Question Video: Finding the Angle of Inclination of the Point on Limiting Equilibrium of a Ladder Resting Against a Rough Floor and a Rough Wall Mathematics

Video Transcript

A uniform rod rests with one end against a rough vertical wall, and the coefficient of friction between the rod and wall is three-tenths. The other end of the rod sits on a rough horizontal floor, and the coefficient of friction between the rod and the floor is one-third. Calculate the angle of inclination 𝜃 between the rod and the floor when it is in limiting equilibrium, giving our answer to the nearest minute.

Before we perform any calculations here, we’re going to simply begin by drawing a free-body diagram. Remember, this is a really simple sketch that just shows all of the key forces we’re interested in. Here is a rigid rod inclined at an angle of 𝜃 degrees to the horizontal. We’re going to call end 𝐵 the end that sits against the wall and end 𝐴 the end that sits on the floor. We’re also told the rod is uniform. This means the downwards force of its weight will act at the point exactly halfway along the ladder.

For ease, let’s define then the length of the ladder to be two 𝑥 meters. And so, the weight force acts at a point 𝑥 meters from point 𝐴. Then, we consider Newton’s third law of motion. This tells us that if the rod exerts a force on the wall and the floor, the wall and the floor must themselves exert a normal reaction force on the rod. We’re going to call the normal reaction force of the floor on the rod 𝑅 sub 𝐴 and the normal reaction force of the wall on the rod 𝑅 sub 𝐵.

Then, there are two further forces that we’re interested in. We’re told that the wall and the floor are rough. This means there must be frictional forces at play. We’re also told that the rod is in limiting equilibrium. That means it’s on the point of sliding. Remember, the frictional force will act against the direction in which the body is trying to travel. So, the frictional force at 𝐴 will act towards the wall as shown. And the frictional force at 𝐵, let’s call that Fr sub 𝐵, must be acting upwards.

And at this point, we can recall a formula to help us calculate the friction. Friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the normal reaction force at that point. We’re told the coefficient of friction between the rod and the floor, so at point 𝐴, is one-third. So, 𝜇𝑅 here would be a third 𝑅 sub 𝐴. Similarly, the coefficient of friction between the rod and the wall is three-tenths. So, 𝜇𝑅 at point 𝐵 must be three-tenths times 𝑅 sub 𝐵.

We’ve now extracted all the relevant information we need from this question. So, let’s clear some space. So, once we have all the forces labeled, what do we do next? Well, we recall that the body was in limiting equilibrium. We discussed what the limiting part meant. That meant it’s on the point of sliding. But for a body to be in equilibrium, there are two conditions we need to consider. Firstly, the sum of all the forces acting on the body must be equal to zero. We tend to break these down directionally and say that the sum of the horizontal forces is zero and the sum of the vertical forces is zero.

We also know that the sum of all the moments acting on the body must also be equal to zero, where moment is 𝐹𝑑. 𝐹 is the force, and 𝑑 is the perpendicular distance of the line of action of this force from the point about which the body is trying to pivot. And so, what we’re going to do here is begin by resolving the forces on our diagram. Then, we’ll consider moments.

We’re going to resolve both horizontally and vertically. Let’s begin by resolving vertically, and we’re going to take upwards to be positive. Acting upwards, we have the reaction force at 𝐴 and the frictional force at 𝐵. And then acting downwards, we have the weight of the body. The sum of these forces is, of course, equal to zero. Now, we’re actually going to replace the frictional force at 𝐵 with the expression we developed earlier, three-tenths 𝑅 sub 𝐵.

So, we get 𝑅 sub 𝐴 plus three-tenths 𝑅 sub 𝐵 minus 𝑤 equals zero. Next, we add 𝑤 to both sides. Eventually, we want to be getting rid of all 𝑤’s in our equation. So, we get 𝑅 sub 𝐴 plus three-tenths 𝑅 sub 𝐵 equals 𝑤.

Next, we resolve horizontally. And we’re going to take the direction in which the frictional force at 𝐴 is acting to be positive. In this direction, we have frictional force of 𝐴. And in the other direction, we have the reaction force at 𝐵. So, their sum is Fr sub 𝐴 minus 𝑅 sub 𝐵 and that, of course, is equal to zero. Now, we’re going to replace the frictional force at 𝐴 with the expression we developed earlier, a third 𝑅 sub 𝐴.

So we get one-third 𝑅 sub 𝐴 minus 𝑅 sub 𝐵 equals zero. We’re going to make 𝑅 sub 𝐴 the subject here. And eventually, what we’re going to do is try and find an expression purely in terms of 𝑅 sub 𝐵. If we add 𝑅 sub 𝐵 to both sides of our equation and then multiply by three, we find that 𝑅 sub 𝐴 is three times 𝑅 sub 𝐵. Now, this is really useful because we can now replace this 𝑅 sub 𝐴 in our earlier equation. And so, notice we have an equation just in terms of two variables now. We have 𝑤 equals three times 𝑅 sub 𝐵 plus three-tenths 𝑅 sub 𝐵 or 𝑤 equals 33 over 10 𝑅 sub 𝐵.

We’re now going to calculate moments. We begin by finding the point about which we want to calculate those moments. It doesn’t really matter which point we choose. We should get the same answer either way, but we tend to choose the foot of the ladder, since there’s usually more forces acting here. We’re also going to define the counterclockwise direction to be positive. And so, we’ll begin with the weight force.

Now, remember, when we calculate a moment, we want to find the component of each force that acts perpendicular to the body. We add a right-angled triangle in, and we need to find the value of 𝑥. Now, 𝑥 is the component of the weight force we’re interested in. It’s not the same as the length of the ladder. The included angle here is 𝜃, 𝑤 is the hypotenuse, and 𝑥 is the adjacent. So, we can use the cosine ratio to find some sort of expression for 𝑥.

When we do, we get cos of 𝜃 is 𝑥 over 𝑤. And then when we multiply by 𝑤, we get 𝑥 equals 𝑤 cos 𝜃. Now, of course, we said 𝑤 is thirty-three tenths 𝑅 sub 𝐵 And so, we have an expression for 𝑥 in terms of 𝜃 and 𝑅 sub 𝐵. The moment is going to be negative since this is trying to turn the rod in a clockwise direction. Force times distance is thirty-three tenths 𝑅 sub 𝐵 cos 𝜃 times 𝑥.

There are two further moments we need to consider; these are the moments to do with 𝑅 sub 𝐵 and the frictional force at 𝐵. Once again, we’re going to resolve 𝑅 sub 𝐵 to find the component of this force that acts perpendicular to the rod. This time, 𝑅 sub 𝐵 is the hypotenuse, but we’re trying to find the value of 𝑦, which is the opposite. So, we’re going to use the sine ratio. When we do, we find that sin 𝜃 is 𝑦 over 𝑅 sub 𝐵. And we can multiply both sides of this equation by 𝑅 sub 𝐵. So, 𝑦 is 𝑅 sub 𝐵 sin 𝜃. This time, this moment is going to be positive since it’s acting in the counterclockwise direction. And force times distance is 𝑅 sub 𝐵 sin 𝜃 times two 𝑥.

There’s one more force we’re interested in, and that’s the frictional forces at 𝐵. Once again, we can add a right-angled triangle here. And we’re trying to find the value of 𝑧; this is the component of the frictional force that’s perpendicular to the rod. It’s a little bit trickier to find the angle we’re interested in here. In fact, though, this angle, the angle at 𝐵, is 90 minus 𝜃. We can convince ourselves this is true because the angle between 𝑅 sub 𝐴 and the frictional force at 𝐴 must be 90. Then, we have a pair of corresponding angles. And so, this angle up here must itself be 𝜃, since angles in a triangle add up to 180.

We know the hypotenuse and the adjacent in this triangle then. So, we go back to the formula cos of 𝜃 is adjacent over hypotenuse. So, cos of 𝜃 is 𝑧 over the frictional force at 𝐵. We multiply both sides of this equation by the frictional force at 𝐵. But remember, we have an expression for that; it’s three-tenths 𝑅 sub 𝐵. And we found the component of the frictional force at 𝐵 that’s perpendicular to the rod; it’s three-tenths 𝑅 sub 𝐵 times cos 𝜃. And so, we can find the moment of this force. Once again, it’s positive. It’s three tenths 𝑅 sub 𝐵 cos 𝜃 times two 𝑥.

The body is in equilibrium. So, we set this equal to zero. Now, we want to solve for 𝜃, but we actually have three unknowns here. We have 𝑅 sub 𝐵, 𝜃, and 𝑥. However, we defined two 𝑥 to be the length of the ladder, so 𝑥 was half that length. And there’s no way 𝑥 can be equal to zero. We’re therefore able to divide through by 𝑥. Similarly, we know the reaction force at 𝐵 cannot be equal to zero either. So, we divide through by 𝑅 sub 𝐵. And so, we’re left with negative 33 over 10 cos 𝜃 plus two sin 𝜃 plus three-fifths cos 𝜃 equals zero.

Notice now we’re working with just one variable. We’re going to gather together the cos 𝜃s first. So, two sin 𝜃 minus 27 over 10 cos 𝜃 equals zero. And then, we’re going to recall one of our key trigonometric identities; that is, tan 𝜃 is equal to sin 𝜃 over cos 𝜃. We want our equation to look a bit like this. So, we’re going to add 27 over 10 cos 𝜃 to both sides. Next, we’re going to divide through by cos 𝜃. So, two sin 𝜃 over cos 𝜃 equals twenty-seven tenths. But of course, we can replace sin 𝜃 over cos 𝜃 with tan 𝜃, and now we’re going to divide through by two.

We’re nearly done. We know that tan 𝜃 is equal to twenty-seven twentieths. So, 𝜃 must be equal to the inverse or arc tan of this value. That gives us 53.471 degrees. Now, of course, we were told to give our answer to the nearest minute. And so, we use the relevant button on our calculator. And we find that 𝜃, which is the angle of inclination of the rod to the floor, is 53 degrees and 28 minutes.