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Motion on Inclined Planes

Motion on Inclined Planes. A tilted coordinate system is convenient, but not necessary!. Inclined Plane Problems. a. Its important to understand ∑ F = ma & How to resolve it into x,y components in the tilted coordinate system!!. You MUST understand this case to understand

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Motion on Inclined Planes

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  1. Motion on Inclined Planes

  2. A tilted coordinate systemis convenient, but not necessary! Inclined Plane Problems a • Its important to understand∑F = ma& How to resolve it into x,y components in the tilted coordinate system!!

  3. You MUSTunderstand • this case to understand • the case with friction!! • By geometry, the 2 • angles marked θ are • the same! a By Trigonometry: FGx= FGsin(θ) = mgsin(θ) FGy= -FGcos(θ) = -mgcos(θ) FG = mg Both Angles = !

  4. Example: Sliding Down Incline • A box of mass m is placed on a smooth (frictionless!) • incline that makes an angle θwith the horizontal. • Calculate: a)The normal force on the box. • b) The box’s acceleration. • c) Evaluate both for m = 10 kg & θ = 30º Free Body Diagram

  5. Example: The Skier • A skier descends a 30° slope, at constant • speed. What can you say about the • coefficient of kinetic friction? Is the normal force FNequal & opposite to the weight??? NO!!!!!!!

  6. Summary of Inclines:An object sliding down an • incline has 3 forces actingon it: the normal forceFN, • gravity FG = mg, & friction Ffr. FN is always • perpendicular to the surface & is NOTequal & • opposite to the weightmg. The friction force Ffris • parallel to the surface. Gravity FG = mg points down. • If the object is at rest, the • forces are the same except • that we use the static • frictional force, & the • sum of the forces is zero.

  7. Problem FN Ff Newton’s 2nd Law ∑F = ma x: mgsinθ – Ff = ma y: FN - mgcosθ = 0 Friction: Ff = μkFN NOTE!!!  FN = mgcosθ FN  mg mg sinθ mg cosθ FG = mg THE NORMAL FORCE IS NOT EQUAL TO THE WEIGHT!!!

  8. Example: A ramp, a pulley, & two boxes • Box A, mass mA = 10 kg, rests on a surface inclined at • θ = 37°to the horizontal. It’s connected by a light cord, • passing over a massless, frictionless pulley, to Box B, which • hangs freely.(a) If the coefficient of static friction is s= 0.4, • find the range of values for mass B which will keep the • system at rest. (b) If the coefficient of kinetic friction is k= • 0.3, and mB = 10 kg, find the acceleration of the system.

  9. Example: A ramp, a pulley, & two boxes • (a) Coefficient of static friction s= 0.4, find the range of values • for mass B which will keep the system at rest. • Static Case (i): Small mB << mA: mAslides down the incline, so • friction acts up the incline. • Static Case (ii): Larger mB > mA: mAslides up the incline, so • friction acts down the incline. Static Case (ii): Larger mB > mA mAslides up incline Ffr acts down incline Static Case (i): mB << mA mAslides down incline Ffr acts up incline

  10. Example: A ramp, a pulley, & two boxes (b) The coefficient of kinetic friction is k= 0.3, & mB = 10 kg. find the acceleration of the system & the tension in the cord. Motion: mB = 10 kg mAslides up incline Ffr acts down incline

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