Bodies Moving Together

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Three blocks are tied in a line by light strings, on a smooth floor. You pull the front block with twelve newtons. So, what is the tension in each string? Twelve newtons in both? Most students say yes, straight away. Hold that thought; we will settle it in the first example. Today's idea is short but powerful: when bodies move together, treat them as one system first, then cut, and look at one part. Connected bodies come up often in JEE mechanics. So, let's begin.

0:36Step 1: the whole system

So, think of bodies that move together: blocks tied by strings, or blocks pushed while they touch. Moving together means one common acceleration, small a, for all of them. The double arrow on screen means implies. Okay? Now, take two blocks tied by a light string, and pull the front one. For the whole system, should the tension in that string appear in the equation? No, it should not! The string pulls the back block forward, and the front block backwards, equally. Inside the system, those two pulls cancel. So, for the whole system, only external forces appear: forces from outside, like your pull. Right? So, small a equals F external, over sigma m. F with the small subscript ext is the net external force along the acceleration. The Greek capital sigma means sum of, so sigma m is the total mass. Total outside push, over total mass. Really simple. Clear?

1:38Step 2: cut, and look at one part

But wait. The tension cancelled out, so the system equation can never give it to us. Step two, then. To find an internal force, a tension or a contact force, we isolate one body. Cut the string in your mind, and look at one side only. With me? Now, which side should you look at? The side you pull, or the other side? The other side, exactly! The part beyond the cut, away from your pull. On a smooth floor, the only force along the motion on that part is the string. So the string alone gives that part its acceleration. Internal force equals the mass beyond the cut, in brackets, times small a. The cross is the times sign. Got it? And one more piece. If an outside force also acts on that part, like its weight when it is being lifted, or a rope pulling it, that force goes into the equation for that part too, with its proper sign. So, a force opposite to the acceleration adds to the internal force needed; a force along the acceleration reduces it. Clear?

2:41The trap, and a topper's shortcut

Now, the classic trap. You pull the front block with a force, capital F, and the tension in the string is capital T, okay? The string is tied right there, so T equals F, right? No, no, careful! Your pull F must accelerate every block. The string only accelerates the blocks beyond it. So the tension is less than your pull. The crossed equals sign means is not equal to. Got it? Now put small a equals F over sigma m into step two. T equals m beyond over sigma m, times F, okay? Here m with the small subscript beyond is the mass beyond the cut. So the pull is shared out in proportion to mass. Remember, this needs a smooth floor and a single pull. Nice shortcut, isn't it? And see what it means. Pull a different block, and the mass beyond the cut changes. So the tension changes too, with the same force and the same blocks! Always ask: what lies beyond the cut? That is the key.

3:42Example 1: tensions when 12 N pulls three blocks in a line

Now, remember our puzzle from the start? Let's settle it. Blocks of one, two and three kilograms are tied in a line by light strings, on a smooth floor. On screen, k g is kilogram, and capital N is the newton. Twelve newtons pull the three kilogram block. Find both tensions. Pause the video, and try it first. Done? Okay, let's check. Step one, the whole system. The strings are internal, so only the twelve newtons appear. Total mass: one plus two plus three, six kilograms. So small a equals twelve over six: two metres per second squared. That is what m s to the power minus two means. Okay? Now cut the string next to the three kilogram block. What lies beyond this cut? The one and the two kilogram blocks together, yes! Three kilograms. So this tension is one plus two, times two: six newtons. Right? And the other string pulls only the one kilogram block. One times two: two newtons. So, six newtons and two newtons. Not twelve, and not equal! That was our puzzle. The string nearer the pull carries more, because more mass lies beyond it. Got it?

5:01Example 2: man pulling himself up on a platform

Now a harder one, JEE level. Read it with me. Here, a sixty kilogram man stands on a forty kilogram platform. He pulls on a rope that goes up over a pulley fixed to the ceiling, and comes down to be tied to the platform. With what force must he pull, so that he and the platform rise with an acceleration of one metre per second squared? And what is the force between his feet and the platform? Pause and try. Hint: count the rope ends carefully. Done? Okay, let's check. Take g, the acceleration due to gravity, as ten metres per second squared. Our system is the man plus the platform: sixty plus forty, one hundred kilograms. Clear? Now, let capital T be the tension in the rope. He pulls the rope down with T, so the rope pulls the system up with just T, right? No, no, careful! Count the rope ends. One end pulls his hands up with tension T, and the other end, tied to the platform, pulls it up with the same tension. The rope runs over the ceiling pulley, outside our system, so both pulls are external, two T in total. The system's weight is one hundred times ten, one thousand newtons. So, two T minus one thousand equals one hundred times one. With me? Two T is eleven hundred, so T is five hundred and fifty newtons. That is how hard he must pull. We used the same T at both ends, which needs a light rope over a light, frictionless pulley. Okay? Now, the force between his feet and the platform. That is internal, so we cut, and look at the man alone. Which forces act on him? Three forces, yes! The rope pulls him up with T, and the platform pushes his feet up with italic N, the normal force. Careful: the upright N after a number means newtons. And his weight, six hundred newtons, pulls him down. The rope is an external force on this part, so it goes in, just as step two said. T plus N minus six hundred equals sixty times one. Right? So N equals six hundred and sixty, minus five hundred and fifty: one hundred and ten newtons. He pulls with five hundred and fifty newtons, and the platform pushes his feet with one hundred and ten. System first, then one part, and a scary problem is done in five short lines. Got it?

7:29Quick revision: whole system first, then cut for tension

So, quick revision. One: bodies moving together share one acceleration. Treat them as one system, and small a equals F external, over sigma m. Two: for a tension or a contact force, cut, and take the mass beyond the cut, times the acceleration. Three: count every external force on the part you choose, like both ends of a rope. Now go and try the practice problems on this page. You will enjoy them. All the best!