Apparent weight of a body with upward acceleration

AI Thread Summary
The discussion revolves around calculating the apparent weight of an astronaut during upward acceleration. Initially, a misunderstanding led to an incorrect calculation by simply multiplying mass by an assumed acceleration. The correct approach involves recognizing that apparent weight is the sum of gravitational force and the force due to upward acceleration. A free body diagram helped clarify the forces at play, leading to the realization that the apparent weight is effectively the true weight plus the additional force from acceleration. Ultimately, when upward acceleration is 5g, the apparent weight is six times the astronaut's rest weight.
TH02
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Homework Statement
What is the apparent weight during take-off of an astronaut whose actual weight is 750N if the resultant upward acceleration is 5g?
Relevant Equations
F=ma
Weight = mg
I initially attempted to get the answer by multiplying the mass by 50, as I assumed if the upward acceleration was 5g then the weight could be found by just multiplying the mass by the acceleration of 50ms^-2. However that resulted in an answer of 3750 which was far below the correct answer of 4500. After much trial and error I only got the correct answer when I multiplied the mass of 75kg by an acceleration of 60ms^-2, however I do not understand why this works.
 
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TH02 said:
Homework Statement:: What is the apparent weight during take-off of an astronaut whose actual weight is 750N if the resultant upward acceleration is 5g?
Relevant Equations:: F=ma
Weight = mg

I initially attempted to get the answer by multiplying the mass by 50, as I assumed if the upward acceleration was 5g then the weight could be found by just multiplying the mass by the acceleration of 50ms^-2. However that resulted in an answer of 3750 which was far below the correct answer of 4500. After much trial and error I only got the correct answer when I multiplied the mass of 75kg by an acceleration of 60ms^-2, however I do not understand why this works.
What's the weight if the upward acceleration is zero?
 
PeroK said:
What's the weight if the upward acceleration is zero?
Wouldn't it just be 750N?
 
TH02 said:
Wouldn't it just be 750N?
Yes. And what if the upward acceleration is ##1g##?
 
PeroK said:
Yes. And what if the upward acceleration is ##1g##?
75N?
 
TH02 said:
75N?
How do you get that?
 
PeroK said:
How do you get that?
My thinking was if I have a gravitational field strength of 10ms^-2 acting downwards and I also have an upwards acceleration of 1g (I think this is 10ms^2 in this context) acting in the opposite direction, they would cancel out? I have a feeling I'm making a big mistake here
 
TH02 said:
My thinking was if I have a gravitational field strength of 10ms^-2 acting downwards and I also have an upwards acceleration of 1g (I think this is 10ms^2 in this context) acting in the opposite direction, they would cancel out? I have a feeling I'm making a big mistake here
You should be thinking in terms of forces. When there is more than one force, you must add them together to get the unbalanced force (*). See the Khan academy page. You should also learn about free body diagrams if you haven't already.

(*) Note that force is a vector quantity.
 
PeroK said:
You should be thinking in terms of forces. When there is more than one force, you must add them together to get the unbalanced force (*). See the Khan academy page. You should also learn about free body diagrams if you haven't already.

(*) Note that force is a vector quantity.
Thank you, I'll have a look
 
  • #10
Have you drawn a free body diagram of the astronaut showing the forces acting on him, or do you think you have advanced beyond the need to do that?
 
  • #11
Chestermiller said:
Have you drawn a free body diagram of the astronaut showing the forces acting on him, or do you think you have advanced beyond the need to do that?
I drew a free body diagram but I'm not sure if I did it correctly
 
  • #12
TH02 said:
I drew a free body diagram but I'm not sure if I did it correctly
Let’s see it and the associated force balance equation.
 
  • #13
Chestermiller said:
Let’s see it and the associated force balance equation.
Chestermiller said:
Let’s see it and the associated force balance equation.
I'm not sure what you mean by force balance equation
But here is a rough free body diagram
IMG_20200703_163354130.jpg
 
  • #14
TH02 said:
I'm not sure what you mean by force balance equation
But here is a rough free body diagram
View attachment 265763
That clearly shows a downward force and an upward acceleration!

However, perhaps the real question in this case is what is weight anyway?
 
  • #15
I don’t see a. gravitational force acting on him, and I don’t see an upward force from the seat he is in.
 
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  • #16
PeroK said:
That clearly shows a downward force and an upward acceleration!

However, perhaps the real question in this case is what is weight anyway?
Isn't weight the force acting on the mass of a body due to the gravitational attraction of the earth?

Chestermiller said:
I don’t see a. gravitational force acting on him, and I don’t see an upward force from the seat he is in.
Ah yes, maybe I should step back for a moment and learn how to properly do FBD's
 
  • #17
TH02 said:
Isn't weight the force acting on the mass of a body due to the gravitational attraction of the earth?

Yes, then I guess the question is what is the apparent weight you are asked to calculate?
 
  • #18
PeroK said:
Yes, then I guess the question is what is the apparent weight you are asked to calculate?
I'm not quite sure, would it be a sum of the true weight and the force caused by the body's mass and acceleration?
 
  • #19
TH02 said:
I'm not quite sure, would it be a sum of the true weight and the force caused by the body's mass and acceleration?
Note that acceleration does not cause a force. An unbalanced force causes acceleration.

In this case I would take apparent weight to be the force between the astronaut and the seat she is sitting in.
 
  • #20
F-mg = ma
 
  • #21
TH02 said:
Isn't weight the force acting on the mass of a body due to the gravitational attraction of the earth?Ah yes, maybe I should step back for a moment and learn how to properly do FBD's
Obviously
 
  • #22
I've figured out how to do the FBD's and I think I've got it now, thanks for the help and sorry for being a bit thick
##F-mg=ma##
##F=mg+ma##
##F=m(g+a)##
##F=75(10+50)=4500N##
 
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  • #23
TH02 said:
I've figured out how to do the FBD's and I think I've got it now, thanks for the help and sorry for being a bit thick
##F-mg=ma##
##F=mg+ma##
##F=m(g+a)##
##F=75(10+50)=4500N##
Yes, but you did not need to assume a value for g.
If the upward acceleration is 5g then the apparent weight is 6 times the rest weight.
 
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