Torque and Couples

Transcript

0:00Introduction

Hello everyone! Let's start with a door. Push it at the handle, and it swings open easily, right? Now push with the same force, but right next to the hinges. It hardly moves! Same push, same door. So what changed? By the end of this video, you will answer that with one simple formula. Today: torque and couples, the turning effect of a force. It runs through all of rotational motion and comes up often in JEE, so let's begin.

0:31Torque: the turning effect of a force

So, look at the figure. On the left is the point O, our reference point. The force F acts at some point, and r is the position vector of that point, drawn from O to it. The little arrows on top mean both are vectors, okay? The torque about O is the Greek letter tau, also with an arrow. Tau vector equals r cross F, with r first and F second, in that order. In this figure, as written on the right, tau points out of the page. And since r is measured from O, always ask: torque about which point? Right? Now, you already know cross products from vectors. So tell me, what is the magnitude of r cross F? r F sine theta! Yes, exactly. Theta, the Greek letter next to the little arc in the figure, is the angle between r and F, okay? So tau equals r F sine theta. Clear? Now look at the dashed line through the force. That is its line of action. The perpendicular distance from O to this line is r perp: r with a small upside down T below it, the perpendicular sign. We call it the moment arm. And r perp is just r sine theta, so tau also equals F times r perp. Force times moment arm, okay? Now, back to our door. Let O be the hinge. If you push close to the hinge, what happens to the moment arm? It becomes tiny! So the torque is tiny, and the door hardly turns. At the handle, the moment arm is large, so the same push gives a big torque. Mystery solved! And a topper's trick: only the line of action matters. Slide the force anywhere along that dashed line, and r perp, and so the torque, stays the same. With me? Last, the units. Torque is measured in newton metre. Its dimensions, written in square brackets, are M L squared T to the power minus two. Hmm, the same as energy! So, can we write torque in joules? No, no, careful! Torque is a vector, and it is never written in joules. Always newton metre. Got it?

2:39Torque about an axis

Now, a door does not turn about a single point. It turns about an axis, the line through its hinges. So we need the torque about an axis. It is simply the component of tau vector along that axis, okay? And to find tau vector here, take O at any point on the axis. The component along the axis comes out the same. Right? Now tell me. Push the door straight upwards, parallel to the line of the hinges. Does it turn? Not at all! So a force parallel to the axis has zero torque about that axis. Keep this in mind for example two. Clear? One more. Now push the door straight towards the hinges, along the door itself. Any turning this time? Again, no! The line of action of this force cuts the axis, so the moment arm is zero. Zero torque about that axis. Got it?

3:34Couples: zero net force, but still a torque

Now, the second word in our title: couple. Think of your two hands on a steering wheel: one pushes up, the other pulls down. So, a couple is a pair of equal and opposite forces, acting along different lines of action. With me? First question. What is the net force of a couple? Zero! Yes, exactly. Equal and opposite, so they cancel. Now, careful. The net force is zero. So is the torque also zero? No, no! That is the trap. The steering wheel clearly turns, right? The torque of a couple is F times d, where d is the distance between the two lines of action. Zero net force, but a real torque. Okay? And here is the beautiful part. This torque, F d, is the same about every point. So for a couple, you never need to ask about which point. Just F times d. That's it!

4:31Example 1: torque of a spanner pulled at 30 degrees

Now, example one, an easy one. Take a spanner, zero point five metre long, pulled at its end by a ten newton force. The force makes thirty degrees with the spanner. Find the torque about the nut. Pause the video, and try it first. Done? Okay, let's check. Take the nut as O; then r is zero point five metre, along the spanner. F is ten newton, and theta, the angle between r and F, is thirty degrees. Okay? Now, many students get stuck here. Sine thirty, or cosine thirty? Sine thirty! Theta is the angle between r and F, and the force makes thirty degrees with the spanner itself, which lies along r. So tau equals r F sine theta: zero point five times ten times sine thirty degrees. Right? Sine thirty degrees is one half. Zero point five times ten is five, and five times one half is two point five. So the torque is two point five newton metre. And remember, newton metre, not joules!

5:42Example 2: torque about the origin and the z-axis

Now a slightly harder one, with vectors. Here, a force F equals five k hat newton, acting at the point three, four, zero, in metres. k hat, with that little cap on top, is the unit vector along the z axis, okay? Find the torque about the origin, and about the z axis. Pause the video, and try it. Done? Let's check. The origin is our O, so r is just the position of the point: three i hat plus four j hat metres. i hat and j hat are the unit vectors along x and y. And F is five k hat newton. With me? So tau equals r cross F: three i hat plus four j hat, in brackets, cross five k hat. Multiply out: fifteen times i hat cross k hat, plus twenty times j hat cross k hat. Okay? Now, i hat cross k hat. Is it plus j hat, or minus j hat? Minus j hat! Remember the cycle: i, j, k, and back to i. Forwards gives plus; backwards gives minus. From i to k is backwards, so minus j hat. And j hat cross k hat is forwards, so it is plus i hat. Clear? So tau equals twenty i hat minus fifteen j hat, newton metre. Its magnitude, mod tau, with the two vertical bars, is the square root of twenty squared plus fifteen squared. That is the root of six hundred and twenty five: twenty five newton metre about the origin. With me? Now, the second part. The origin lies on the z axis. So the torque about the z axis is also twenty five newton metre, right? No, no, careful! That is the trap. Torque about an axis is only the component of tau along that axis, here the k hat component. And our tau has no k hat part at all! So the torque about the z axis is zero. Of course! The force is along k hat, parallel to the z axis, just like pushing the door straight up. Got it?

7:58Quick revision: r x F, moment arm, axis and couple

So, quick revision. Torque is r cross F; its size is r F sine theta, or F times the moment arm, in newton metre. About an axis, take the component of tau along that axis. And a couple: net force zero, torque F d, the same about every point. Now go and try the practice problems on this page. All the best!