How much does a battleship “weigh” when it’s floating on the ocean?

It depends on what you mean by “weigh”. Ship masses are described in the term “displacement”, i.e. the mass of water a ship displaces in order to float.

For example, HMS Dreadnought displaced between 18,000 and 21,000 tons depending on load. This was the same as her weight, because this was the downward force she experienced due to gravity. The water exerted an equal upward force.

RE: the elevator question. Have you considered all the possibilities that down may have to offer?

Ah, I wonder if you had a hand in designing the Syrius Cybernetics Corporation’s Happy Vertical People Transporters?

Noen (25) said:

Cindy said:
“If the elevator is accelerating down, then gravity must be exerting more of a force on you than the elevator, so the scale will read light.”

Um, no… I don’t think that’s right or maybe it’s worded confusingly. If you and the elevator are both traveling at a constant speed and both are in free fall then the scale will read your weight as zero.

Noen, you seem to have missed part of what Cindy was saying. She was talking specifically about acceleration, which does not mean moving at constant speed.

Also, if you are in free fall, you are by definition accelerating. Assuming falling through an atmosphere, once you have achieved terminal velocity, you are no longer accelerating and therefore no longer in free fall. Constant speed only occurs in free fall under the special circumstance that is an orbit – and even then it is a constant angular velocity (assuming a circular orbit), not a constant velocity (velocity being a vector).

Joseph G (27) said:

The “Vomit Comet” plane flies a parabola which is basically identical to the path that would taken by a projectile fired from a gun in a vacuum – that is, with its trajectory determined by gravity with no other forces acting on it. Another way to look at it is a segment of an extremely elliptical orbit that intersects the ground (or would, if the pilot didn’t pull up )

I think that’s termed a ballistic trajectory.

But if you’re in free fall, you’re not traveling at a constant speed, you’re accelerating (unless you’re in a perfectly circular orbit).

At speeds less than that needed for free fall the scale will read less than your rest weight. That is the principle behind the “vomit comet” plane.

There isn’t a magic speed for free fall, though. It’s all about having your vertical acceleration precisely equal to that of gravity. The “Vomit Comet” plane flies a parabola which is basically identical to the path that would taken by a projectile fired from a gun in a vacuum – that is, with its trajectory determined by gravity with no other forces acting on it. Another way to look at it is a segment of an extremely elliptical orbit that intersects the ground (or would, if the pilot didn’t pull up )

@20 Trebuchet You seem to be in agreement that the scale will only change while accelerating. The ad says you “weigh less while going down.” It’s wrong. Not also that downward acceleration occurs whether the elevator is rising or descending, just at opposite ends of the ride.

Hehe, yeah, that bugged me too, but heck, devil’s advocate here, maybe we can call it a question of semantics? Does “going down” mean the entire ride, or just the part where you’re accelerating from a stop (the “going” part)? Maybe you do weigh less when “going down” but your weight returns to normal when “continuing down”?

From a safe distance.

Um, no… I don’t think that’s right or maybe it’s worded confusingly. If you and the elevator are both traveling at a constant speed and both are in free fall then the scale will read your weight as zero. The force gravity is exerting on you and the elevator remains the same. At speeds less than that needed for free fall the scale will read less than your rest weight. That is the principle behind the “vomit comet” plane. The passengers and the plane are essentially in free fall and that simulates weightlessness. Well, actually, it *is* weightlessness.

Astronauts sometimes practice their maneuvers in a large pool and take measures to neutralize their buoyancy and simulate the zero gravity of space.

The moment you put a scale under something you are making a practical measurement and the results you get will depend on the conditions under which you are weighing the object. The force of gravity will always be F=ma but the apparent weight of an object will vary a lot.

How much does a battleship “weigh” when it’s floating on the ocean? I’d say the same as a block of wood floating on water. A battleship in a really really big elevator in free fall will have a weight of zero. You won’t be able to push it around due to it’s mass but it won’t weigh anything.

Actually, I agree with DanM: I’m not afraid of falling or clowns individually, but the thought of falling and being incapacitated while surrounded by clowns, now that’s scary!

My apologies, I thought it said “Amy”. Didn’t read it carefully enough – too many late nights grading lately.

We used to do a lab at my school where we would bring an accelerometer into the elevator. Tobin Dax is correct about the results – we also could only go up two floors at most.

I would love to have a smartphone app that could record some accelerometer readings and then take it into a high-rise building.

RE: the elevator question. Have you considered all the possibilities that down may have to offer?

You’ve got it right. (I’m trying to remember how I put it a few weeks ago.) You weigh less at the top of the ride and more at the bottom*, no matter which way you are going. In between, while moving at a constant velocity, your weight is the same as if you were at rest.

*This was in a 2-story building, so there was no possibility of additional stops between the bottom floor and the top floor with the elevator I had in mind when I said that.

Who’s Amy?

You seem to be in agreement that the scale will only change while accelerating. The ad says you “weigh less while going down.” It’s wrong. Not also that downward acceleration occurs whether the elevator is rising or descending, just at opposite ends of the ride.