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Video: Determining the Point at Which a Force Acts to Produce a Particular Torque

Parth Gharfalkar

A seesaw consists of a horizontal plank of wood 5 m long, with its mass equally distributed along its length. The fulcrum of the seesaw is positioned at the center of the plank. A child of mass 28 kg sits at the right-hand end of the seesaw, and a larger child with a mass of 35 kg sits at the left-hand end. The seesaw tips down on the larger child’s end, and the larger child shuffles towards the seesaw’s fulcrum until it starts to rise again. At what distance of the larger child from the seesaw’s fulcrum does the seesaw start to rise?

08:35

Video Transcript

A seesaw consists of a horizontal plank of wood five meters long with its mass equally distributed along its length. The fulcrum of the seesaw is positioned at the center of the plank. A child of mass 28 kilograms sits at the right-hand end of the seesaw and a larger child with a mass of 35 kilograms sits at the left-hand end. The seesaw tips down on the larger child’s end, and the larger child shuffles towards the seesaw’s fulcrum until it starts to rise again. At what distance of the larger child from the seesaw’s fulcrum does the seesaw start to rise?

Okay, so this is a very long and complicated question so let’s go through a bit by bit and breakdown all the information that we’ve been given. So the first thing that we’ve been told is that we’ve got a seesaw which consists of a horizontal plank of wood, which is five meters long. So here’s our five meters long horizontal plank of wood. Now we’ve also been told that in this plank of wood the mass is equally distributed along its length.

Well if the mass is evenly distributed along its length, then the center of mass must be at the center of the plank. In other words, the weight of the plank will act from its center. The reason for this is that we have a uniform mass distribution, because we’ve been told that the mass of the plank is equally distributed along its length. So it’s not like there’s more mass over here and less mess here or vice versa or anything like that.

If that was the case, then that would result in the center of mass shifting towards the region where there was more mass in the plank. However this is not the case, we’ve been told that the mass is equally distributed along its length. So the center of mass is going to be at the center of the plank. Now we’ve been told that the fulcrum of the seesaw is positioned at the center of the plank. So of course the seesaw consists of a fulcrum and a plank, and the fulcrum in this case is positioned at the center.

So this triangle represents the fulcrum. Now coincidentally, this is positioned just below the center of mass of the plank. So if we would just to have the seesaw by itself without worrying about the kids sitting on either end, then because the weight is acting through the center of mass and the center of mass is right above the fulcrum, the weight would be acting through the fulcrum as well. Now because the weight of the plank acts at the same point along the length of the plank as the fulcrum, this means that there isn’t a rotational force on the seesaw.

In other words, it’s perfectly balanced at this point. But all of this is about to change, because we’ve been told that firstly a child of mass 28 kilograms sits at the right-hand end of the seesaw. So we’ve got a child here who has a mass of 28 kilograms sitting on the right-hand end of the seesaw. And we’ve also been told that a larger child with a mass of 35 kilograms sits at the left-hand end. So here’s our child with a mass of 35 kilograms.

Now naturally, both of these kids are going to be exerting a force on the plank. That force is their weight. Now the child at the right-hand end is going to exert a force, which we’ll call 𝑊 sub 𝑟 for the weight of the child on the right-hand end. And the weight of the child on the left-hand end we’ll call 𝑊 sub 𝑙. Now what are the weights of the children on the right- and left-hand end? Well to find this out, we can recall that the weight of any object 𝑊 is given by multiplying the mass of the object, 𝑚, by the gravitational field strength of the Earth, 𝑔.

Now 𝑔 of course is just a constant: 9.8 meters per second squared. And in the case of both the children, we’ve been given their masses. So we can work out the weight of each of these kids. Let’s start with the weight of the right-hand side child, 𝑊 sub 𝑟. We say that the weight is equal to the mass of the child, which is 28 kilograms, multiplied by the gravitational field strength of the Earth 𝑔, but we won’t put 9.8 meters per second squared. Instead, we’ll just say say this is 28 times 𝑔.

This is because it keeps it simpler. We don’t have to actually do any calculations at this point. And also as we’ll see later, 𝑔 will cancel out in our calculations. Okay, so similarly, we can say that the weight of the left hand side child, 𝑊 sub 𝑙, is equal to the mass of the child, which is 35 kilograms, multiplied by 𝑔. So at this point, we can clearly see that the weight of the left-hand side child is larger than the weight of the right-hand side child.

Therefore, the force exerted by the left-hand side child trying to turn the plank counterclockwise is larger than the force exerted by the right-hand side child trying to turn the plank clockwise. So the plank is going to tip down on the left-hand side. In other words, this is what our seesaw looks like now. And we’ve been told that this happens in the question. We’ve been told that the seesaw tips down on the larger child’s end.

However, to counteract this, we’re told that the larger child shuffles towards the seesaw’s fulcrum. We’re told that the larger child, which is the one on the left, starts moving towards the fulcrum of the seesaw. So the child is no longer here, but is rather here at some position closer to the fulcrum. What we’ve been asked to do is to find at what distance of the larger child from the seesaw’s fulcrum, in other words at what distance between the child and the fulcrum, does the seesaw start to rise.

In other words, at what point does the seesaw start to level out again? So we need to find out the value of this question mark here when the seesaw starts to level out again. Now of course the weight of the child on the left still is the same; it’s 𝑊 sub 𝑙. And the weight of the child on the right is still the same, 𝑊 sub 𝑟. However, what we need to consider is the torque exerted by each child. We can recall that torque is the force applied on an object multiplied by the perpendicular distance between the point at which the force is applied and the fulcrum.

So we can work out the torque or the turning force applied by the right-hand side child, which we’ll called 𝑇 sub 𝑟 for torque sub right hand side, and we can work out the torque applied by the left hand side child, which we’ll call 𝑇 sub 𝑙. Of course, as we’ve said earlier, the right-hand side child is trying to turn the plank clockwise and the left-hand side child is trying to turn it counterclockwise.

It’s when these two torques 𝑇 sub 𝑟 and 𝑇 sub 𝑙 are equal that the seesaw will be perfectly balanced again. So let’s first clear some space by writing 𝑊 sub 𝑙 over here. And now let’s try and work out the value of 𝑇 sub 𝑟 first. Well 𝑇 sub 𝑟 is equal to the force applied by the right-hand side child on the seesaw, which of course is the weight of the child so that’s 28𝑔, multiplied by the perpendicular distance between where the force is applied and the fulcrum. In other words, that’s this distance here, because the force is applied at this point and the fulcrum is at this point.

Now that distance is naturally going to be half the length of the plank, because remember we’ve been told that the fulcrum is positioned at the center of the plank. So this distance that we’re looking for is five meters divided by two cause it’s half the length of the plank. And five meters divided by two is 2.5 meters. So coming back to our torque calculation, we’ve got the force applied by the child on the right multiplied by the distance, which is 2.5 meters.

Now we won’t evaluate this any further. We’ll leave it as it is for now. Let’s try and work out the value of 𝑇 sub 𝑙 instead. Well 𝑇 sub 𝑙 is the force applied by the child on the left-hand side, which we know to be the weight of the child which is 35𝑔, times the distance that we’re trying to find out. Let’s call that distance 𝑥 as we’ll label it on the diagram. We don’t know what 𝑥 is so in our torque calculation we just say 35𝑔 times 𝑥, at which point we have an expression for 𝑇 sub 𝑙 now.

Okay, so as we said earlier, 𝑇 sub 𝑟 and 𝑇 sub 𝑙 have to balance — in other words, their sizes or magnitudes have to be equal — in order for the seesaw to start rising again so that it balances out. So let’s clear some more space and set 𝑇 sub 𝑟 to be equal to 𝑇 sub 𝑙. In other words, 𝑇 sub 𝑟, which is 28𝑔 times 2.5, is equal to 𝑇 sub 𝑙, which is 35𝑔 times 𝑥. Now at this point, the first thing we can do is to divide both sides of the equation by 𝑔. This way the 𝑔s on the left cancel out and the 𝑔s on the right-hand side cancel out. And so as we’ve said earlier, the 𝑔s in our weight calculation are going to cancel out.

Now what this leaves us with is 28 times 2.5 is equal to 35 times 𝑥. Now if we divide both sides of the equation by 35, then the 35 on the right-hand side cancel. And so we’re left with 28 times 2.5 in the numerator divided by 35 in the denominator is equal to 𝑥. So all that’s left for us to do now is to evaluate the fraction on the left-hand side. When we do this, we find that 𝑥 is equal to two meters, at which point we found our final answer. We’ve just found out that the child on the left-hand end of the seesaw has to be two meters away from the fulcrum for the seesaw to start to rise again.