The weight of an astronaut during take off?

shabby_goblin

1. The problem statement, all variables and given/known data
A spaceship accelerates with constant acceleration to 120 m / s within 10 seconds after launch from Earth.
Imagine that one of the astronauts who weighs 77 kg sits on a bathroom scale while the Launch takes place.
What does the scale show during the launch ?

2. Relevant equations
120m/s : 10 s = 12 m/s2

3. The attempt at a solution

My teory is:
F=m(g+a)
F=77 kg (9,81 m/s2+ 12 m/s)
F=1679,37N
1679,37N / 9,81 = 171 kg
So the scale show 171 kg

darkxponent

that is right. What is the problem?

shabby_goblin

Well i didnt state my problem O_o.
It is that we are 4 peeps arguing on what's right.
Some just want to take F = m*g which is F=77 kg * 12 m/s2 = 924N
But i mean that since this is on earth gravity has to play a part in this.
So everything is right in what i stated above ?

darkxponent

How do you get this: F=m(g+a). Can you show me?

shabby_goblin

If you look on the drawing ive added then the forces in work here should be
ƩF = G + a
ƩF = m x g + a
ƩF = m x (g+ a)

Attached Thumbnails

 

connor02

Do not confuse your N and kg.

The bathroom scale shows weight, not mass. Why they refer to it as kg and not N i do not know.

Mass : 77/9.81 = 7.849kg

Weight with acceleration of 12m/s = 7.849 x (12 + 9.81) = 171.2 N

shabby_goblin

Ok so what does the scale show? that should be in kg.
Do you just divide 171.2 N with 9,81 m/s2 = 17,45 kg ???

connor02

No your answer is right, the scale shows 171.2. However, the unit in your answer should be N not kg.

darkxponent

The scale shows niether weight nor mass. It gives the value N/9.8. Where N is the normal.

connor02

The common bathroom scale does not divide by 9.81. It shows N. And for someone standing on the scale, this is the person's weight.

darkxponent

No scale is made to give the weight of a body. All scales are designed to give mass. That is N/9.8.

Whovian

Scales are basically designed to find the weight, and, assuming we're near the surface of the Earth, under the influence of only gravity, normal force from the scale, horizontal forces, and relatively negligible forces, calculate mass. What they give, technically, is mass.

Filip Larsen

Assuming the scale has been calibrated to show 77 kg when the astronaut sits on the pad it is perfectly valid to answer either that the scale reads around 171 kg or 1680 N, depending on the exact value of g used. Or in other words, the scale shows around (g+a)/g = 2.22 times the calibrated weight of items when accelerated like that.

D H

Nonsense. Google "bathroom scale newton" and you will see a bunch of physics homework questions, some right here at this site, but nothing to buy. To find a bathroom scale that displays a metric result you need to google "bathroom scale metric", "bathroom scale kg", or "bathroom scale kilograms" -- and those scales will display your "weight" in kilograms, not newtons.

That's much closer to the mark.

What a bathroom scale measures is the displacement of a spring. A scale that registered spring displacement, however, would be absolutely worthless. "Oh! I only weight 1.2 millimeters today!"


If the scale is analog, the displacement of the spring rotates a dial. The scale under the dial will inevitably have units of mass (kg or lb). This will be off if the zero point is not right (which it isn't), if the spring doesn't exactly obey Hooke's law (which it doesn't), if the spring is a little stiffer or spongy than nominal (which it is), if gravitational acceleration is something other than 9.80665 m/s² (which it is), if the load isn't perfectly distributed on the spring pan (which it isn't), ...

A digital scale removes the zero point problem but adds even more sources for error. Instead of rotating a dial, a digital spring uses something such as a strain gauge to convert pressure to an electrical signal, and an A/D converter to convert this electrical signal to a number. Now the strain gauge can be non-linear, have a scale factor error, have a bias, as can the A/D converter.

fluidistic

I agree with darkxponent and D H. When I passed my final exam of introductory mechanics my professor corrected me. I thought a scale measured weight and that my weight would change if the reading of the scale would change. When close to the Earth's surface, weight is a force that always equal to "mg" in magnitude where [itex]g \approx 9.81 m/s^2[/itex].
When you are in an elevator your weight doesn't change despite what a scale under your shoes would show.

shabby_goblin

Thanks for all the answers people :)
This was really good help

Whovian

Wait. Typo, and I can't edit this post now. What I mean is that they give, technically, weight, or rather, the force pushing down on them, which, under the circumstances most people measure themselves with, is going to be their weight, and the scale may or may not calculate mass from that.

I wanted to bring general relativity into account, but by the time I typed up the layman's explanation of everything we need to know what I wanted to say, I'd forgotten what I wanted to say.