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Just starting with friction, question

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 25° and P = 750N

    μs[\SUB] = .35
    μk[\SUB] = .25


    Part 1.png Part 2.png

    2. Relevant equations

    Fs = μs * N
    Fk = μk * N

    3. The attempt at a solution

    Attachment 1 up there is the drawing of the scenario.
    Attachment 2 is my addition to it to solve this problem.

    Using the axis shown, I have

    ΣFy = Ncosθ - Fssinθ - 1.2kN = 0
    and apparently
    ΣFx = Nsinθ - Fscosθ - .750kN = 0

    otherwise that similar triangle I drew there wouldn't make sense, just odd to think of the force of friction acting in the SAME direction as the force...regardless, continuing...

    Solving Top equation for Fs[\SUB] I get

    Fs[\SUB] = (Ncosθ - 1.2kN)/(sinθ)

    And subbing that into the second equation for F I receive

    Nsinθ - ((Ncosθ - 1.2kN)/(sinθ))cosθ - .750 = 0
    Nsinθ - (Ncos^2 θ - 1.2kNcosθ) - .750 = 0
    N(sinθ - cos^2θ) + 1.2kNcosθ - .750 = 0
    N = (-1.2kNcosθ + .750)/(sinθ - cos^2 θ)
    N = 0.8465kN

    So Fs[\SUB] = .35*0.8465kN = .2963kN


    Seeing as how this doesn't match up to the answer given, I'm not seeing where I went off...can anyone guide me in the right direction? Should I use a slanted axis?
     
    Last edited by a moderator: Dec 4, 2014
  2. jcsd
  3. Dec 4, 2014 #2

    BiGyElLoWhAt

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    Gold Member

    I don't think anyones gonna see where you went wrong.
    Sooooo many unnecessary "subs"
     
  4. Dec 4, 2014 #3

    jbriggs444

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    Science Advisor

    You can compute N much more easily by looking at the components of the 1.2 kN and the 0.75 kN normal to the surface. The the value of .8465 kN is not correct.

    Yes, switching to a coordinate system aligned with the surface makes things easy.
     
  5. Dec 4, 2014 #4

    BiGyElLoWhAt

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    Gold Member

    [qoute]... And subbing that into the second equation for F I receive
    ##Nsin(\theta ) - \frac{Ncos(\theta ) - 1.2}{sin(\theta )}cos(\theta ) - .750 = 0 ##[/quote]
    yes
    ##Nsin(\theta ) - Ncos^2(\theta ) - 1.2cos(\theta ) - .750 = 0##
    try again.
     
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