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Find the Magnitude of the Normal Force on the Slider and v'

  1. Feb 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A 2.4-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to have a speed of 11.3 ft/sec as it passes position B, determine (a) the magnitude N of the force exerted by the fixed rod on the slider and (b) the rate at which the speed of the slider is changing (positive if speeding up, negative if slowing down). Assume that friction is negligible.


    2. Relevant equations



    3. The attempt at a solution

    For part a I drew a FBD of the slider and assigned the normal and tangential axes. I can also see that the normal force and the weight of the slider are the only forces acting on the slider.

    ƩFen = man = N - mgcos(31)

    m(v2/r) = N - mgcos(31)

    I then solved for N and I get 6.383 lbf

    I'm unsure about part b. Any suggestions?
     

    Attached Files:

  2. jcsd
  3. Feb 15, 2013 #2

    Doc Al

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    Staff: Mentor

    What's the direction of the radial acceleration? (Check your signs.)

    What force acts tangentially?
     
  4. Feb 16, 2013 #3
    When you say radial acceleration are you referring to the normal acceleration? I don't quite see the point you're making.

    The force acting tangentially would be:

    F=ma

    so it would be the tangential acceleration*mass.
     
  5. Feb 16, 2013 #4

    Doc Al

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    Staff: Mentor

    You can call it the normal acceleration if you like. Which way does it point? Towards the center or away? Which way does the radial component of gravity point?

    Well, sure. But the tangential acceleration is what you need to find. So what is the tangential force?
     
  6. Feb 16, 2013 #5
    Oh, the positive normal axis points towards the center of the curve, so it should be:

    ƩFen = mgcos(31) - N = mv2/r

    N = mgcos(31) - mv2/r

    N = -2.2688 lbf

    Correct me if I'm wrong:

    ƩFet = mat = -mgsin(31)

    Solving for at = -gsin(31)

    at = -32.2*sin(31)

    at = -16.58 ft/sec2

    I've just checked my answers with the system and it says they're correct.

    Thanks for your help Doc Al. The second part was a lot more straight forward than I thought.
     
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