Video: Studying the Equilibrium of a Ladder Resting between a Rough Floor and a Rough Vertical Wall

𝐴𝐡 is a uniform ladder weighing 51 N whose end 𝐴 is resting on rough horizontal ground, while its end 𝐡 is against rough vertical wall so that the ladder is inclined to the horizontal at an angle of 45° The coefficients of friction of 𝐴 and 𝐡 are 7/9 and 1/4 respectively. If the end 𝐴 of the ladder is pulled by a horizontal force 𝐹 to bring the ladder to the brink of moving away from the wall, find the magnitude of 𝐹.

08:43

Video Transcript

𝐴𝐡 is a uniform ladder weighing 51 newtons whose end 𝐴 is resting on rough horizontal ground, while its end 𝐡 is against rough vertical wall so that the ladder is inclined to the horizontal at an angle of 45 degrees. The coefficients of friction of 𝐴 and 𝐡 are seven-ninths and one-quarter, respectively. If the end 𝐴 of the ladder is pulled by a horizontal force 𝐹 to bring the ladder to the brink of moving away from the wall, find the magnitude of 𝐹.

To answer this question, we’re going to begin by drawing a diagram. Here is our ladder resting on the wall at an angle of 45 degrees to the floor. We’re told that the ladder is uniform and then it weighs 51 newtons. The fact that it’s uniform means that its weight is evenly distributed throughout the ladder. And so, we’re able to model this as the weight acting exactly halfway along the ladder. Then, we don’t know how long the ladder is. So, let’s define that to be equal to 𝐿 units so that the weight is acting a half 𝐿 units away from 𝐴.

We’re told that the ground and the wall are both rough. And so, there are frictional forces acting on the ladder at both 𝐴 and 𝐡. We’ll call them 𝐹 sub 𝐴 and 𝐹 sub 𝐡, respectively. In fact, we also know the coefficient of friction here as well. Let’s call that πœ‡, where πœ‡ sub 𝐴 is the coefficient of friction at 𝐴. πœ‡ sub 𝐴 is seven-ninths and then πœ‡ sub 𝐡 is one-quarter.

There are three further forces we need to consider. There are the normal reaction forces at 𝐴 and 𝐡; these act perpendicular to the ground and perpendicular to the wall, respectively. The final force we’re going to consider is this force 𝐹. The ladder is being pulled at 𝐴 away from the wall. In fact, it’s on the brink of moving away from the wall. So, we know that this system is in limiting equilibrium.

Let’s begin by resolving the forces in a vertical direction. We know the system is in limiting equilibrium; it’s on the point of moving, but not quite. So, the sum of the forces in a vertical direction β€” I’ve called that 𝐹 sub 𝑦 β€” must be equal to zero. Let’s take upwards to be positive. And we know we have 𝑅 sub 𝐴 acting in this direction. We also have the friction at 𝐡 acting in this direction. Remember, friction is acting in the opposite direction to which the object wants to move. And then, we have the weight, 51 newtons, acting in the opposite direction. So, when we find the sum of these forces, it’s 𝑅 sub 𝐴 plus 𝐹 sub 𝐡 minus 51. And of course, that’s equal to zero.

But we also have a formula for friction. We know it’s πœ‡π‘…, that’s the coefficient of friction, times the reaction force at this point. We know the coefficient of friction at 𝐡 is one-quarter and the reaction force here is 𝑅 sub 𝐡. So, the friction here, 𝐹 sub 𝐡, is a quarter times 𝑅 sub 𝐡. And so, replacing the frictional force with a quarter 𝑅𝐡, and we find we have an equation linking 𝑅 sub 𝐴 and 𝑅 sub 𝐡. Let’s repeat this process, but this time, looking at the horizontal direction.

In this direction, we’ll assume that going to the right is positive. And once again, we know that the sum of these forces is equal to zero. We have the frictional force at 𝐴 acting in this direction, but the force that’s pulling the ladder away from the wall is 𝐹. This acts in the opposite direction. Similarly, 𝑅 sub 𝐡 acts in the opposite direction. So, the sum of our forces is 𝐹 sub 𝐴 minus 𝐹 minus 𝑅 sub 𝐡, and this is equal to zero. We can replace 𝐹 sub 𝐴, the frictional force at 𝐴, with πœ‡π‘…. And we know πœ‡ here, the coefficient of friction, is seven-ninths. So, we have two equations in 𝑅 sub 𝐴 and 𝑅 sub 𝐡. And our second equation is also in terms of 𝐹, so we’ll deal with that in a moment.

Our next job is to take moments about 𝐴. Now, in fact, we can take moments about any point on this ladder, but there’s usually more forces at the base of the ladder than anywhere else. So, it can be a sensible starting point. We’ll say that the anticlockwise or counterclockwise direction is positive. And then, we recall that we find the moment by finding the product of the force and the distance. Don’t forget that 𝑑, distance, is the perpendicular distance from the pivot, so here, 𝐴, to the line of action of the force. So, let’s begin by looking at the moment of the weight of the ladder.

This force 51 newtons is acting downwards. It’s not acting perpendicular to the ladder, and so we add a right-angled triangle. Our job is to find the component of this force that’s perpendicular to the ladder. The included angle here is 45 degrees, and let’s call the side that we’re looking to find π‘Ž newtons. π‘Ž is the adjacent side in the triangle where the hypotenuse is given. It’s 51 newtons. So, we can link these sides using the cosine ratio. cos of πœƒ is adjacent over hypotenuse. So, cos 45 is π‘Ž over 51. And if we multiply through by 51, we find that π‘Ž is equal to 51 times cos of 45. In fact, cos 45 is root two over two. So, the component of the weight that we’re interested in is 51 root two over two newtons.

We look at the moment. Now, this is trying to push the ladder in a clockwise direction. So, the moment is going to be negative. And we’re going to multiply this value 51 root two over two by the distance from 𝐴, which we define to be a half 𝐿. We’ll now look for some further forces. There are two forces at 𝐡. The first we’re going to look at is the reaction force at 𝐡. We’re going to find the component of this force, which is perpendicular to the ladder. So once again, we add a right-angled triangle.

This time, we’re looking for the opposite. I’ve labeled that 𝑏. And we have an expression for the hypotenuse. To link these sides, we use the sine ratio. So, sin 45 is 𝑏 over 𝑅 𝐡. We multiply through by 𝑅 sub 𝐡. And we find that 𝑏 is 𝑅 sub 𝐡 times sin of 45. And since sin 45 is root two over two, this is root two over two 𝑅 sub 𝐡. This force is trying to move the ladder in a counterclockwise direction, so its moment is positive. And so, we’re going to multiply root two over two 𝑅 𝐡 by the distance this point is away from 𝐴, which we define to be 𝐿.

There’s one more force we’re going to look at, and that’s the frictional force. Once again, we want the component of this force that’s perpendicular to the ladder, so we drop in another right-angled triangle. The included angle is 45 degrees. We have an expression for the hypotenuse, and we’re looking to find the adjacent. And so, we go back to the cosine ratio. cos 45 is 𝑐 over 𝐹 𝐡. And multiplying through by 𝐹 sub 𝐡, and we get 𝑐 is equal to 𝐹 sub 𝐡 times cos of 45. We can write this as root two over two 𝐹 sub 𝐡. And since this is trying to move the ladder once again in a counterclockwise direction, its moment is positive. It’s root two over two 𝐹 sub 𝐡 times 𝐿.

Since the ladder is in equilibrium, the sum of these moments is equal to zero. Notice we didn’t worry about any of the forces acting at 𝐴 and this is because each of these three forces is zero units away from 𝐴. So, if we were to find the moment, we’d be multiplying it by zero anyway, and it would be zero. So, what next? Well, firstly, we know that the length of the ladder 𝐿 cannot be equal to zero and that allows us to divide through by 𝐿. And so, our equation simplifies as shown.

What we’re now going to do is to look to replace 𝐹 sub 𝐡 with an expression for 𝑅 sub 𝐡. We go back to our equation for friction. And we see we can replace this with a quarter 𝑅 sub 𝐡, where quarter is the coefficient of friction here. We simplify our terms in 𝑅 sub 𝐡 and add 51 root two over four to both sides. Next, we divide through by root two. And now, we’re going to divide by five-eighths to find the value of 𝑅 sub 𝐡. 51 over four divided by five-eighths gives us 20.4 newtons. And we found the value of 𝑅 sub 𝐡.

We’re going to clear some space and use these two earlier equations to find 𝐹. We’ll substitute it into the equation we found by resolving vertically to find the value of 𝑅 sub 𝐴. We get 𝑅 sub 𝐴 plus a quarter times 20.4 minus 51 equals zero. That gives us 𝑅 sub 𝐴 minus 45.9 equals zero or 𝑅 sub 𝐴 equals 45.9 newtons. We’re now ready to replace 𝑅 sub 𝐴 and 𝑅 sub 𝐡 in our second equation, the one that we found by resolving horizontally. We get seven-ninths times 45.9 minus 𝐹 minus 20.4 equals zero. Seven-ninths times 45.9 minus 20.4 is 15.3. We add 𝐹 to both sides, and we’ve solved and found the value of 𝐹. 𝐹 is 15.3 newtons.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.