝑅 𝐵. And so the component of this reaction force that acts perpendicular to the ladder must be 𝑅 𝐵 times sin 60. And therefore, the moment at 𝐵 of this force is 𝑅 𝐵 sin 60 times the distance from 𝐴, which is 𝐿 or 𝐿 meters.

But we also have this other force. We have the weight acting downwards at 𝐵. So we need the component of the weight that acts perpendicular to the ladder. This time, if we draw this triangle a little bit bigger, we see its hypotenuse is 𝑤 or 𝑤 newtons. Once again, the included angle is 60 degrees. And so when we label our triangle, we see we want to relate the adjacent side and the hypotenuse. Let’s call that adjacent side 𝑦 or 𝑦 newtons. And we, therefore, use the cos ratio. cos 60 equals 𝑦 over 𝑤. And if we multiply by 𝑤, we see that 𝑦, which is the component of 𝑤 which acts perpendicular to the ladder, is 𝑤 cos 60. This force is trying to turn the ladder in a clockwise direction, so we subtract the moment, that’s force times distance. And we subtract 𝑤 cos 60 times 𝐿.

There’s one more moment that we need to calculate. And that’s the moment halfway along the ladder. This time, if we add a right-angled triangle here, we see that we still know the hypotenuse and we’re looking to find the adjacent. This time, I’ve called that 𝑧. We use the cosine ratio again. So cos 60 is 𝑧 over 177 or 𝑧 is 177 cos 60. This is trying to turn the ladder in a clockwise direction, so we subtract the moment.

It’s halfway along the ladder, so the distance is a half 𝐿. And we’re going to subtract 177 cos 60 times a half 𝐿. This is all of the moments. And we know that their sum is equal to zero. So what are we going to do with this equation? Well, firstly, we see that we have a constant factor of 𝐿. So we’re going to divide this equation by 𝐿. We’re allowed to divide through by this unknown here because we know that this value of 𝐿 cannot be equal to zero. The ladder can’t be zero meters in length.

But we’re also able to replace 𝑅 sub 𝐵 with an expression in terms of 𝑤. When we do, our first term becomes root three over four times 177 plus 𝑤 times sin 60. We can also simplify our third term to get 177 over two cos 60. But we know that sin 60 is root three over two and cos 60 is equal to one-half. So we can simplify a little further.

Let’s distribute our parentheses in our first term. Root three over four times root three over two becomes three-eighths. And three-eighths times 177 gives us 531 over eight. Then three-eighths times 𝑤 gives us three 𝑤 over eight. Our next two terms become negative 𝑤 over two and negative 177 over four. We need to now solve for 𝑤. Three 𝑤 over eight minus 𝑤 over two gives us negative 𝑤 over eight. So we add 𝑤 over eight to both sides. Similarly, adding 531 over eight to negative 177 over four gives us 177 over eight.

And we see that 177 over eight must be equal to 𝑤 over eight. And we can either use observation or multiply both sides of this equation by eight. When we do, we get 𝑤 is equal to 177 or 177 newtons. The maximum weight that can be suspended from 𝐵 without the ladder slipping is 177 newtons. Remember, we considered limiting equilibrium, where the frictional force was exactly equal to the forces acting in the opposite direction. If the forces acting in the opposite direction are any greater, then they get greater than the frictional force and the ladder will move.

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