Free-body Diagrams

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Picture a man standing on a platform. There is a rope tied to the platform. It goes up over a pulley on the ceiling, and back down to his hands. He pulls, and keeps himself and the platform hanging at rest. So, how hard must he pull? The full weight of both, right? Hmm, we will see. The tool that cracks this is the free-body diagram. Laws of motion problems in JEE usually start with one. So, let's begin.

0:34Steps 1 and 2: isolate the body, list its forces

So, what is a free-body diagram? It is a picture of one body, alone, with every force that acts on it. Step one: isolate. Pick one body and draw it by itself, or pick a group of bodies that move together, and treat them as one. Okay? Now tell me: once the body is alone, which force is always there, on every body near the Earth? Its weight, yes, exactly! Step two: draw every force on the body, and start with its weight, m g, mass times g, the acceleration due to gravity, acting vertically down. Right? Next, the contacts. For each surface it touches, one normal force, capital N, the push of the surface, perpendicular to it. If the surface is rough, friction too, small f, along the surface. And for each string attached, one tension, capital T, pulling along the string, away from the body. Clear? Finally, any spring forces, and any applied force, like a push from your hand. That is the full list. With me? Now a small test. Here, a block rests on a floor and presses down on it. Should that push go into the block's diagram? No, no, careful! That push acts on the floor, not on the block. Never draw a force the body exerts on others; draw only the forces acting on it. This is a very common mistake.

2:00Step 3: choose axes along the acceleration

Okay? Step three: choose your axes. Usually one along the acceleration, and one perpendicular to it, across it. Then the acceleration appears in only one equation. Okay? Now look at the figure. Here, a block sits on an incline at an angle theta, the Greek letter near the bottom corner. Its weight m g points straight down, the normal force N in blue is perpendicular to the slope, and friction f in red acts up the slope, only if it is rough. So, which way can this block accelerate? Along the slope, yes! While it stays on the surface, it can only slide along it. So, axes along and across the incline. But m g is along neither axis, so we split it. Along the slope, the dashed arrow: m g sine theta, down the slope. Right? And across the slope, the other dashed arrow: m g cos theta, cos for cosine, pointing into the surface. Clear? Now the normal force. Many students write N equals m g, every single time. Is that right here? No, no, careful! Across the slope there is no acceleration, as long as the block stays on the surface. So N balances only the across part: N equals m g cos theta, less than m g. It is written on the figure too. Got it?

3:23Step 4: Newton's second law on each axis

Step four: Newton's second law along each axis. The big Greek letter sigma means sum. Sum of F x, the forces along x, equals m times a x; the small x below means the x part. The same for y, and for each body separately. Okay? On our incline, along the slope, taking down the slope as positive: m a equals m g sine theta, minus f. Here friction acts up the slope, as in the figure, opposing the sliding, so it comes with a minus sign. Right? Now, with two bodies joined by a string, you get more unknowns: two accelerations, a tension, maybe more. So what extra fact closes the equations? Yes, a constraint! For example, a light string has the same tension all along it, and bodies moving together have the same acceleration. Clear?

4:17Example 1: block on a smooth 30 degree incline

Alright, example one. Here, a two kilogram block slides down a smooth incline at thirty degrees; the small circle means degrees, and k g is kilogram. Find the normal force and its acceleration. Pause the video, and try it first. Done? Okay, let's check. Smooth means no friction, so f is zero: the figure without the red arrow, just weight and normal force. And take g as ten metres per second squared, the m s to the minus two on screen. Okay? Now, across the incline. Does the block accelerate in that direction? No! It stays on the surface. So across, the forces balance: N equals m g cos thirty degrees. m g is two times ten, twenty, and cos thirty is root three over two. With me? So N equals ten root three newtons, about seventeen point three. The wavy equals sign means approximately, and capital N after the number is newton, the unit. Less than the weight of twenty newtons, just as we said. Right? Now along the slope. No friction, so only m g sine thirty acts; the mass cancels, and a equals g sine thirty. Ten times one half: five metres per second squared, down the slope. The mass does not matter at all! That's it!

5:40Example 2: man holding himself and a platform up

Now, remember our puzzle? Here it is, as example two. So, a sixty kilogram man stands on a forty kilogram platform. The rope tied to the platform passes over a pulley fixed to the ceiling, and the man holds the other end. Find the pull he must apply to keep himself and the platform at rest, and the force between his feet and the platform. Hint: draw two free-body diagrams. Pause and try. Done? Okay, let's check. With g as ten, the man weighs six hundred newtons, and the platform four hundred. First, isolate the man. Which forces act on him? Three forces, very good! His weight, six hundred newtons, down. The rope pulls his hands up with tension T, and the platform pushes his feet up with the normal force N, from below. He is at rest, so T plus N equals six hundred. Right? Now isolate the platform. Taking the rope as light and the pulley as frictionless, the rope pulls it up with the same tension T, okay? Its weight, four hundred newtons, acts down. And the normal force N: which way does it act on the platform? Down, exactly! The man's feet press the platform down, the third-law partner of the push on his feet. So T equals N plus four hundred. Clear? Now the trap. Together they weigh one thousand newtons, so he must pull with one thousand newtons, right? No, no, wait! Here is a topper's trick: isolate the man and the platform together, as one system, just as step one allowed. Then N acts inside the system, so it drops out. And the rope pulls the system up in two places: his hands and the platform. So two T equals one thousand; the double arrow means implies: T equals five hundred newtons. Got it? Then, from the man's equation, N equals six hundred minus five hundred: one hundred newtons. Puzzle solved! He pulls with only five hundred newtons, half the total weight, because the rope holds them up at two places. That's the power of a good free-body diagram!

7:54Quick revision: the four steps of a free-body diagram

So, quick revision. One: isolate the body, and draw only the forces acting on it, never those it exerts. Two: weight first, then a normal force, and friction if rough, for each surface, and a tension for each string. Three: axes along and across the acceleration, sum of forces equals m a for each body, plus a constraint. Now go and try the practice problems on this page. All the best!