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Inclined Planes and Net Force Renate Fiora

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Representing Forces When we’re representing forces for an object on an inclined plane, we don’t really do much differently than we normally represent forces in diagrams. We use arrows to depict the magnitude and direction of the forces acting on the object. To get the clearest understanding of what’s happening, it’s best to make both a situational diagram and a free-body diagram.

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Two Diagrams Situation Diagram Free-body diagram FAFA FNFN FfFf FWFW +y / +x / FAFA FNFN FfFf FWFW

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Two Situations, Generally a = 0 There are two possibilities for when there is no acceleration, v = 0 and v 0. In either case, because there is no acceleration, we know there is no net force on the object, so when we write Newton’s second law (F net =ma) we get (F A + ) F – F f = 0 (F A + ) F = F f a 0 In this situation, the object is accelerating as it moves along the inclined plane, and so we know there must be a net force acting on the object. Writing out Newton’s second law, we get (F A + ) F – F f = ma The applied force is in parentheses as there is not always one present.

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46 30° W W || W FAFA F || FF FNFN Equations from Diagram: W || = F || F N = F + W m = 5 W = 49 W = 42.4 W || = 24.5 cos30 = F || F A cos30 =

46 30° W W || W FAFA F || FF FNFN Equations from Diagram: W || = F || F N = F + W m = 5 W = 49 W = 42.4 W || = 24.5 cos30 = F || F A cos30 =

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