### Video Transcript

A uniform ladder with length ๐ฟ and weight 35 kilogram-weight is resting with its upper end against a rough vertical wall and lower end on a rough horizontal floor. The floor and wall both have the same coefficient of friction of one-half with the ladder. If the ladder is inclined to the horizontal at an angle whose tangent is three-quarters, find the maximum distance in terms of ๐ฟ that a man weighing 78 kilogram-weight can ascend the ladder without it slipping.

Before we do anything, letโs begin by drawing a diagram. Hereโs a ladder inclined at an angle whose tangent is three-quarters to the horizontal. If we call that angle ๐ผ, then tan ๐ผ is equal to three-quarters. Now, what weโre not going to do is solve this equation for ๐ผ by finding the inverse tan of it. Instead, in a little while, weโll construct a right-angled triangle that will help us find exact values for sin ๐ผ and cos ๐ผ. For now, though, letโs continue with our diagram.

The ladder is uniform with length ๐ฟ and a weight 35 kilogram-weight. The fact that itโs uniform means that its mass is evenly distributed throughout the ladder. And we can say that the force of weight must act exactly halfway along the ladder. And so we have a force of 35 kilogram-weight acting downward half ๐ฟ along the ladder. If we say the base of the ladder is ๐ด and the top of the ladder is ๐ต, we know we have reaction forces at these points.

These reaction forces are perpendicular to the surface that the ladder is resting on, as shown. We also know though that both the wall and the floor are rough. And that means we need to consider a frictional force. The frictional force will act against the direction in which the ladder is trying to move. So if someone stands on the ladder, it wants to slide downwards. And so, at point ๐ด, the frictional force will act towards the wall. At point ๐ต, the frictional force will act upwards away from the floor.

We know that friction is equal to ๐๐
, where ๐ is the coefficient of friction. And weโre told that the coefficient of friction of both the floor and the wall with the ladder is one-half. So the frictional force at point ๐ด must be one-half times the reaction force there. So thatโs one-half times ๐
sub ๐ด. Similarly, the frictional force at ๐ต must be a half ๐
sub ๐ต. We want to find how far up the ladder the man can get before it slips. So we donโt know exactly how far up the ladder he can get, so weโll call that ๐ฅ๐ฟ, where ๐ฅ will be a fraction of the total distance. His downward force that he exerts on the ladder is 78, as shown.

So weโre going to do two things. Firstly, weโre going to begin by resolving forces horizontally and vertically. After weโve done that, weโll be able to calculate moments about a point on the ladder. Letโs begin by resolving forces in a vertical direction. We know that the sum of these forces is equal to zero. The ladder isnโt slipping, but it will be on the point of slipping. So itโs in equilibrium. If we take upwards to be positive, we see we have ๐
sub ๐ด acting upwards. And then we have the frictional force at ๐ต half ๐
sub ๐ต also acting in that direction.

In the opposite direction, we have the weight of the ladder and the weight of the man. So weโre going to subtract both of those forces. And we know the sum of all of these forces is equal to zero. So ๐
sub ๐ด plus a half ๐
sub ๐ต minus 35 minus 78 is equal to zero. Negative 35 minus 78 is negative 113, so we can simplify this equation a little. Next, weโre going to see if we can create another equation linking ๐
sub ๐ด and ๐
sub ๐ต by resolving horizontally.

Letโs call the forces in this direction ๐น sub ๐ฅ. Then, the sum of these forces is once again equal to zero. If we take the direction right to be positive, we know we have a half ๐
sub ๐ด acting in this direction and ๐
sub ๐ต acting in the opposite direction. The sum of these forces is zero, so a half ๐
sub ๐ด minus ๐
sub ๐ต is equal to zero. And we can rearrange to say that a half ๐
sub ๐ด is equal to ๐
sub ๐ต.

Weโre going to multiply through by two to find an expression for ๐
sub ๐ด in terms of ๐
sub ๐ต. We get ๐
sub ๐ด is two times ๐
sub ๐ต. And this is really useful because we can replace ๐
sub ๐ด with two ๐
sub ๐ต in our earlier equation. Simplifying this becomes five over two ๐
sub ๐ต minus 113 equals zero. And by adding 113 to both sides, we find five over two ๐
sub ๐ต equals 113. Weโll solve this equation for ๐
๐ต by dividing through by five over two or 2.5. And we find ๐
๐ต to be equal to 45.2 or 45.2 kilogram-weight.

Letโs replace all instances of ๐
sub ๐ต in our diagram with this value. The normal reaction force of the wall on the ladder then is 45.2, and then a half of this is 22.6. So the frictional force acting at the wall is 22.6 kilogram-weight. In fact, if we go back to this equation, we can also find the value of ๐
๐ด. Itโs two times ๐
sub ๐ต, so two times 45.2, which is 90.4 kilogram-weight. Once again, weโll replace ๐
sub ๐ด in our diagram. The normal reaction force of the floor on the ladder then is 90.4 and the frictional force is half of this; itโs 45.2.

Now weโve replaced all of our forces with numbers, weโre going to clear some space and take moments. Now we can take moments about any point on the ladder if we want to. However, weโve measured our distances from point ๐ด. And also generally, thereโs more going on at the foot of the ladder, so thatโs where we tend to take moments about. Letโs say counterclockwise is positive. And then we use the formula moment is equal to force times distance, where the distance is the perpendicular distance from the pivot to the line of action of the force. So letโs begin by considering the moment of the weight of the ladder. Remember, weโre interested in the component of this force, which is perpendicular to the ladder. So we add a right-angle triangle as shown.

The included angle is ๐ผ. And we want to work out this side. Letโs call that ๐ฅ. Its units will be kilogram-weight. This is the adjacent side in our triangle, and we know the hypotenuse is 35 kilogram-weight. Since cos ๐ is adjacent over hypotenuse, here, we can say that cos of ๐ผ must be ๐ฅ over 35 or ๐ฅ is equal to 35 times cos ๐ผ. Now we can use the fact that tan of ๐ผ is equal to three-quarters to find an exact value for cos of ๐ผ. By drawing a right-angled triangle and recognizing that tan of ๐ is opposite over adjacent, we can label the opposite side in our triangle as three units and the adjacent as four.

The Pythagorean theorem or the fact that we recognize this as a Pythagorean triple tells us that the hypotenuse is five units. We can therefore say that cos of ๐ผ, which is adjacent over hypotenuse, must be four-fifths. Similarly, we can form an expression for sin of ๐ผ. Itโs opposite over hypotenuse, which is three-fifths. This means that ๐ฅ can be rewritten as 35 times four-fifths, which is 28. The component of the weight of the ladder that is perpendicular to the ladder is therefore 28 kilogram-weight.

We know that the moment is force times distance. But this force is trying to push the ladder in a clockwise direction, so its moment is going to be negative. Itโs negative 28 multiplied by a half ๐ฟ. Letโs repeat this process with the component of the weight of the man thatโs perpendicular to the ladder. This time, weโll call that ๐ฆ. Once again, weโre trying to find the adjacent, and we know the hypotenuse. But the hypotenuse here is 78, so we get cos ๐ผ is equal to ๐ฆ over 78 or ๐ฆ is 78 times cos ๐ผ. cos ๐ผ, of course, is four-fifths, so we get 78 times four-fifths. And the component of the weight of the man thatโs perpendicular to the ladder is therefore 62.4 kilogram-weight. This moment is once again negative, so itโs negative 62.4 times ๐ฅ๐ฟ.

There are two more moments we need to consider, and those are the moments of the forces at ๐ต. Letโs consider the component of the reaction force at ๐ต thatโs perpendicular to the ladder. Iโve called that ๐ and drawn another right-angled triangle. The included angle is still ๐ผ. We want to find the opposite. And this time, the hypotenuse is 45.2. So we can say sin ๐ผ is ๐ over 45.2 or ๐ is equal to 45.2 times sin ๐ผ. Now, remember, we calculated that sin ๐ผ was equal to three-fifths. And so the component of this reaction force that acts perpendicular to the ladder is 27.12 kilogram-weight. Now this force is trying to turn the ladder in a counterclockwise direction. So its moment is positive. Itโs ๐ฟ times 27.12.

Now that weโve considered that force, we clear some space to look at the component of the friction thatโs perpendicular to the ladder. We once again have an included angle of ๐ผ and weโre trying to find the side labeled ๐. This is the adjacent side in our triangle, and we know the hypotenuse. So the cosine ratio says cos ๐ผ is ๐ over 22.6, and so ๐ is 22.6 times cos ๐ผ or 22.6 times four-fifths. Thatโs 18.08 kilogram-weight. And so the moment of this force is ๐ฟ times 18.08. Now, the ladder is in equilibrium, so the sum of these moments must be equal to zero. And then we notice before we solve this equation for ๐ฅ, we can actually divide through by ๐ฟ.

Now the reason weโre allowed to divide through by an unknown is because we know the length of the ladder cannot be equal to zero. Negative 28 times one-half is negative 14. So our equation becomes negative 14 minus 62.4๐ฅ plus 27.12 plus 18.08 equals zero. This simplifies to negative 62.4๐ฅ plus 31.2 equals zero. And if we add 62.4๐ฅ to both sides, we get 62.4๐ฅ equals 31.2. We then divide through by 62.4. Now, 31.2 divided by 62.4 is one-half. So we find that ๐ฅ is equal to one-half.

This means the man can get a halfway up the ladder before it slips, so we can get a half ๐ฟ distance along the ladder without it slipping. The answer is a half ๐ฟ.