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Atwood Machine Lab calculations

by ~christina~
Tags: atwood, calculations, machine
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Oct15-07, 01:09 AM
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P: 824
1. The problem statement, all variables and given/known data
My results are coming out funny and I was wondering if I was doing something wrong.
I have to find acceleration due to gravity.

one of my values ....

m2 (decending mass) = 60g
m1(ascending mass)= 50g

distance traveled= 95.3cm or .953m

time traveled average= 19.60s

total mass (m1 + m2)= 110g

2. Relevant equations
am (measured)= 2y/t^2

at(theoretical) = (m2-m1)g/(m1 + m2)

3. The attempt at a solution

well I try to find g from the am (or measured acceleration) by using the am and plugging into the theoretical accleration (at) equation but find g instead of a.

am= 2(.953m)/(19.60s)^2
am= 0.00496m/s^2

then plugging into the at equation..

am(m1+m2)/ (m2-m1)= g

[0.00496m/s^2 (110g)]/ 10g= 0.54m/s^2 ===> this is so not 9.8m/s^2..

Basically that's it..except I also find the theoretical acelleration from using 9.8m/s^2 which would be the ideal and find that but I get...

at= (m2-m1)g/ (m1 + m2)

at= (10g)(9.8m/s^2) / (110)= .891m/s^2 for acceration.

I'm supposed to ignore the friction..

I really don't know why it comes out like this..Am I doing anything incorrectly??
I have to get this right or explain why it went wrong since this is for a lab report.

Thank You .
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Oct15-07, 03:53 AM
Sci Advisor
PF Gold
P: 4,980
You seem to be doing the analysis correctly. I'd guess that there was some sort of unwanted resistance in the pulley system, because 19.6 seconds is a long time to travel 1 meter.
Oct15-07, 07:19 AM
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~christina~'s Avatar
P: 824
Oh..okay Thank Kurdt! =)

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