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A uniform ladder 𝐴𝐵 having a length 𝐿 and weighing 40 kilogram weight is resting with one of its ends on a smooth floor and the other against a smooth vertical wall.
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The ladder makes an angle of 45 degrees with the horizontal, and its lower end 𝐴 is attached to a string that is fixed to a point at the junction of the wall and floor.
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Given that the maximum tension the string can withstand is 60 kilogram weight, find how far up the ladder a man of weight 140 kilogram weight can go before the string breaks.
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There’s quite a lot of information here.
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So what we’re going to begin with doing is sketching out the scenario.
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Here is our uniform ladder, resting against a smooth vertical wall and a smooth floor.
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Now, the fact that it’s uniform means that its weight is evenly distributed across the ladder.
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We can therefore say that the downward force of the weight must act exactly halfway along the ladder.
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So it acts half 𝐿 away from 𝐴.
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We don’t have any units for 𝐿, so that’s absolutely fine.
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We see that it makes an angle of 45 degrees with the horizontal but also that the lower end 𝐴 is attached to a string.
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Now, that string is fixed to a point at the junction of the wall and floor, here.
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And so there must be tension acting at the point 𝐴.
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Essentially, it’s the tension in the string that keeps the ladder in place.
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Now, the maximum tension the string can withstand is 60 kilogram weight.
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So we’re actually going to use tension equals 60 kilogram weight in this question.
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And then we have this man that weighs 140 kilogram weight going up the ladder.
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Now, we don’t know exactly how far up the ladder he goes.
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He might make it less than halfway; he might make it more than halfway.
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We’ll add this to our diagram and say he makes it 𝑥𝐿 up the ladder.
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𝑥 will be a fraction.
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We’re working out what fraction of 𝐿 he will be able to get up the ladder.
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What other forces have we got, though?
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Well, there must be a reaction force of the floor on the ladder and the wall on the ladder.
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These forces are perpendicular to the surface.
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So we have a reaction force at 𝐴 acting directly up and a reaction force at 𝐵 acting to the left.
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There are no other forces.
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Remember, the wall and the floor are smooth, so there’s no frictional force.
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So what do we do next?
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Our next job is to resolve forces both horizontally and vertically.
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Once we’ve done that, we’ll be able to consider the moments acting about a point.
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Let’s begin by resolving our forces vertically.
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The string is on the point of breaking when its tension hits 60 kilogram weight.
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So we’re assuming that it’s still in equilibrium or a bit limiting equilibrium.
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It’s on the point of breaking.
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So in a vertical direction, we can say that the sum of forces must be equal to zero.
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That’s the sum of 𝐹 sub 𝑦 is equal to zero.
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So what forces have we got acting in a vertical direction?
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Well, we’ve got the reaction force at 𝐴 acting upwards.
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So let’s take upwards to be positive.
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Then we’ve got the weight of the ladder acting in the opposite direction.
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So we’re going to add negative 40.
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And we’ve got the weight of the man acting in the opposite direction to the reaction force.
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So that’s a force of negative 140.
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So we say that the sum of the forces acting in the vertical direction is 𝑅 sub 𝐴 minus 40 minus 140.
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And this must be equal to zero.
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This simplifies to 𝑅 sub 𝐴 minus 180 equals zero.
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So 𝑅 sub 𝐴, if we add 180 to both sides, we see must be equal to 180 or 180 kilogram weight.
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Now, don’t worry if you’re used to measuring forces in newtons kilogram weight; it’s just another way of doing so.
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Next, we’re going to resolve forces in a horizontal direction.
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Once again, the sum of these forces will be equal to zero.
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This time, let’s take the direction to the right to be positive.
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And we have tension acting in that direction.
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Then we have the reaction force at 𝐵 acting in the opposite direction.
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So tension minus 𝑅 sub 𝐵 must be equal to zero.
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And if we add 𝑅 sub 𝐵 to both sides, we find tension must be equal to 𝑅 sub 𝐵.
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But remember, we said that we’re taking the case where tension is equal to 60 kilogram weights, since that’s the maximum tension it can take before it breaks.
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So 𝑅 sub 𝐵 must be equal to 60 or 60 kilogram weight.
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We add this force to our diagram, and we’re ready to start taking moments.
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Remember, the moment is the turning effect of a force.
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We’re going to take moments about the point 𝐴.
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Now, we could take moments about any point on this ladder.
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But generally there’s more going on at the point where the ladder hits the floor.
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So if we take moments about 𝐴, we’d basically have less to calculate.
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We’re going to assume that a counterclockwise direction is positive.
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And then we recall that we calculate the moment of a force by multiplying that force by the perpendicular distance of the line of action of the force from the point where we’re calculating.
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We’ll see what that looks like in a moment.
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So let’s consider all of the forces acting on our ladder.
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Let’s begin by looking at the reaction force at 𝐵.
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We calculated that to be 60 kilogram weight.
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We need to work out the component of this force that is perpendicular to the ladder.
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This is acting in a counterclockwise direction.
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So its moment is going to be positive.
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And so let’s enlarge this triangle a little bit.
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We should see that it’s a right-angled triangle with a hypotenuse of 60 kilogram weight.
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The included angle is 45 degrees.
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And this is because the reaction force at 𝐵 is parallel to the floor.
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And we know alternate angles are equal.
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We want to work out the component of this force that acts perpendicular to the ladder.
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So let’s call that 𝑎 or 𝑎 kilogram weight.
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This is the opposite side in our triangle.
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And we know the hypotenuse is 60.
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So we can link these using the sine ratio.
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sin 𝜃 is opposite over hypotenuse.
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So sin of 45 is 𝑎 over 60.
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By multiplying through by 60, we see that 𝑎 is equal to 60 sin 45.
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But in fact, we know that sin of 45 degrees is root two over two.
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So 𝑎 is 60 times root two over two or 30 root two kilogram weight.
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So we found the component of this reaction force that acts perpendicular to the ladder.
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We need to calculate its moment.
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Now remember, it’s acting in a counterclockwise direction, so that’s positive.
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The moment is this force multiplied by the distance away from 𝐴.
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So that’s 30 root two multiplied by 𝐿.
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And what about our other forces?
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Let’s look at the force which is the weight of the man.
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This time, the included angle is 45 degrees.
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And the hypotenuse is 140 kilogram weights.
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We’ll call the side that we’re looking to calculate, which is the component of this force which acts perpendicular to the ladder, 𝑏.
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Now, this time, we’re looking to find the adjacent and we know the hypotenuse.
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And so we use the cosine ratio.
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cos of 45 degrees is 𝑏 over 140.
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So 𝑏 is 140 times cos of 45.
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But once again, cos of 45 is root two over two.
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So we find that 𝑏 is 140 times root two over two or 70 root two kilogram weights.
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Let’s now calculate the moment of this force.
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We’re going to subtract it since it’s acting in a clockwise direction.
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We said that this was 𝑥𝐿 away from 𝐴.
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So the moment is negative 70 root two times 𝑥𝐿.
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There’s one more force to consider, and that’s the force of the weight of the ladder.
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If we add a right-angled triangle to this force, it looks a lot like our previous right-angled triangle.
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This time, though, the hypotenuse is 40.
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And we’ll call the length we’re looking to find 𝑐.
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cos of 45 is 𝑐 over 40.
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And so if we rearrange, we get 𝑐 equals 20 root two or 20 root two kilogram weights.
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And so we’re ready to find the moment.
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Once again, it’s acting in a clockwise direction.
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So it’s negative and its distance is a half 𝐿.
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So we have 20 root two times a half 𝐿.
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We know that this is an equilibrium.
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So the sum of these moments is zero.
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And now, we’re going to look to solve for 𝑥.
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But there’s an extra variable here; there’s 𝐿.
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Luckily, we know that 𝐿, the length of the ladder, cannot be equal to zero.
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So we can divide through by 𝐿.
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And our equation becomes 30 root two minus 70 root two times 𝑥 minus 20 root two over two equals zero.
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But of course, 20 root two over two is 10 root two.
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So this equation simplifies further to 20 root two minus 70 root two 𝑥 equals zero.
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In fact, we can also divide through by root two.
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And then adding 70𝑥 to both sides, our equation simplifies even further to 20 equals 70𝑥.
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To solve for 𝑥, we divide through by 70.
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So 𝑥 is 20 over 70, or 𝑥 is equal to two-sevenths.
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So we’ve said that the man can get two-sevenths of the way up the ladder.
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Since the length of the ladder is 𝐿, we say that the man can get two-sevenths 𝐿, length units, up the ladder before the string breaks.
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Now, notice that we actually didn’t need to calculate the reaction force at 𝐴.
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This is because when we take moments about 𝐴, we’re multiplying each of the forces at this point by zero.
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So they’re actually zero.
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It’s often required, though, to resolve these forces in a vertical direction.
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So it’s always sensible to do this for completion.