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    galveston's Avatar
    galveston Posts: 451, Reputation: 60
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    #1

    Nov 25, 2006, 02:17 PM
    Inclined plane
    I need to know how to calculate force transferred from the perpendicular to horizontal assuming zero friction. Example: 10 pounds of verticle force applied to a 30 degree plane will move the plane horizontally with how much force? Or to put it another way, how much force is lost in the example, and what formula is used to calculate it?
    Thanks.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Nov 25, 2006, 03:44 PM
    This is simple trigonometry, you want to work out how much of the force is directed along the plane, and how much is directed perpendicular to it. (the force you are applying is the hypotenuse).
    You can now ignore the force along the plane, since friction is 0

    Then, do the same again using the perpendicular force, split it into horizontal and vertical components (the perpendicular force is now the hypotenuse). The answer you get for the horizontal force is the answer you are looking for.

    If you need further help just ask! :)
    galveston's Avatar
    galveston Posts: 451, Reputation: 60
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    #3

    Nov 26, 2006, 01:11 PM
    I never took trig, but have used it for years to calculate dimensions and angles, but am fuzzy about using it for forces. Can you explain the steps necessary in more detail? I am desiging an engine using lever and plane to produce rotation without a crankshaft and would like some mathmatical estimate of efficiency.
    Thanks for your reply.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Nov 26, 2006, 01:23 PM
    Well, if you imagine a right angled triangle. (sketch it)

    Your force is pointing downwards and is the hypotenuse (long edge)
    For a 30 degree plane, you have another edge of the triangle pointing from the top of this force in a 30 degree direction perpendicular to your plane, this edge is the longer of the 2 edges.

    Then your parallel force is from the tip of the perp force, to the tip of the force you're applying, and completes the 90 degree triangle

    In this way, you have split the force into Parallel and perpendicular components (you can do this as long as your components are at 90 degrees).

    The only calculations you need are to work out the lengths of the sides, one will be F*cos30 and the other will be F*sin30, using SohCahToa, you can work out which.

    Then you ignore the parallel force (since it's frictionless) and do the same splitting the perp force into horizontal and vertical components.

    The vertical component doesn't factor in 0 friction, it just pushes the plane towards the table, the horizontal is what you're after.

    This should give a fairly good approximation for a low friction situation.. like plastic or something.

    If you're still having trouble, I'll break out paint and do a picture for you.
    galveston's Avatar
    galveston Posts: 451, Reputation: 60
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    #5

    Nov 27, 2006, 08:21 PM
    OK. I guess I'm about as smart about this as a box of rocks. Visualize a ramp at x degrees with a 10 pound weight on it. I can get an applet off the net that shows how much force is required to move the weight up the ramp. What escapes me is this: if the weight does not move up the ramp, but instead the ramp moves horizontally, wouldn't it take less force to move the ramp horizontally than to move the weight up the ramp? Is the difference in the force required the same ratio as the difference in the hypotenuse and the base of the triangle? Thanks again for your patience.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #6

    Nov 28, 2006, 06:01 AM
    Here is a picture of how I figure you would work it out. I hope it's easier to follow

    Apologies for size, it wasn't very legible any smaller.


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