1. Not finding help here? Sign up for a free 30min 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!

Tension acceleration homework question

  1. Nov 9, 2004 #1
    I'm just practicing some tension questions. Say there were two weights free hanging on a pully. One side hangs a 9kg weight and the other side hands a 6kg weight. So, the acceleration would be: a=F/m, a=(3kg*10m/s^2)/15kg, a=2m/s^2

    Now, I want to find the tension force on the 9kg block. It’s acceleration down at 2m/s^2
    T=m(g-a)
    T=9kg(8 m/s^2)
    T=72 N

    Did I do this correct?
     
  2. jcsd
  3. Nov 9, 2004 #2
    i'm confused by your problem

    consider the situation. there's a pulley, so the forces are related. there is mg force being exerted on each side. one object weights 3 kg more than the other. what's the net force on each block?

    what forces are there acting on the large block?
     
  4. Nov 9, 2004 #3
    I'm several steps a head of you hehe. Look, I wrote:
    a=F/m, a=(3kg*10m/s^2)/15kg, a=2m/s^2

    this means, net force=(3kg*10m/s^2)
    then, I found the acceleration of the whole system to be 2m/s^2

    so, Tension on the larger block would equal F=ma, T=m(a), however, it is accelerating downwards so it looses gravitational force...therefor (g-a) so...

    T=m(a-g)
    T=9kg(8 m/s^2)
    T=72 N

    I dont know if I did this correct, but I think so. Need a check on the "T=m(a-g)" to know if I can do that.
     
  5. Nov 9, 2004 #4
    does that make sense? 72 N acting on the block?

    what force is pulling the block down? what force is pulling the block up?

    obviously the force pulling down on the block is from gravity. so is the force pulling up on the block, right? (how is that force propagated?)

    to me, it looks like neither one of your answers are correct. i could always be wrong though, i'm a ways from my phD.

    edit: 2 m/s^2 is the difference between the two accelerations, if down is the positive direction
     
    Last edited: Nov 9, 2004
  6. Nov 9, 2004 #5
    mg-T=ma
    (9kg*10m/s^2)-T=(9kg*2m/s^2)
    T=72N
     
    Last edited: Nov 9, 2004
  7. Nov 9, 2004 #6
    Got 72N also. This is the tension in the rope so it acts up in both cases.
    using 9kg as mass 1, and towards the area under 9kg as positive.
    [tex]
    m_1 a = m_1 g - T
    [/tex]
    [tex]
    m_2 a = T - m_2 g

    [/tex]
    combine to solve for a you get

    [tex]
    a = \frac{m_1 g - m_2 g}{m_1 + m_2}
    [/tex]

    2m/s^2

    [tex]
    m_1 a = m_1 g - T
    [/tex]
    [tex]
    18 = 90 - T
    [/tex]
    [tex]
    T= 72N
    [/tex]
     
    Last edited: Nov 9, 2004
  8. Nov 9, 2004 #7
    They are accelerating on the same rope, why would there be any difference in acceleration?
     
  9. Nov 9, 2004 #8
    that was my problem, tension wasn't uniform in the rope the way i calculated it. i wasn't considering that the tension is higher than the weight on the side of the lighter mass.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Tension acceleration homework question
Loading...