1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Compute acceleration and tension in the cord

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    A cord passing over a frictionless, massless pulley has a 4.0 kg object tied to one end and a 12 kg object tied to the other. Compute the tension and acceleration in the cord.

    Right now, I'm just trying to calculate the acceleration. It is supposed to be 4.9 m/s^2

    2. Relevant equations


    3. The attempt at a solution

    For the 4kg block-

    T1-mg = ma

    T1-39.2 = 4a
    Solving in terms of T1 to get
    T1= 4a+39.2

    For the 12kg block-
    T2-mg = ma

    T2-117.6= 12a
    Solving in terms of T2-
    T2= 12a+ 117.6

    Now, because T1=T2, I can say

    12a+ 117.6 = 4a+ 39.2
    But then I end up with a=9.8.

    The answer is 4.9, which, incidentally, is 9.8/2.

    Do I have to divide by 2 because I have two cords? Or is there some other reason I am missing?
  2. jcsd
  3. Dec 8, 2011 #2
    Assuming the following is correct can you solve for T and a?

    Attached Files:

  4. Dec 9, 2011 #3
    Yeah, you'd set T=4a + 4g , plug that into the other equation for T, to get

    12g+(-4a-4g) = 12a
    to get
    8*9.8 = 16a


    Ah. I must have screwed up somewhere in my algebra.

    Thank you!
  5. Dec 9, 2011 #4


    User Avatar
    Homework Helper

    co-incidence almost.

    One way to view this: the weight of a 4 kg mass is pulling one way, and the weight of a 12 kg mass is pulling the other way.

    If you consider the weight of the 4 kg mass cancells out [balances if you like] 4 of the other 12, we have a net force equal to the weight of an 8kg mass to accelerate this system - which has a total mass of 16 kg.

    Now the weight of a 16 kg mass will accelerate a 16kg mass at 9.8 ms-2, so we would expect that half that weight would achieve only half that acceleration → 4.9 ms-2
  6. Dec 9, 2011 #5
    Another way to solve this problem is to use just two directions for forces. It may sound crazy, but I have seen this procedure in two or more textbooks.
    The positive direction would the direction of motion (the direction of falling of the 12-kg block). This way, if m1 is the 12-kg block, and m2 is the 4-kg block, we would have the following system of equations (each equation for a separate free-body diagram):

    -T + m1g = m1a
    T - m2g = m2a

    I got 58.8 N for tension, and 4.9 m/s^2 for acceleration ;)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook