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Homework Help: Resolving force

  1. Jul 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Particle P has a mass of 1kg and is confined to move along the smooth vertical slot due to the rotation of the arm Ab. Assume that at the instant as shown, the acceleration of the particle P is 2.46 m/s^2 upward. Determine the normal force N(rod) on the particle by the rod and the normal force N(slot) on the particle by the slot in the position shown. Frictions forces are negligible.

    ----N(rod) ---- N(slot)
    A. 14.2 N ---- 7.1 N
    B. 10.5 N ---- 3.6 N
    C. 7.1 N ---- 14.2 N
    D. 3.2 N ---- 1.6 N

    2. Relevant equations F= ma

    3. The attempt at a solution
    F = ma = 2.46 N
    The force acting on the particle can be resolved into 2 components
    N(rod) = F x cos(30) = 2.13 N
    N(slot) = F x sin(30) = 1.23 N

    The answer does not match any of the choices provided. Can someone help me figure out what I'm doing wrong? Thanks
    Last edited: Jul 16, 2011
  2. jcsd
  3. Jul 15, 2011 #2
    Hi drawar :smile:
    Welcome to PF!

    a diagram would be nice.
  4. Jul 16, 2011 #3
    Hi cupid.callin :)
    I've already drawn a vector diagram showing my work

    Attached Files:

  5. Jul 16, 2011 #4
    Your N(slot) force is in the wrong direction. N(rod) is correct. If you resolve N(rod) onto forces in the x and y directions you will see that N(slot) must act in the positive x-direction (to the right). Draw a free body diagram with all forces. Identify direction of acceleration. Use F=ma for x and y directions.
  6. Jul 16, 2011 #5
    let AP= l
    and vertical position of particle = y
    now: y= l*sine(theeta)

    velocity of particle= v= dy/dt= l*cos(theeta)*d(theeta)/dt
    v = Vr*cos(theeta)

    acc of paticle = a= dv/dt = (-)*l* sine(theeta) *w + l*cos(theeta)* q {q=angular acc of rad, w= angular vel of rad}

    acc of particle = -Vr*sin(theeta) + Ar*cos(theeta)

    {Vr , Ar = velocity of rod, Accc of rod at particles point}

    this means :: DATA insufficient...either posotion or velocity particle is required
  7. Jul 16, 2011 #6


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    Staff: Mentor

    How is the gravitational force being accounted for (the particle has a mass, therefore a weight)? I don't see it in your FBD or calculations. :smile:
  8. Jul 16, 2011 #7
    Am i the only one who still cant see the pic?
  9. Jul 16, 2011 #8
    Thanks for all your kind help :-)
    I totally forgot about the gravitational force. I've just give it another try and here is the result, hope someone might have a check at this:

    Resolving vertically upward:
    [tex]N_{rod} \times \cos 30^o - mg = ma \Rightarrow N_{rod} = \frac{{m(g + a)}}{{\cos 30^o }} = 14.2[/tex](1 d.p)

    Resolving horizontally rightward:
    [tex]N_{slot} - N_{rod} \times \sin 30^o = 0 \Rightarrow N_{slot} = \frac{{N_{rod} }}{{\sin 30^o }} = 7.1[/tex]

    Hence choose A

    edit: @cupid.callin: I'm afraid so. Seems like your browser is having trouble reading image from imgur.com. I'll upload it to another host
    Last edited: Jul 16, 2011
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