Equilibrium of Concurrent Forces

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Picture a ten kilogram mass hanging at rest from two strings tied to the ceiling. Taking g as ten, its weight is a hundred newtons. So the pulls of the two strings must add up to a hundred newtons, right? Not quite! Today we will meet two such strings whose tensions, added as plain numbers, come to much more. More than the weight they hold? How? Hold that thought. This idea, the equilibrium of concurrent forces, comes up often in JEE problems on strings. So, let's begin.

0:38Equilibrium: the forces add up to zero

So, imagine a few strings all pulling on one small knot. Every one of these forces acts through one point, the knot. Such forces are called concurrent forces. With me? Now, think. If this knot is to stay at rest, in equilibrium, what must all these forces add up to? Zero, yes, exactly! And we check it in components. This capital sigma, the Greek letter like a sideways M, means the sum of. F with a small x below is the x component of a force. So, the x components add up to zero, and so do the y components. Two equations, okay? And in three dimensions, there is a third equation: the z components also add up to zero. But remember the condition: all the forces act through one point. Got it? Now, exactly three forces in equilibrium. Draw them head to tail: the tail of each arrow at the head of the one before. They add up to zero. So, what shape must they make? Yes, a closed triangle! Correct. Total zero means the last arrow ends exactly where the first one started, so the triangle must close. Clear?

1:53Lami's theorem: which angle goes with which force

Here is the picture. On the left, three forces, P, Q and R, and each has a small arrow on top, because force is a vector. All three act through one point, this black dot, so they are concurrent. On the right, the same three drawn head to tail: P, then Q, then R, and we land back where we started. The closed triangle, just as we said. With me? And from this triangle comes a lovely result, Lami's theorem. Alpha, beta and gamma, these Greek letters, are the three angles around the black dot, and s i n is short for sine. P over sine alpha equals Q over sine beta equals R over sine gamma. So each force is proportional to the sine of one angle. Okay? Now, the key point: which angle goes with which force. Many students pair P with an angle right next to it, touching it. No, no, careful! It is the angle between the other two forces. For P, that is alpha, between Q and R; the line under the figure says the same, alpha opposite P, and so on. So beta goes with Q, and gamma goes with R, okay? And the condition: exactly three forces keeping a point in equilibrium. Right?

3:09Topper's trick: two strings at right angles

Now, a trick that toppers use. Take a mass m hanging from two strings that are at ninety degrees to each other; the small circle means degrees. Right angles: that is the condition for this trick, okay? Then each tension, capital T, is m g times the cosine of that string's angle with the vertical. Here m g is the weight: mass times g, the acceleration due to gravity, and cos means cosine. One line, no equations to solve. Nice, isn't it? Why does it work? Take your x and y axes along the two strings; they are perpendicular, so we may. Then along each string, its tension alone balances that string's part of the weight, which is m g times the cosine of the angle. We will use this as a check in example two. With me?

3:58Example 1: ratio P:Q:R by Lami's theorem

Alright, example one. Three forces, P, Q and R, keep a point in equilibrium. The angle between Q and R is one hundred and fifty degrees, between R and P it is one hundred and twenty, and between P and Q it is ninety. Find P is to Q is to R; the colons mean a ratio. Pause the video, and try it first. Done? Okay, let's check. Exactly three forces in equilibrium, so Lami's theorem applies. It says P is to Q is to R equals sine alpha is to sine beta is to sine gamma. Okay? Now, the ninety degree angle, between P and Q, has P as one of its arms. So alpha is ninety degrees, right? No, no, wait! That is the trap. Alpha is the angle between the other two, Q and R: one hundred and fifty degrees. Beta, between R and P, is one hundred and twenty, and gamma, between P and Q, is ninety. Clear? Now the sines. Sine one fifty is one half, sine one twenty is root three over two, and sine ninety is one. The tick-shaped sign is the square root. Got it? Multiply every term by two, and we get one is to root three is to two. That's it! And notice: R, the largest force, goes with the largest sine. Simple, right?

5:21Example 2: tensions in two strings holding 10 kg

Now example two, and our puzzle from the start! Here, a ten kilogram mass hangs from two strings that make thirty degrees and sixty degrees with the horizontal ceiling. On screen, k g is kilogram. Find the tensions in the two strings. Pause the video, and try it. Done? Okay, let's check. Taking g as ten, the weight m g is ten times ten, a hundred newtons; capital N is the newton. Call the tension in the string at thirty degrees T one, and in the string at sixty degrees T two; the one and two are small subscripts. Clear? Now, think. The mass does not move sideways at all. So, what must the horizontal pulls of the two strings do? Cancel each other, exactly! T one times cos thirty equals T two times cos sixty. Cos thirty is root three over two, and cos sixty is one half, so T two equals root three times T one. The double arrow means implies. Okay? Vertically, the two upward pulls hold up the weight: T one times sine thirty, plus T two times sine sixty, equals a hundred. Put in T two equals root three times T one: T one over two, plus three T one over two, equals a hundred. Right? The two terms add up to two T one, so T one is fifty newtons. Then T two is fifty root three, about eighty six point six newtons; the wavy equals sign means approximately. Got it? Now, our puzzle. Fifty plus eighty six point six is far more than a hundred! But tensions are vectors, and only their vertical parts hold up the weight. Fifty times sine thirty is twenty five, and fifty root three times sine sixty is seventy five. Twenty five plus seventy five: exactly a hundred! Puzzle solved! Now, let's check this with the topper's trick. The string for T one makes thirty degrees, so T one is a hundred times cos thirty, right? Oh no, careful! First, the condition: one eighty minus thirty minus sixty is ninety, so the strings are at right angles. But the trick needs the angle with the vertical, and that string is sixty degrees from the vertical. So T one is a hundred times cos sixty, fifty newtons, and T two is a hundred times cos thirty, fifty root three. Both match! Clear?

7:50Quick revision: equilibrium, Lami, strings at 90 degrees

So, quick revision. One: concurrent forces are in equilibrium when the x components and the y components each add up to zero. Two: three forces in equilibrium close a triangle, and by Lami's theorem, each force over the sine of the angle between the other two is the same. Three: for two strings at right angles, each tension is m g times the cosine of its angle with the vertical. Now go and try the practice problems on this page. All the best!