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Math Help - Need to find G on a ramp

  1. #1
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    Need to find G on a ramp

    I have a block of ice sliding down a FRICTIONLESS ramp with an accelleration of 5.3 m/s.s.. If the ramp makes an angle of 32.7 degrees with the horizontal , find g, the acceleration due to gravity.

    This is using the application of vectors, the only way that I can solve this.
    But I am not sure. what to do, help get me started please.
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  2. #2
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    Draw a figure. Always when you have a problem in classical kinematics and kinetics, draw a figure. There's a reason every single textbook on classical mechanics include a section about free body diagrams.

    In this case, you have an inclining plane and a block on it. There are two forces involved - identify which they are and add them into the figure in the appropriate place. Gravity acts at the center and so on.
    Also, don't forget to specify a set of coordinates. The convention for this type of problem is to have the positive x-direction in the direction of motion along the incline of the plane, and the positive y-direction then becomes in the direction of the normal force from the plane on the block.
    After that it's application of Newton's second, F = ma all the way.
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  3. #3
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  4. #4
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    How do I do that calculation if all I have is a degree and an Acceleration of 5.3 m/s.s.
    That is what I am not sure about??
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  5. #5
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    Quote Originally Posted by bradycat View Post
    How do I do that calculation if all I have is a degree and an Acceleration of 5.3 m/s.s.
    That is what I am not sure about??
    what is Newton's 2nd law of motion and what does it mean?
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  6. #6
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    well it's F=ma. Sorry still lost on this one..
    Do I take 5.3 and / by 9.8 to find the velocity?
    Which is -5.4 m/s.
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  7. #7
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    if you look at the force diagram, the normal force , N , is equal and opposite to the weight component perpendicular to the incline, mg\cos{\theta} ... in essence, those two forces cancel.

    ... which leaves the parallel component of weight, mg\sin{\theta} , alone as the net force.

    2nd law ...

    F_{net} = ma

    mg \sin{\theta} = ma

    g\sin{\theta} = a

    ... the standard formula for the acceleration of an object down a frictionless incline with elevation angle \theta.
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  8. #8
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    So is the answer 9.81 g??
    5.3/sin32.7 ?
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  9. #9
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    The answer is 9.81 m/s^2, not 9.81 g, lol.
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  10. #10
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    Well I have to find g. So why would it be 9.81m/s.s.?? if it's the weight?
    Or do I have it confused somehow?

    On never mind, re read the question, and its g, the acc due to gravity.
    GOT IT, thanks.
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