Equilibrium of Rigid Bodies

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Picture a painter leaning a ladder against a smooth, polished wall. The wall cannot grip the ladder at all. So the only thing that stops the ladder from sliding down is the grip of the floor. But how rough must that floor be? By the end of this video, you will calculate the exact answer. Today: the equilibrium of rigid bodies. Two simple conditions, and they handle see-saws, beams, hinges and ladders. This comes up often in JEE, so let's begin.

0:33Two conditions for equilibrium

So, take a rigid body: a body that does not bend or stretch. When is it in equilibrium? Condition one, on screen: sigma F ext equals zero. This big Greek letter, sigma, means add them all up. The arrow on F says force is a vector, and the small ext below means external, the forces from outside the body. So, all the external forces add up to zero. This is the translational condition: no net push to move the body as a whole. Okay? Now you tell me. If the net force is zero, is the body surely in equilibrium? No, no, not enough! Remember the couple from our last lesson? Two equal and opposite forces, along different lines. They add up to zero, and yet the body turns, like a steering wheel. So a zero net force alone does not give equilibrium. Right? That's why we need condition two: sigma tau ext equals zero. Tau, this Greek letter, stands for torque, and again the arrow means a vector. So, all the external torques add up to zero. This is the rotational condition. And look at the bracket: about any point. You choose the point! Hold on to that, because it is the secret behind today's trick. Clear? Now, in a plane, these two vector conditions become three ordinary equations. Sigma F x equals zero, and sigma F y equals zero, with small x and y below for the components along x and y. And sigma tau equals zero. Three equations. That is the whole toolkit. Got it?

2:12Topper's trick: choose the point wisely

Now look at this figure: a ladder against a smooth wall, standing on a rough floor. Theta is its angle with the floor, okay? W is the weight, acting at the middle of the uniform ladder. N w, that is N with a small w below, is the push of the wall. The wall is smooth, so it pushes straight out, with no friction. At the foot, the floor pushes up with N f, N with a small f below, and friction, small f, acts along the floor, towards the wall. The working written on the right is this same method in symbols; we will do it with real numbers soon. With me? So, three forces here are unknown: N w, N f and f. And we may take torques about any point we like. Which point would you choose? The foot of the ladder! Yes, exactly. Here is the topper's trick: take torques about the point that the most unknown forces pass through. Okay? For example, a hinge, a support, or the foot of a ladder. These are exactly the points where the unknown forces act. Why does it work? Any force through the point has zero moment arm, so its torque about that point is zero. It simply drops out of the torque equation. Right? So, for our ladder: N f and f both act at the foot, so they drop out. Only N w and W remain: one equation, one unknown. That's how clean it gets!

3:37Example 1: where the second child sits on a see-saw

Now, Example one, an easy one. One child, of thirty kilograms, sits two metres from the pivot of a see-saw. Where must a child of forty kilograms sit to balance it? Pause the video, and try it first. Done? Okay. First, a quick sense check. The heavier child: closer to the pivot than two metres, or farther? Closer! Correct. The heavier the weight, the shorter the moment arm it needs. Now, which point? The pivot, of course. The pivot's own force passes right through it, so that force drops out. Weight thirty g at two metres on one side, weight forty g at x metres on the other, where g is the acceleration due to gravity. For balance, the two torques must be equal. Okay? Now, g is on both sides, so it cancels. Thirty times two equals forty times x. So x is sixty over forty: one point five metres, on the other side of the pivot. Closer than two metres, just as we guessed. Clear?

4:44Example 2: ladder on a smooth wall, least friction

Now, our painter's ladder, with real numbers. Read it with me. The uniform ladder is five metres long and weighs two hundred newtons. It leans against a smooth wall, with its foot three metres from the wall, on a rough floor. Find the forces on the ladder, and the least coefficient of friction needed. Pause and try. Hint: take torques about the foot. Done? Okay, let's check. First, the geometry. The ladder is five metres long, and the foot is three metres out. So the top is at the square root, the sign you see on screen, of five squared minus three squared: four metres up the wall. Yes, our old friend, the three, four, five triangle. Okay? Now, torques about the foot. The weight acts at the middle, two and a half metres along the ladder. So its moment arm is two and a half metres, right? No, no, careful! This is the JEE trap. The moment arm is the perpendicular distance from the foot to the line of the force. The weight acts straight down, so we need the horizontal distance: half of three, one point five metres. And N w is horizontal, so its arm is the height, four metres. So, N w times four equals two hundred times one point five. With me? That gives N w equals three hundred over four: seventy five newtons. Now the force equations. Horizontally, there are only two forces: the wall's push and friction. So f equals N w, seventy five newtons. Vertically, the floor holds up the whole weight: N f equals W, two hundred newtons. Okay? Last, friction can be at most mu times N f. Mu is the Greek letter for the coefficient of friction, and the small min below means its least value. So, mu min equals f over N f: seventy five over two hundred, zero point three seven five. So the floor needs mu of at least zero point three seven five. That's our puzzle solved!

6:56Quick revision: net force and net torque both zero

So, quick revision. Equilibrium of a rigid body needs two things: the external forces add up to zero, and the external torques add up to zero. In a plane, that gives three equations. And take torques about the point where the most unknown forces act. Now go and try the practice problems on this page. You are ready for them. All the best!