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Starter -Motion Along an Inclined Plane

Starter -Motion Along an Inclined Plane. N = ( 0, N) W = ( mgsin q , -mgcos q ) . Find an expression for the normal force in terms of m, g and q . S F y = 0 ; N + ( -mgcos q ) = 0 ; . N = mgcos q. Acceleration Down a Frictionless Inclined Plane. N = ( 0, N)

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Starter -Motion Along an Inclined Plane

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  1. Starter -Motion Along an Inclined Plane N = ( 0, N) W = ( mgsinq , -mgcosq ) Find an expression for the normal force in terms of m, g and q. SFy = 0 ; N + ( -mgcosq ) = 0 ; N = mgcosq

  2. Acceleration Down a Frictionless Inclined Plane N = ( 0, N) W = ( mgsinq , -mgcosq ) Find an expression for the acceleration down the plane in terms of g and q. SFx = max ; 0 + mgsinq = max ; ax = gsinq

  3. Example A mass m is held at rest on a frictionless inclined plane by a string. Find the tension in the string and the normal force. m = 2kg and q = 60o.

  4. Sketch the forces……….. Make the free-body diagram. N = ( 0, N) T = (-T, 0) W = ( mgsinq , -mgcosq ) Apply the 2ndLaw for Statics: SFx = 0, SFy = 0 0 – T + mgsinq = 0 ; T = mgsinq = 2(9.8)sin60 = 17.0 Newtons N + 0 - mgcosq = 0 ; N = mgcosq =2(9.8)cos60 = 9.8 Newtons

  5. Example – Inclined Plane A mass m is held at rest on a frictionless inclined plane by a string. I f the string breaks, how long does it take to move 2m down the incline? m = 2kg and q = 60o.

  6. Example – Inclined Plane N = ( 0, N) W = ( mgsinq , -mgcosq ) SFx = max ; 0 + mgsinq = max ; ax = gsinq = (9.8)sin60 = 8.48 m/s2 xf =2m vf = ? a = 8.48 m/s2 xi = 0 vi = 0 t = ? xf = xi + vit + 1/2at2 , 2 = .5(8.48) t2 t = ( 2 / (.5)(8.48) ) ½ = .687 s

  7. EXIT Can an object be moving and still have zero acceleration? Explain.

  8. Newton’s Third Law If object A exerts a force on object B, then B exerts a force of equal magnitude and opposite direction on A.

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