Lifts and Apparent Weight

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You stand on a weighing machine in a lift. The lift is at rest, and the machine reads five hundred newtons. Now the lift starts going up, speeding up, and the reading jumps to six hundred newtons! Did you suddenly get heavier? Hmm. And if the cable snapped, what would this machine read then? Strange, right? So what does a weighing machine really measure? Today: lifts and apparent weight. In JEE, lift questions come up often. So, let's begin.

0:35What does the weighing machine read?

So, look at the figure: three lifts, each with a person, drawn as a block, on a weighing machine, the grey strip on the floor. The blue arrow, capital N, is the normal force, the push of the machine on the person. The green arrow, m g, is the weight: mass m times g, the acceleration due to gravity. And the dark arrow, small a, is the lift's acceleration. Now tell me: what does the machine actually measure? Your mass? Your weight? Something else? Something else! It measures how hard you press on it, and that equals how hard it pushes back on you. So it reads the normal force, capital N, and not m g. Okay? The same goes for a spring balance: if a body hangs from it, it reads the tension, capital T, in its spring. Clear? Now the lift accelerates upwards at a, and you stand still on its floor, so you move with it. Work in the ground frame, an inertial frame, so Newton's second law holds with just the real forces. Net upward force, N minus m g, equals m times a. Right? So, N equals m times, in brackets, g plus a. The greater-than sign says this is more than m g. The machine pushes harder, so you feel heavier. Your mass has not changed at all; only the reading has. Got it?

1:53Accelerating down, constant velocity, free fall

Alright, now flip it. The lift accelerates downwards. Will the machine read more than m g, or less? Less, yes! Now the net force points down, so the weight is bigger than the normal force. m g minus N equals m times a. Okay? So, N equals m times, in brackets, g minus a, and the less-than sign says it is below m g. You feel lighter: that sinking feeling when a lift starts going down. Right? Now the lift moves at constant velocity, up or down. Then a is zero, so N equals m g, the true weight, like the middle lift in the figure. Clear? Now the scary one. The cable snaps, and the lift falls freely, with acceleration g downwards. What does the machine read now? Zero! Put a equals g in m times, g minus a, and N equals zero. You and the machine fall together, so you do not press on it at all. This is called weightlessness. But gravity still pulls you with m g; only the reading is zero. With me?

2:58The trap: acceleration, not motion

Now, the most common trap in this topic. The lift is moving up, but slowing down, about to stop at your floor. Heavier or lighter? Many students say heavier, because the lift goes up. What do you say? No, no, careful! The velocity points up but is getting smaller, so the acceleration points downwards. The little arrow on top of a means it is a vector, so its direction counts. So N equals m times, g minus a: lighter, even though the lift goes up. Got it? And the opposite case: moving down and slowing down. Now the acceleration points upwards, so N equals m times, g plus a: heavier, even though the lift is going down. Right? So, only the direction of the acceleration matters, not the direction of motion. Before you write any formula, first find which way a points. Clear?

3:51Topper's trick: effective gravity

Now a topper's trick. Inside the lift, everything behaves as if g were replaced by g eff, the effective gravity; the small e f f written after the g stands for effective. g eff equals g plus or minus a: plus when the acceleration is up, minus when it is down. With me? So the reading is simply m times g eff. And not only that: inside the lift, every effect of gravity uses g eff, not g. Okay?

4:20Example 1: 50 kg person, lift up, down, steady, free fall

Now, remember our puzzle from the start? Let's settle it. The person has a mass of fifty kilograms; k g is kilograms. Find the reading, in newtons, when the lift, part a, accelerates up at two metres per second squared, the m s to the power minus two; part b, accelerates down at the same rate; part c, moves at constant speed; and part d, falls freely. Pause the video, and try it first. Done? Okay, let's check. Take g as ten metres per second squared. Then the true weight, m g, is fifty times ten: five hundred newtons. The upright N after the number is the unit, newtons. Right? Part a: the acceleration is upwards, so plus. N equals fifty times, ten plus two: six hundred newtons. See, that is the six hundred from our puzzle! The machine pushes harder to give the person an upward acceleration. Clear? Part b: the acceleration is downwards, so minus. Fifty times, ten minus two: four hundred newtons. Lighter. With me? Part c: constant speed means zero acceleration. So the machine reads the true weight, five hundred newtons. Okay? And part d, free fall. What is the acceleration of the lift now, and what does the machine read? Yes, exactly! The acceleration is g, downwards. Fifty times, ten minus ten: zero. That is the snapped cable from our puzzle: weightlessness. Got it?

5:53Example 2: scale readings over three stages of a lift ride

Now a longer one, in three stages. The lift starts from rest and accelerates up at two metres per second squared for two seconds; the small s is seconds. Then it moves uniformly for four seconds, and then slows at two metres per second squared, to rest. Find the readings of a sixty kilogram man's weighing machine in the three stages, and the height gained. Pause the video, and try it. Done? Let's check. Stage one: the lift speeds up while going up, so the acceleration is upwards. N equals sixty times, ten plus two: seven hundred and twenty newtons. Okay? Stage two: uniform motion, so zero acceleration. The reading is just m g: six hundred newtons. Right? Now stage three. The lift is still going up. So, do we use g plus a again? No, no, careful! It is going up, but slowing down, so the acceleration points downwards. N equals sixty times, ten minus two: four hundred and eighty newtons. Direction of acceleration, not of motion. Clear? Now the height. First, the speed after stage one. v equals a times t: two times two, four metres per second; that is the m s to the power minus one. With me? Stage one: half, times two, times two squared: four metres. Stage two: four metres per second for four seconds: sixteen metres. Stage three: four squared, over two times two: four metres again. Got it? So the height gained is four plus sixteen plus four: twenty four metres. And the only trap was the direction in stage three. Okay?

7:39Quick revision: N = m(g + a) or m(g - a), use direction of a

So, quick revision. The machine reads the normal force: m times g plus a when the acceleration is up, and m times g minus a when it is down. Constant velocity gives m g, and free fall gives zero, weightlessness. And always use the direction of the acceleration, never the direction of motion. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!