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Distribution of Forces

  1. Jan 11, 2016 #1
    Hello, we were given homework, which is perfectly described by this picture. We should figure out F1 and F2 (it's number). I know that if the both angles are equal, it is easy to solve. However, now i should solve a situation when first angle (alpha) = 50 grades, the second (beta) = 25 grades - they are not the same
    Furthemore, i know that weight of the man is 70 kg. We should find out F1 and F2.
    2. Relevant equations
    of course Fg = m.g so Fg = 700N (approximately)

    3. The attempt at a solution
    Now i know that if the 2 parts of rope are equals, it can be solve in this way : Fg/2 = F1 . sin alpha and both forces (F1 and F2) would be equal. However, i do not know what will happen if the angles are not equal

    Thank you really much for your help!
    Last edited by a moderator: Jan 11, 2016
  2. jcsd
  3. Jan 11, 2016 #2
    Fg/2 thats so .. lets say shady.. the drawing is really helping.. the dotted lines are a fantastic help, trust me... Have you tried the net force approach?
  4. Jan 11, 2016 #3

    Simon Bridge

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    There is no angle labelled beta in your diagram.

    To handle this sort of problem, you realise that the sum of the forces is zero.
    Start by resolving components for the forces along the rope sections.

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