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!

Monkeys on ropes- mechanics

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Two monkeys of masses 10kg and 8kg are moving along a vertical rope (There's just a single vertical rope in the diagram, attached to a fixed support (no Atwood's machine kind of thing), with two monkeys drawn on it.)
    The first monkey (this monkey is above the other monkey, but obviously below the fixed support), is climbing up with an acceleration of 2m/s^2, while the second monkey is coming down with a uniform velocity of 2m/s. Find the tension in the rope at the fixed support.

    2. Relevant equations

    3. The attempt at a solution
    So, basically, the solution says that you add the tensions due to both the monkeys, ie
    T1-100=20 (from first monkey)
    T2=80 (Second monkey)
    Adding T1 and T2, you get 200 N.
    How can the tensions possibly be different in a mass-less string at different points? And why are we adding both the tensions to get the tension at the fixed support?
  2. jcsd
  3. Nov 8, 2015 #2


    User Avatar
    Gold Member

    The rope below the first monkey only have to take the forces due to the second one. The rope below the second monkey only has to zake its own weight (which is zero). So different tensions can be found between the points of application.

    It's not that the tensions in the rope are summed up, but all the loads affecting the rope, which have to be held by the support.
  4. Nov 8, 2015 #3
    Could you please expand on why we're adding the acceleration (ie, 20) to the total load, as opposed to subtracting it, since you mentioned that it's the loads affecting the rope that are being summed up?
  5. Nov 8, 2015 #4


    User Avatar
    Gold Member

    Monkey 1 (m1) is in accelerated movement upwards of a1=2 m/s2 + the gravition (g). The load due the monkey 1 is F1 = m1 ⋅ (a1 + g) = 10 ⋅ (2 + 10) = 120 N

    Monkey 2 (m1) is in non-accelerated movement (constant velocity) so a2 = 0 + the gravition (g). The load due the monkey 2 is F2 = m2 ⋅ (a2 + g) = 8 ⋅ (0 + 10) = 80 N

    Below monkey 2 the tension in the rope T3 = 0
    Between monkey 1 and monkey 2 the tension in the rope T2 = F2 = 80 N
    Above monkey 1 the tension in the rope T1 = F1 + F2 = 120 + 80 = 200 N
  6. Nov 8, 2015 #5
    stockzahn is right.

    As for your question as to how tensions in different parts of the same massless rope can be different, all you need to know is that net force on every element is to be zero. If external forces act on one part, tension will change accordingly, as shown by this question.
  7. Nov 8, 2015 #6
    Thanks! Got it :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Monkeys on ropes- mechanics
  1. Monkey Climbing a Rope (Replies: 2)

  2. Monkey Climbing Rope (Replies: 1)