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!

Homework Help: Forces in Equilibrium.

  1. Apr 26, 2014 #1
    Hi All- Hope you can help me :|
    1. The problem statement, all variables and given/known data
    The person in the drawing is standing on crutches. Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing indicate. If the coefficient of static friction between a crutch and the ground is 0.90, determine the largest angle θMAX that the crutch can have just before it begins to slip on the floor.


    2. Relevant equations
    Fup= Fdown
    Fright= Fleft

    3. The attempt at a solution
    I figured the for each crutch
    Fy = N= 1/2mg + FcosΘ
    Fx= μsN=FsinΘ
    But I end up with mg and no way to eliminate it?

    I know I am missing some important detail- but as with many force problems, I can't visualize it.

  2. jcsd
  3. Apr 26, 2014 #2
    It appears from the picture that the person is standing on one of his legs in addition to the two crutches. So wouldn't there be 3 upward forces: normal force on his leg, and 2F cos ##\theta##?
  4. Apr 26, 2014 #3
    Does that mean that because he is standing- his weight isn't a factor in the upwards forces- because it gets canceled out? and the only net force is the vertical component of the crutch?
  5. Apr 26, 2014 #4
    Upon thinking some more, assuming that he's supporting himself on one leg plus 2 crutches doesn't seem to help. For the moment, let's assume that he's supporting himself entirely on the two crutches only. With this assumption, let's take a look at the vertical net force:

    Fnet(y) = 0 = 2F cos ##\theta## - mg

    Now, I'm unsure how you derived the following: Fy = N= 1/2mg + FcosΘ. Can you explain your thought process?
  6. Apr 26, 2014 #5
    I realized my mistake. SO I was assuming that the F of the crutches are like an external force. But its just a component of the mg. So

    I set it up as μmgsinθ=mgcosθ -> .9 tan-1= θ

  7. Apr 26, 2014 #6
    Just a sec, ##.9 tan^{-1}= θ## isn't a valid equation. Do you mean ##tan^{-1}(.9)=\theta##?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted