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!

Equilibrium & Friction Problem

  1. Dec 29, 2013 #1
    The person in the system above has a mass of 60 kg. If the system is in equilibrium, what is the coefficient of friction between the person and the ground, to two decimal places?

    I need help on this question , I have done some work on it which is given below:

    Converted masses into weights/forces:

    for weight of man : 60*98.1 = 588.6 N

    for weight of the hanging mass = 10 * 98.1 = 98.1 N

    I have done the tension in the string , the man is pulling the weight.

    T* COS60° = 98.1N

    T = 98.1N/COS60°

    T = 196.2 N

    IF THE SYSTEM IS IN EQUILIBRIUM SUM OF THE FORCES SHOULD BE ZERO.

    I HAVE ALSO FOUND THE OTHER COMPONENT OF T WHICH I THINK WILL BE EQUAL AND OPPOSITE THE FRICTIONAL FORCE. ITS GIVEN AS

    Ty = 196.2* sin 60° = 169.9144 N

    My understanding is that i need to use F = μ * W

    But i am not sure about what will be W

    Could someone explain this question and guide me if i have done this right.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 29, 2013 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You cannot determine the coef of static friction, but you can determine the min coef required to keep the mans feet from sliding, which perhaps is what the problem is asking? Your tension calc is wrong, based on a free body diagram of the hanging object, what is T?? Then look at the forces on the man, break up T into its horiz and vert components and solve for the friction force and normal force using equilibrium in x and y directions separately, then friction force = coef of friction times normal force at the
    limiting value of static friction.
     
  4. Dec 29, 2013 #3
    Solution for friction equilibrium

    THANKS FOR GUIDING ME - I NEED FURTHER HELP

    I have solved this according to your advice. Confusion i have about angle. i did 90-60=30

    x Component of T = 98.1* COS 30° = 84.95N
    y component of T = 98.1 SIN 30° = 49.05


    SUM X forces is only 84.95N (equal and opposite to friction force).

    SUM of Y forces is = weight of man + y component of T
    ∑ Y forces = 588.6N + 49.05N = 637.65N

    Friction Force = μ * normal reaction

    μ = 84.95 N / 637.65N

    μ = 0.13322
     
  5. Dec 29, 2013 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    .
    Ok so far, good.
    the weight of the man acts down on him. In which

    direction does the y comp of T act on him? What about the normal force?
     
  6. Dec 30, 2013 #5
    the weight of the y component acts down on him and the y component of T acts upwards
    so ∑ Y forces = 588.6N - 49.05N = 539.55

    normal force will be equal in magnitude t the y foces

    μ = 84.95 N / 539.55N = 0.1574



    HAVE I UNDERSTOOD IT CORRECT LY??
     
  7. Dec 30, 2013 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    YES, I think so, considering that the problem is not worded correctly or you copied it down wrong. The value of the static friction coefficient that you have calculated, which should be rounded to 0.16, is the minimum value required to prevent the man from sliding rightward. The actual value could be much greater than that, but the friction force would be the same 84.95 N.


    -The Phantom
     
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: Equilibrium & Friction Problem
Loading...