# Question Video: Analysis of the Equilibrium of a Ladder Resting between a Smooth Wall and a Rough Ground While an Inclined Force Is Acting on Its End Mathematics

A uniform ladder of weight 72 N is resting with its upper end against a smooth vertical wall and its lower end against a rough horizontal ground, where the coefficient of friction between the ladder and the ground is √3/5. When a force of magnitude 12 N acts on the lower end of the ladder trying to move it away from the wall in a direction upward of the horizontal, where the force makes an angle of 30° with the horizontal, the ladder is about to slide. Determine the tangent of the angle that the ladder makes with the horizontal ground.

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### Video Transcript

A uniform ladder of weight 72 newtons is resting with its upper end against a smooth vertical wall and its lower end against a rough horizontal ground, where the coefficient of friction between the ladder and the ground is root three over five. When a force of magnitude 12 newtons acts on the lower end of the ladder trying to move it away from the wall in a direction upward of the horizontal, where the force makes an angle of 30 degrees with the horizontal, the ladder is about to slide. Determine the tangent of the angle that the ladder makes with the horizontal ground.

We’re going to begin by adding all the forces and any information that we know about this ladder to the diagram itself. First, we’re told that the ladder is uniform. That means that we can model the downwards force of its weight as occurring at a point exactly halfway along the ladder. Now we’re not actually given the length of the ladder, so we’ll define it to be equal to 𝐿 length units. Then the downwards force of its weight occurrs at a point one-half 𝐿 length units from either end of the ladder. Next, we’re told that the wall is smooth, so there’s no frictional force between the wall and the ladder.

The ground, however, is rough. This means that there is a frictional force between the ladder and the ground. This frictional force will be acting against the direction in which the ladder is trying to move. So on our diagram, the frictional force is acting to the left.

Now we’re also told that the coefficient of friction between the ladder and the ground is root three over five. We’ll define that as 𝜇. And we’ll come back to that in a moment. The force of magnitude 12 newtons that acts on the lower end of the ladder is already drawn on our diagram. And we see it makes an angle of 30 degrees with the horizontal. And at this point, the ladder is about to slide. This means the ladder is going to be in limiting equilibrium. And so the vector sum of the forces that act on the ladder is going to be equal to zero. But similarly, if we take moments about a given point on the ladder, the sum of those moments must also be equal to zero.

The question is asking us to determine the tangent of the angle that the ladder makes with the horizontal ground. So let’s add that angle to our diagram. And we’ll call that 𝜃. Now, because the ladder will be exerting some sort of force on the ground and the wall, we know that there must be a normal reaction force at each of these points. We’ll define them to be 𝑅 sub 𝐺 and 𝑅 sub 𝑤, respectively — the reaction force of the ground on the ladder and the reaction force of the wall on the ladder. We now have enough information to get started, so let’s clear some space.

We notice that there are a whole bunch of forces on our diagram, so we’re going to begin by resolving forces in both the vertical and horizontal direction. This will allow us to gain some additional information about, for example, the frictional force at the base of the ladder. We’ll start with the vertical forces. And let’s take the upward direction to be positive. Remember, we said that the sum of these forces is equal to zero; the ladder is in limiting equilibrium. In an upward direction, we have 𝑅 sub 𝐺. We also need to think about the component of the 12-newton force that acts in that vertically upward direction. So adding a side to this and creating a right triangle, we want to calculate the measurement 𝑥.

This is the opposite side to our included angle. And we know that the hypotenuse is equal to 12. So we can use the sine ratio, where sin of 30 is 𝑥 over 12 or 𝑥 equals 12 sin 30. But of course, sin 30 is equal to one-half, so 12 sin 30 is six or six newtons. So the sum of the forces acting vertically upwards is 𝑅 sub 𝐺 plus six. We subtract the downwards force of the weight of the ladder. And we now have the sum of all forces acting in this direction. It’s 𝑅 sub 𝐺 plus six minus 72, which simplifies to 𝑅 sub 𝐺 minus 66.

But remember, the sum of these forces is going to the equal to zero, so 𝑅 sub 𝐺 minus 66 equals zero, meaning 𝑅 sub 𝐺 equals 66 or 66 newtons. Now this is really useful because it’s going to help us calculate the value of the frictional force. In turn, we’ll then be able to calculate the value of the reaction force at the wall.

Let’s resolve in a horizontal direction. And now we’ll take the direction to the right to be positive. We have 𝑅 sub 𝑤 acting in this direction, but there’s a second force acting towards the right. It’s the horizontal component of the 12-newton force. Let’s call that 𝑦. We can either use right-triangle trigonometry or the Pythagorean theorem to calculate this value.

Let’s use right-triangle trigonometry. We get cos of 30 is 𝑦 over 12. So 𝑦 is 12 cos 30, but in fact cos of 30 degrees is root three over two. So 𝑦 is six root three newtons. So the sum of the forces acting to the right is 𝑅 sub 𝑤 plus six root three. But then we have the frictional force acting towards the left. So the sum of all of our forces in this direction is 𝑅 sub 𝑤 plus six root three minus the frictional force.

But we know a formula that will help us rewrite this. The frictional force is equal to 𝜇𝑅. It’s the coefficient of friction multiplied by the normal reaction force at that point. We are given that the coefficient of friction is root three over five. And we just calculated the normal reaction force at that point. It’s 66 newtons. So the frictional force here is root three over five times 66. And this simplifies to 𝑅 sub 𝑤 minus 36 root three over five.

Once again, of course, the sum of the forces acting in this direction is zero. So the normal reaction force at the wall must be equal to 36 root three over five newtons.

So what now? We still need to work out the tangent of the angle that we defined to be equal to 𝜃. Well, remember we said that the sum of the moments about a given point is also going to be equal to zero. So we’re going to take the moments about a specific point on the ladder. Let’s clear some space.

It’s worth noting that we can take moments about any point on the ladder. But since the moment is the product of the force and its perpendicular distance from the pivot, it often make sense to choose the point about which we take the moments to be the point which has the most forces acting at it. Well, here, that’s the foot of the ladder. There are one, two, three forces acting at this point. So we’re going to take moments about the foot of the ladder. So we’ll take moments about the ground, so moments about 𝐺. And we’ll take the counterclockwise direction to be positive.

First, let’s consider the moment of the weight of the ladder. Since the force and distance need to be perpendicular to one another to calculate the moment, we’re going to find the component of this force that acts perpendicular to the ladder. Let’s call that 𝑎. And drawing a right triangle, we see we have an included angle of 𝜃. Now 𝑎 is the adjacent in this triangle. And we know the hypotenuse. So we’re going to use the cosine ratio. When we do, we find that this component is 72 cos 𝜃. So the moment of this force about the base of the ladder is 72 cos 𝜃 multiplied by that distance, which we said is half 𝐿.

So we’ve dealt with this force. We now need to consider the reaction force of the wall. Once again, this reaction force makes an angle of 𝜃 with the ladder. And we need to calculate the component of this force that acts perpendicular to the ladder. Labeling that side 𝑏 and we see that it’s the opposite side in a triangle for which we know the hypotenuse. So we use the sine ratio, where 𝑏 is 𝑅 sub 𝑤 sin 𝜃. But we calculated 𝑅 sub 𝑤 to be 36 root three over five. And so we have a finalized version for 𝑏.

We notice that this force is trying to move the ladder in a clockwise direction, so its moment is going to be negative. It’s negative 36 root three over five sin 𝜃 times 𝐿. Remember, that’s the length of the ladder in length units. The sum of these moments, of course, is equal to zero. So we have a finalized equation. It’s 72 cos 𝜃 times a half 𝐿 minus 36 root three over five sin 𝜃 times 𝐿 equals zero. Now we know the length of the ladder cannot be equal to zero. So in fact we can neaten things up a little bit by dividing through by 𝐿. 72 times one-half is 36. So our equation simplifies to 36 cos 𝜃 minus 36 root three over five sin 𝜃 equals zero. In fact, we can then once again divide by 36.

Now notice we’re trying to find the tangent of the angle. And so we can use the identity tan 𝜃 equals sin 𝜃 over cos 𝜃 to find the tan of our angle 𝜃. We need to find a way to achieve sin 𝜃 divided by cos 𝜃. So let’s first add root three over five sin 𝜃 to both sides of our equation. That gives us cos 𝜃 equals root three over five sin 𝜃. Then if we divide through by cos 𝜃, we’re left with one on the left-hand side. But of course, on that right-hand side, sin 𝜃 over cos 𝜃 is tan 𝜃. So our equation becomes one equals root three over five tan 𝜃.

There’s just one more step. We need to divide both sides of this equation by root three over five. One divided by root three over five is five over root three. We’ll simplify this expression by rationalizing the denominator. And we do that by multiplying both the numerator and denominator by root three. Root three times root three is three, so the denominator becomes three. And we found the tangent of the angle that the ladder makes with the horizontal ground. It’s five root three over three.

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