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!

Tension & Pulley system

  1. Jul 11, 2017 #1
    tension ropes bigger.jpg

    1. The problem statement, all variables and given/known data
    M = M
    There is another mass "m" attached to mass "M" with a rope
    There is an acceleration (clockwise)
    Does Tension2 smaller,bigger or equal to Tension1?

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    So this is what I understand just by the logic:
    In any case - T1 > T2 ,

    I tried to put it into equations:
    T2 = mg - ma
    T1 = T2 +Mg - Ma

    →→ T1 = mg - ma + Mg - Ma = m⋅(g-a) + M⋅(g-a)

    "(g-a)" is constant value. Hence no matter what is the mass correlation between M & m , T2 will always be smaller than T1

    Am I right?

    Thanks a lot.
  2. jcsd
  3. Jul 11, 2017 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    I think you essentially have it, except for maybe one loose end. As you said, (g-a) is a constant. What must be true about this constant in order for your argument to hold?
  4. Jul 11, 2017 #3
    If g=a we get 0 . which means there is no tension in the ropes?
  5. Jul 11, 2017 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Yes, but could a = g?

    Maybe I'm being too picky. But is seems to me that your argument requires that (g-a) be a positive constant.
    Your equation T1 = T2 + Mg - Ma can be written as

    T1 = T2 +M(g-a).

    So, T1 is greater than T2 as long as (g-a) is positive. It's not hard to find a reason why (g-a) is positive.
  6. Jul 11, 2017 #5
    If "m" is big enough to create 9.8 m/s2 acceleration. so a=g . Logically,I think, it could be the case.
    Actually I can't find a reason why "(g-a)" must be positive when "a" is bigger than g... maybe that's the direction of the accelaration? if it minus this term will be always positive. but we already defined the direction of the acceleration within the equation.
    so I dont know :confused:
  7. Jul 11, 2017 #6


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Tension in a string cannot be negative. So, your equation T2 = mg - ma implies g - a cannot be negative. So, you just need to rule out a = g which would imply T2 = 0.

    If T2 = 0, then the free body diagrams for the two blocks of mass M would be identical. Yet you know that one block must accelerate upward while the other accelerates downward.

    [EDIT: Another approach is to treat the system as a whole. Taking clockwise motion around the pulley as positive, the net external force acting on the system is Fnet = mg + Mg - Mg = mg. Thus the acceleration is Fnet divided by the total mass of the system. You can easily check that this is less than g.]
    Last edited: Jul 11, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Tension Pulley system Date
Pulley-block-spring system Feb 15, 2018
Finding Tension and theta in a pulley system Dec 7, 2017
Find acceleration of a pulley system Mar 21, 2017
Tension of a Pulley System Nov 4, 2016