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WEIGHT and ACCELERATION as you FALL and CLIMB UP

  • Thread starter sousou_88
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1. An airplane flying horizontally encounters a low pressure and drops, and you feel lighter. If your apparent weight during that time is 60% of your true weight, what is the acceleration of fall of the plane? If the plane now climbs back with an acceleration equal and opposite to the acceleration of fall, what will be your apparent weight at that time?



2. ??? Net Force = mass x acceleration ??? Weight = (-)mass x acceleration of free fall of 9.8 ???



3. ???????Not sure how to start this. I was thinking maybe the acceleration of the fall of the plane would be acceleration of free fall -g = -9.8. But there is no way that it could be this easy. There is a vital piece of information I am missing out on! And if -9.8 is the acceleration of the fall, then 9.8 would be acceleration of it climbing back up. So if i did use the weight = mass x acceleration equation. Apparent weight would be 140% of true weight???????

I'm very confused, so any assistance will be very much appreciated!
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

1. An airplane flying horizontally encounters a low pressure and drops, and you feel lighter. If your apparent weight during that time is 60% of your true weight, what is the acceleration of fall of the plane? If the plane now climbs back with an acceleration equal and opposite to the acceleration of fall, what will be your apparent weight at that time?

There is a vital piece of information I am missing out on!
Hi sousou_88! Welcome to PF! :smile:

The vital piece of information you are missing is R, the reaction force on you (from the airplane) …

good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" won't work without it. :wink:

(and R is also your apparent weight)
 
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  • #3
Answer# 1. as we know that the equation relating the Apparent and the Real weight is given as
Fnet=W(real)-W(apparent)...............1

but
W(real)=Mg...................................................2
W(app)=60% X W(real)...................................3
and F(net)=Ma...............................................4
where M is the mass of the aeroplane

so Ma =Mg-0.60xMg

this gives us that "a" is = gx(0.40) = 3.92m/s^2

this is the acceleration with which the aeroplane must fall...
 
  • #4
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A good way to approach this problem would be with fictitious forces. A free body diagram from the accelerated point of view, would reveal three forces acting on you, one of them is the fictitious force, and the other two are the force of gravity, and the reaction force from the airplane. From the accelerated frame, you are at rest.

For a positive y axis pointing down, the direction of the acceleration is also down, and positive. Remember that the fictitious force is opposite the direction of the acceleration of the frame of reference.

Fd = -mA
Fg = mg
The reaction force is your apparent weight (Negative sign, as it is pointing up)

I think that a point that's confusing you is what your weight during free-fall would be. In free-fall, you are weightless!
Gravity has an equal effect on everything around you, you are not accelerating relative to the scale you're standing on, so you don't enact any force on it.
If you were being accelerated at a constant acceleration g, upwards, your apparent weight would be twice your actual weight.
 
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