I recently had a friend ask me a long line of questions about weight, the sort of questions I was sure I knew the answers to until he started really interrogating me. I just couldn’t work it out in my head, so I turn to you.
The premise is this: a really big sealed airtight box is on some really, really big and really sensitive scales. At the start of each situation, the scales read X.
There is a helicopter in the box. The pilot starts the engine, takes off, climbs a few metres, and hovers there.
What do the scales read when
1a. The helicopter is still landed, but the rotors are spinning.
1b. The helicopter takes off.
1c. The helicopter hovers.
I’m in the box holding a bowling ball. I hurl it into the air as hard as I can, and then catch it.
What do the scales read when
2a. The ball leaves my hands.
2b. The ball reaches its maximum height.
2c. I catch the ball.
I’m in the box holding a heavy barbell at waist height. I jerk it into the air and hold it above my head. I then jerk it down to waist height again.
What do the scales read when
3a. The barbell starts to move upwards.
3b. The barbell is moving upwards.
3c. The barbell stops.
3d. The barbell starts to move downwards.
3e. The barbell is moving downwards.
3f. The barbell stops.
I’m in the box holding a bowling ball again. I drop it.
What do the scales read when
4a. The ball leaves my hands.
4b. The ball is falling.
4c. The ball hits the ground.
I’m pretty sure I know the answers to all of these, but for the life of me I cannot figure out why those are the answers, which makes me uncertain that they are the right answers at all.
1 the same at all times. It is a closed system. In each of your examples the system is in steady state. The helicopter’s weight is supported by the scales at all times, either directly or through increased air pressure below the blades
2, 3, 4 - When the ball or barbell is fully supported by the floor or your hands in steady state the scales will read x. When the ball or barbell is not supported by your hands and is in freefall the scales will read x minus the weight of the ball or barbell since they are not being supported by the scales. When the ball or barbell is rising or falling but a force is being applied to it by you or the floor, (to partially support, accelerate or decelarate it), the scale will read x, less the weight of the ball or barbell, plus however much force you or the floor are applying to the ball or barbell.
For all of these questions, keep in mind two simple rules:
The scale must ultimately provide a force to prevent an object from moving in free-fall.
Any force you provide to, say, toss a ball in the air will also register (via repeated application of Newton’s third law) on the scale, but only for the time you are accelerating the object.
The most interesting case is the helicopter. The running helicopter is transferring part of its weight to the scale by blasting air downward. Lift provided by the rotating blades will reduce the amount of weight applied directly by the helicopter’s skids, but this will be balanced by the force of the downward-rushing air.
Specifying that the box is air-tight adds another wrinkle, as I suspect that if the box is small enough, the air will naturally recirculate up the walls of the box and back down into the rotors, which means the helicopter would have to keep cranking the power to maintain altitude. But this is starting to sound like another famous problem that has littered these boards for the past year, and that way madness lies…
I think that as long as the helicopter is not moving up or down the scale will register X. As the helicopter moves up the scale will temporarily read higher, as it moves down it will read lower, as it hovers or lands the scale will return to X.
I am almost certain this is wrong. Princhester gave exactly the answers I expected; I just couldn’t convince myself of it. It somehow felt wrong that the scale would read less while the ball was in the air and so forth.
1b. Higher by (roughly) a factor of m*a where m is the mass of the helicopter and a is the vertical acceleration rate.
1c. Same as if the helicopter were just sitting in the box.
2a. Roughly your weight.
2b. Your weight.
2c. Above the combined weight of you and the ball. The exact value depends on how you catch it.
3a. Higher than your weight plus the weight of the barbell.
3b. What’s important is the acceleration of the barbell. Thus the information in this scenario is insufficient to provide an answer. The weight could be higher, lower, or exactly the same as your weight plus the weight of the barbell.
3c. Less than your weight plus the weight of hte barbell.
3d. Less than your weight plus teh weight of the barbell
3e Same answer as 3b
3f Greater than your weight plus the weight of the barbell.
4a. Your weight
4b. Your weight
4c. Something higher than the sum of your weight and the bowling balls weight.
Let me know if any of those answers are confusing and I will explain the rationale behind them.
Let’s start with the helicopter. Princhester said the scale’s reading didn’t change when the helicopter took off and climbed, which was the answer I was expecting. You’re saying differently. Why does the reading change?
Then, the bowling ball. To propel the ball into the air, I need to exert a force downwards, right? So why does the scale’s reading not increase to reflect that?
The box weighs more when the helicopter climbs because it is accelerating upward and therefore increasing the downward force of the blades on the air. It weighs slightly less when descending because the force on the air is less. Try getting on a scale and lowering your center of mass by bending your knees. You’ll see that the scale registers less weight as you go down and more as you go up.
For the bowling ball, as treis says, acceleration is key. The wording in part 2a is “the ball leaves my hands”, which is exactly when the acceleration stops. While you are “pushing” the ball upwards – while it is still in your hands – that’s when the scale will read higher.
Okay, now I see treis and I were both wrong about the helicopter. The rationale given above would be right if the box weren’t airtight. But since the box is airtight, Princechester is correct. In a closed system, whatever force the helicopter blades apply downwards has to be matched by an equal upward force of air elsewhere in the box. Therefore the box will weigh the same at all times. This wouldn’t change the answers in the other questions though, because somebody is always standing on the bottom of the box transferring forces to the scale.
OK, I understand. I shouldn’t have said when the ball leaves my hands. So the scale reads higher while I’m pushing, and then goes back to “my weight” when I let go of the ball. That’s what I was expecting.
And that explains that.
Well then, we are all in agreement. Niceness all round.
I’m not sure about that - if the system’s centre of mass is moving, I think this will be felt outside the system - it should still net to zero over time, though.
I don’t think that it matters that the box is sealed: if the helicopter is moving upwards, then the center of mass of the system is moving upwards as well, and that will be detectable as a temporary increase in the apparent weight of the box. To rise, the helicopter must exert a force on the floor of the box that is greater than the weight of the helicopter. If the helicopter is hovering, the apparent weight of the system is the same as when the helicopter is resting on the floor. When the helicopter begins to descend, it is exerting a force on the floor of the box that is less than its own weight, so the system’s apparent weight is less for as long as the helicopter is falling.
Consider a gun bolted to the floor of the sealed box, pointed up. If the gun is fired, the recoil force is transmitted to the floor, and would register on the scale, even though the box is sealed.
Yeah, I’m going back and forth on that helicopter question. The more I think about it, the more I can convince myself either way. I look at the air like a pulley; as long as I’m pulling hard enough to raise myself, there has to be a net downward force, right? Similarly there has to be an upward force when I go down.
And now I think I’ve convinced myself back to my original position again. Helicopter climbs – scale reads more. Helicopter descends – scale reads less.
If, on the other hand, we were talking about a neutrally-buoyant fish in a tank, when it swam in any direction, an equal mass of water (actually an equal volume, but also equal in mass if the fish is neutrally buoyant) would be moving in the opposite direction, so there would be nothing to feel from the outside.
On the helicopter question, I’m with RJKUgly, Snarky_Kong, treis, mangetout and brossa : if you’re accelerating the helicopter upwards, you’ve got to be putting more force on the scales than if the helicopter was sitting (and the ‘airtight’ part of the box is important; otherwise the moving air would transmit the force to surfaces far away from the scale). And vice versa, of course.
[Pickyness on/] If we’re using real air (as opposed to ideal physics-land air), it’s compressible and does have mass, so the actual time the force changes on the scale will be slightly different from the actual time the helicopter accelerates; I don’t know if this would be measurable. The time difference will be at least the speed of sound divided by the distance from the helicopter to the scale.
[Pickyness = Maximum] Finally, if the box was tall enough, with the helicopter in steady state at the top of the box, you’d have (very) slightly less force on the scale than with the helicopter in steady state at the bottom, as gravitational force decreases with distance. This would only be measureable as the size of the box starts approaching the size of the planet, though.
Ok, I’ve attempted some drawings. Don’t laugh at the crappiness of my helicopters. There are 8 figures, figures a-c, and figures 1-5. Sorry for the labeling, but I made a-c after 1-5 but, I want to use them first.
Figure A is just a picture of the helicopter sitting on the bottom of the box. It is not moving nor is it accelerating. Figure B is a force diagram on the helicopter. Newton’s second law states that the sum of the forces is equal to mass*acceleration. Since we have no acceleration, the forces must be equal to each other in magnitude, and acting in opposite direction. Thus, in this case Fn is equal to mg.
Let me stop for a second and define some terms here. F[sub]N[/sub] is the normal force, and it is the force from a surface acting perpendicular to that surface. It’s nothing more than the force from, for example, the table that holds the book up. “m” is mass and “g” is the acceleration due to gravity. The product of these terms is the weight of an object.
In figure C I have replaced the helicopter with the force it imparts on the bottom of the box. Remember that Newton’s 3rd law states that reaction forces are equal and opposite. Again, since we know that the box is not accelerating, we know that the sum of mg, i.e. the weight of the box, plus F[sub]N[/sub], which we know is the weight of the helicopter, and Fscale is 0. Thus we know that Fscale’s magnitude is the weight of the box plus the weight of the helicopter.
Figure 1 represents the case where the helicopter is accelerating upward. Figure 2 shows the forces on the helicopter. F[sub]L[/sub] is the lift force from the rotors, and mg is the weight of the helicopter. Since the helicopter is accelerating upwards we know that there is a net force upward on the helicopter from Newton’s second law. Thus we know that F[sub]L[/sub] is greater than the weight of the helicopter. Figure 3 is just replacing the helicopter with the force it imparts on the air. Remember again, the third law states that reaction forces are equal and opposite.
Figure 4 shows the forces on the body of air inside the box (ignoring pressure forces and the weight of the air). Now I’m going to do a bit of hand waving here and say that the net motion of the air is 0. What I mean by this is that individual air molecules might be accelerating, but if you add up all the accelerations you get 0. In other words, for each molecule of air being accelerated downwards by the rotors, there is another molecule of air being accelerated upwards. I think it’s a fair assumption to make, but arguing why is just going to add a layer of complexity that is unnecessary.
Ok, since we assume that the air mass is not accelerating, we know that the net force on that mass is 0. Thus F[sub]N[/sub] is equal but opposite to F[sub]L[/sub]. Remember that F[sub]N[/sub] is the force exerted by the surface of the box on the mass of air. Again by Newton’s 3rd law, we know that there is an equal and opposite force exerted on the box. Figure 5 shows the forces on the box, and from there we can see that the scale will read F[sub]N[/sub] plus the weight of the box. Recall that F[sub]N[/sub] is equal in magnitude to F[sub]L[/sub], which is greater than the weight of the helicopter since it is accelerating upwards. Thus, the scale reads a value higher than the weight of the box plus the weight of the helicopter.
I’m not sure of your question. If you are talking about 2a, I took “When the ball leaves my hand” to mean that you are no longer accelerating the ball. Thus there would be no additional force, and the scale would read just your weight since the ball is now in the air. It is true that the scale would read higher as you accelerate the ball during the throw.
