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Rocket launching - Newton's Second Law

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dandy9
#1
Jan14-11, 04:15 PM
P: 28
1. The problem statement, all variables and given/known data
Suppose that during the launch of a rocket from the surface of the earth, an astronaut onboard the rocket stands on a bathroom scale and measures her weight to be 3.6 times her normal weight. Determine the acceleration of the rocket.


2. Relevant equations
Fnet [tex]\sumF = ma[/tex]

or maybe involving Tension?
Fnet [tex]\sumF = T - Fg = ma[/tex]

3. The attempt at a solution
I tried the second equation but can't get anywhere. I ended up with T - 35 = a.
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rock.freak667
#2
Jan14-11, 05:02 PM
HW Helper
P: 6,205
Your T should be the normal reaction N. So you have N-mg=ma. They told you that N=3.6mg, so that the m will cancel out.
dandy9
#3
Jan14-11, 10:27 PM
P: 28
Thanks for the reply.

Sorry, but I don't see how the "m"'s cancel out...
If T = 3.6mg then I would have the equation:
35m-35=a
I'm still struggling with getting the mass.

Apphysicist
#4
Jan14-11, 10:30 PM
P: 109
Rocket launching - Newton's Second Law

Quote Quote by dandy9 View Post
Thanks for the reply.

Sorry, but I don't see how the "m"'s cancel out...
If T = 3.6mg then I would have the equation:
35m-35=a
I'm still struggling with getting the mass.
No, you'll have:

N-Fg=Fnet

3.6mg-mg=ma. Then you can divide by m so they all cancel. I don't know how you canceled your m's before, and I especially don't know you have a 35 in there.
dandy9
#5
Jan14-11, 10:50 PM
P: 28
Thanks I got it now!

Before, I canceled my "m"'s too early, before I knew that T was N, so when I put in 3.6mg, I was left with that m and didn't know what to do with it.

As for the 35, I used the 3.6 in the wrong way - I used it with Fg because the question stated that the astronaut is 3.6 times her normal weight. Perhaps if you have the time to quickly explain to me why the 3.6 goes with N instead of Fg I would very much appreciate it.

Thank you again!


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