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Inclined Planes

Inclined Planes. Introduction Components Normal Force Whiteboards (No friction) Dealing with friction Whiteboards (With friction) Demo - inclined plane/block. mg. . Gravity is down. But things happen along the plane. So we make components of gravity. . mg.

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Inclined Planes

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  1. Inclined Planes Introduction Components Normal Force Whiteboards (No friction) Dealing with friction Whiteboards (With friction) Demo - inclined plane/block mg 

  2. Gravity is down But things happen along the plane So we make components of gravity  mg Consider an object on an inclined plane:

  3. Meet the components - they are your friends! Fperp - perpendicular to the plane mg F|| - parallel to the plane

  4. FN = mgcos (Causes friction) (FFr = kFN ) And the plane pushes back (It doesn’t break)  Fperp = mgcos mg Since we know the angle, we can calculate the components F|| = mgsin (Acts down the plane)

  5. Whiteboards: Inclined planes with no friction 1 | 2 TOC

  6. 4.50 kg  = 23.5o Find the component of gravity acting into the plane, and the component acting down along the plane: Fperp = mgcos() = (4.5 kg)(9.8 N/kg)cos(23.5o) = 40.44 N F|| = mgsin() = (4.5 kg)(9.8 N/kg)sin(23.5o) = 17.58 N Note that this makes sense - mostly into the plane W 40.4 N, 17.6 N

  7. + - 4.50 kg  = 23.5o The plane is frictionless, so the component of gravity parallel to the plane is unopposed. What is the acceleration of the block down the plane? Fperp = 40.44 N, F|| = 17.58 N, F = ma, <-17.58 N> = (4.50 kg)a a = F/m = F||/m = (-17.58 N)/(4.50 kg) = -3.91 m/s/s W -3.91 m/s/s

  8. Example problem: s = .75 k = .45 + 19.0 kg -  = 32.1o • Will it stay on the plane if u = 0 (static friction)? (force to slide) • What is its acceleration if it is already moving down the plane? • What force in what direction would make it slide down the plane with an acceleration of 1.20 m/s/s? • Dealing with friction on an inclined plane: • Only Fperp causes friction • FFr = FN FN = mgcos • Friction is then one of the forces along the plane • Friction opposes motion (velocity)

  9. Example problem: s = .75 k = .45 + 19.0 kg -  = 32.1o • Will it stay on the plane if u = 0 (static friction)? (force to slide) • What is its acceleration if it is already moving down the plane? • What force in what direction would make it slide down the plane with an acceleration of 1.20 m/s/s? <+/-Fapp - F||+/-Ffr> = m(+/-a) Applied Parallel Friction

  10. Whiteboards: Inclines with friction 1 | 2 | 3 | 4 | 5 TOC

  11. 3.52 kg s = .82 k = .37  = 42.0o Find Fperp, F||, the kinetic and maximum static friction: F|| = mgsin() = (3.52 kg)(9.8 N/kg)sin(42.0o) = 23.08 N Fperp = mgcos() = (3.52 kg)(9.8 N/kg)cos(42.0o) = 25.64 N FFr(kinetic) = kFN = (.37)(25.64 N) = 9.49 N FFr(static)<sFN = (.82)(25.64 N) = 21.02 N W 26 N, 23 N, 9.5 N, 21 N

  12. + 3.52 kg s = .82 k = .37 -  = 42.0o • F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N • Will it stay on the plane if u = 0? • No, it will not stay. The maximum static (FFr(static)<21.02 N) friction is smaller than the gravity parallel to the plane (F|| =23.08 N) W No, Blue

  13. +9.49 N -23.08 N + 3.52 kg s = .82 k = .37 -  = 42.0o • F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N • What will be its acceleration down the plane if it is sliding down the plane? • Down the plane is negative, so we have the parallel force down (-) the plane, and kinetic friction up (+) the plane: • <-23.08 N + 9.49 N> = (3.52 kg)a, a = -3.86 m/s/s = -3.9 m/s/s W -3.9 m/s/s

  14. -23.08 N -9.49 N + 3.52 kg s = .82 k = .37 -  = 42.0o F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N The block is given an initial velocity of 5.0 m/s up the plane. What is its acceleration as it slides up the plane? How far up does it go? As it slides up the plane, the parallel force is down (-) the plane (always), and since the velocity is up the plane, the kinetic friction is now also down (-) the plane: <-23.08 N - 9.49 N> = (3.52 kg)a, a = -9.25 m/s/s = -9.3 m/s/s v = 0, u = 5.0 m/s, a = -9.25 m/s/s, v2=u2+2as, s = 1.35 m = 1.4 m W -9.3 m/s/s, 1.4 m

  15. F = ? -23.08 N -9.49 N + 3.52 kg s = .82 k = .37 -  = 42.0o F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N What force in what direction will make it accelerate and slide up the plane at 6.7 m/s/s? So now we have an unknown force F probably up the plane, and the parallel force as always down (-) the plane, and friction which is down (-) the plane, because the velocity is up the plane, as well as an acceleration that is up (+) the plane: <-23.08 N - 9.49 N + F> = (3.52 kg)(+6.7 m/s/s), F = 56.15 N = 56 N W 56 N

  16. a = -2.5 m/s/s +9.49 N F = ? -23.08 N + 3.52 kg s = .82 k = .37 -  = 42.0o F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static)< 21.02 N What force in what direction will make it accelerate and slide down the plane at 2.5 m/s/s? So now we have an unknown force F , and the parallel force as always down (-) the plane, and friction which is up (+) the plane, because the velocity is down the plane, as well as an acceleration that is down (-) the plane: <-23.08 N + 9.49 N + F> = (3.52 kg)(-2.5 m/s/s), F = 4.797 N = +4.8 N W 4.8 N

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