Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Atwood Machine Lab calculations

  1. Oct 15, 2007 #1


    User Avatar
    Gold Member

    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 .
  2. jcsd
  3. Oct 15, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  4. Oct 15, 2007 #3


    User Avatar
    Gold Member

    Oh..okay Thank Kurdt! =)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook