Weighing a Man on the Equator: Solving the Puzzle

In summary, the man's true weight is 9.81 times the mass of the man, and it is near 0 on the equator.
  • #1
Kernul
211
7

Homework Statement


A man that has an ordinarily weight of 780 N stands on a libra on the equator.

Homework Equations


What is the weight measured by the libra?

The Attempt at a Solution


I thought it was simple, so I just did:
P = m * g
Since it gives me P I can simply do:
m = P / g = 780 / 9.81 = 79.51 kg
But looking at the solution on the book it says that it's 0.
Did I do something wrong?
 
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  • #2
To me it seems you do the right thing. A libra compares two masses and ordinarily we use 9.81 as gravitational acceleration.
Could the book be off ? Misprint or something ?

Otherwise it's a corny interpretation of "weight measured by the libra" ... after all it doesn't literally measure weight
 
  • #3
A "Libra" is a "scale"? Where does he "ordinarily weigh 780 N"? If he "ordinarily" weighs himself on the equator, then his weight on the equator would, of course, be 780 N! You calculate the person's mass but I don't see how that is relevant to the question. You say the books answer is "0" (are there no units) so perhaps I am misunderstanding "libra".

A person's true 'weight', the force the Earth exerts on a person, is 9.81 (approximately) times the mass of the person. What the person weighs is that force minus the mass of the person times the centripetal force on the person due to the rotation of the earth. That will be 0 at the poles and maximum at the equator.
 
  • #4
With "libra" I mean the object. Like the sign.
Where? Wherever on the Earth but the equator, it seems.
Now that I think about it actually calculating the mass doesn't bring me anywhere...
It is 0 on the book. Without anything after.
Anyway this was the third question the the entire problem I didn't thought you could have needed everything since I thought every question was just by itself.
Anyway, in a previous question on the book I calculated the angular velocity so, doing the calculations in the second thing you said we have the centripetal force:
Fcentripetal = m * ω2 * RT
The ω I calculated was because the question before asked what was the ω if the centripetal force was the same as the gravitational force at the equator on the same mass.
So we have:
P - Fcentripetal = 780 - [79.5 * (1.24 * 10-3)2 * 6.371 * 106] = 2
It's not 0 but it is near I think.
 
  • #5
Confusing. What is the complete question + context ?
Seems to me your original post forgot to mention the Earth isn't rotating with the same angular speed that we are used to "ordinarily" :rolleyes:
 
  • #6
Yeah, you are right. Let me write everything.
"What's the value of the angular velocity of rotation of the Earth so that the centrifugal force at the equator is the same as the gravitational force on the mass on the same spot?
If the Earth turn at this speed, how long would it be a day? If a man that weighs ordinarily 780 N would stand on foot on a libra situated on the equator, what would the weight measured by the libra be?"
I'm sorry if something doesn't seem clear but I'm translating the exercise from my language to English language.
 
  • #7
That's somewhat different than what I gathered from post #1 :smile:

If centrifugal force cancels gravitation, a scale (a force measurement, with a spring or something) would show 0 since there is no force left over.
A libra (I interpret that as: a balance) wouldn't work either: nothing is pulling the scales down, so any two masses can be glued to the scales and seem to be "in balance".

On such a fast-rotating planet, wouldn't it be risky to jump ? :rolleyes:
 
  • #8
Hahahahaha! Now I got it!
Okay, okay. Next times I'll write always the full the exercise.
Sorry for not being clear from the beginning!
 

1. How does being on the equator affect a man's weight?

Being on the equator does not directly affect a man's weight. However, the rotation of the Earth at the equator does create a centrifugal force that can slightly reduce a person's weight.

2. What is the puzzle of weighing a man on the equator?

The puzzle is that a man's weight may vary depending on whether he is weighed at the equator or at a different location on Earth.

3. How can the puzzle be solved?

The puzzle can be solved by taking into account the centrifugal force created by the Earth's rotation at the equator. This can be calculated using the formula F = mv²/r, where F is the centrifugal force, m is the mass of the man, v is the velocity due to the Earth's rotation, and r is the distance from the center of the Earth.

4. Does this puzzle have any practical applications?

Yes, understanding the effects of the Earth's rotation on weight can have practical applications in fields such as geology, engineering, and navigation. It can also help researchers better understand the Earth's gravitational forces.

5. Is the puzzle of weighing a man on the equator relevant in modern times?

Yes, the puzzle is still relevant in modern times as it helps us understand the complex interactions between gravity and rotation on Earth. It also demonstrates the importance of considering all factors when making measurements and calculations in scientific research.

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