The Metre Scale and the Principle of Moments

Transcript

0:00Introduction

Hello everyone, here is a small puzzle before we start. You have a uniform metre scale, a knife edge, some thread, and a single mass of sixty grams, with no weighing machine anywhere. Now find the mass of the scale itself. With just one known mass, that sounds impossible, right? But by the end of this video, you will do it yourself. This experiment belongs to experimental skills in JEE, and questions on it do come up. So, let's begin.

0:30The principle of moments

Take a body that is free to turn about an axis, and apply a force to it. Let me ask you this: what decides how strongly that force turns the body? Is the size of the force enough? The size alone is not enough, because the distance from the axis matters too. We call this turning effect the moment. The moment equals the force times the perpendicular distance from the axis, okay? Now, some forces turn the body clockwise, and some turn it anticlockwise. The body is in equilibrium when the sum of the clockwise moments equals the sum of the anticlockwise moments. That line is the principle of moments, and this whole experiment stands on it. Clear?

1:11First, balance the scale alone

Now look at the arrangement on screen. This long strip is the metre scale, with its marks from zero to one hundred centimetres. It rests on a knife edge, the grey triangle, and the label below says the edge is at the centre of gravity. Ignore the two hanging blocks and the arrows for now, we will come to them. Before hanging any mass, we first balance the scale alone on this edge, okay? The point where it balances is its centre of gravity, written capital G, near the fifty centimetre mark. It is only near that mark, so do not assume exactly fifty. Right? So tell me, why do we want the knife edge right under this point? What does it do for our calculation? We want it because the weight of the scale then acts at zero distance from the edge. So the weight has no moment about the edge, and it drops out of the calculation. With me?

2:06Now find the unknown mass

Now we hang the masses by threads. The known mass, small m, hangs on one side, at distance d one. The unknown mass, capital M, hangs on the other, at distance d two. The small one and two written below the d only label these two distances, the arrows in the figure. Then we adjust until the scale is horizontal, okay? Now use the principle. The weight of the known mass is small m times small g, with small g the acceleration due to gravity, not the capital G of the centre of gravity, and it turns the scale anticlockwise. So its moment is small m times small g times d one. The unknown mass turns the scale clockwise, with moment capital M times small g times d two. These two are equal, as the line below the figure says, right? The small g is on both sides, so it cancels. That leaves capital M equals small m times d one, over d two. Write this down, because this is the result. Got it? Now here is a trap. Look at the figure, where the thread of small m hangs at the twenty centimetre mark. So is d one simply twenty centimetres? No, it is not, and this is where students slip. The knife edge is the axis, so the distances are measured from the knife edge to the points where the threads hang. Twenty is only a mark on the scale, and d one runs from that mark to the knife edge, as the arrow shows. Clear?

3:36What if the pivot is away from G?

Now let us change one thing. Suppose the scale is pivoted not at G, but at some other point. Let me ask you this: does the weight of the scale still drop out of the calculation? No, it does not, because the weight of the scale still acts at G, wherever the pivot is, okay? And G is no longer at zero distance from the edge. So the weight now has a moment about the edge, which must be included in the balance. Forgetting it is a common slip, so watch out for it. Right? But see the good side: the balance can now tell us that weight. So this arrangement gives the mass of the scale itself, and that is the key to our puzzle.

4:16Two reasoning questions on this method

Before the examples, here are two quick reasoning questions. First, for the best accuracy, should the masses hang near the knife edge, or far from it? They should hang far from it. The reading error of the scale is fixed, so when the distance is large, that error is a smaller fraction of each distance. Clear? Now the second question is this: what is the advantage of moments over a spring balance for finding a mass? The advantage is that this method compares the unknown mass with a standard mass, okay? So the result does not depend on the local value of small g, which cancelled in our formula. And it does not depend on the calibration of a spring, because there is no spring. Got it?

5:02Example 1: unknown mass on a scale balanced at 50 cm

Now the examples, the easier one first. We have a metre scale balanced at its fifty centimetre mark. Here, a fifty gram mass hangs at the twenty centimetre mark, and an unknown mass at the eighty centimetre mark keeps the scale horizontal. Find the unknown mass. Pause the video and try it first. Done? Okay, let's check. Remember the trap, because twenty is only a mark. The knife edge is at fifty, so d one is fifty minus twenty, which gives thirty centimetres, okay? Now you tell me, what is d two for the unknown mass, hanging at the eighty centimetre mark? Yes, exactly, it is eighty minus fifty, which gives thirty centimetres again. Right? Now balance the moments. Fifty times thirty equals capital M times thirty, so the thirty cancels and capital M is fifty grams. That makes sense, because equal distances need equal masses. Clear?

6:03Example 2: mass of a scale that balances at the 40 cm mark

Now the harder one, and this is our puzzle from the start. Here, a uniform metre scale balances at the forty centimetre mark, when a sixty gram mass is hung at the ten centimetre mark. Find the mass of the scale. Pause the video and try it first. Done? Okay, let's check. The scale is pivoted at forty, away from its centre of gravity, so its weight cannot drop out. So where does the weight of this scale act? It acts at the fifty centimetre mark, because the scale is uniform, so its centre of gravity is at the middle. We call this weight capital W, okay? Now the distances, from the knife edge as always. The sixty gram mass hangs at ten, and the edge is at forty, so forty minus ten gives thirty centimetres. Right? And how far is capital W from the knife edge? Be careful with this one. It is not fifty, because fifty is only the mark. Fifty minus forty gives ten centimetres. With me? Now the sixty gram mass, on the left of the edge, turns the scale anticlockwise, and the weight, on the right, turns it clockwise. The small g cancels as before, so we can work in grams. Sixty times thirty equals capital W times ten, okay? Sixty times thirty is eighteen hundred, and eighteen hundred over ten gives one hundred and eighty. So the mass of the scale is one hundred and eighty grams. That is our puzzle solved, with a single mass of sixty grams and no weighing machine. If you got it on your own, very well done!

7:41Quick revision: moments, the pivot and measuring distances

So, quick revision. In equilibrium, the sum of the clockwise moments equals the sum of the anticlockwise moments. With the edge under G, the weight of the scale drops out, and capital M equals small m times d one, over d two. With the edge anywhere else, include the moment of the weight, which acts at G, in the balance. And measure every distance from the knife edge. Now go and try the practice problems on this page. All the best!