5 additional applications of newton’s laws

Download 5 Additional Applications of Newton’s Laws

If you can't read please download the document

Upload: duyen

Post on 08-Jan-2016

49 views

Category:

Documents


3 download

DESCRIPTION

5 Additional Applications of Newton’s Laws. Friction Drag Forces Motion Along a Curved Path The Center of Mass Hk: 31, 43, 53, 57, 67, 81, 91, 101. Friction. Surface Force opposing relative motion Component of Contact Force (other component is the Normal Force) - PowerPoint PPT Presentation

TRANSCRIPT

  • 5 Additional Applications of Newtons LawsFrictionDrag ForcesMotion Along a Curved PathThe Center of Mass

    Hk: 31, 43, 53, 57, 67, 81, 91, 101.

  • FrictionSurface Force opposing relative motionComponent of Contact Force (other component is the Normal Force)Characterized by coefficients (mu)Static (varies 0 to max)Kinetic (~ constant)Rolling (~ one tenth of kinetic)

  • Surface Dependence of Friction0

  • Normal Force Dependence of FrictionContact area ~ to Normal ForceFrictional Force ~ Normal ForceSummary:Two factors affect frictional force;Surface compositionNormal Force/

  • *Equations for Friction0

  • *Example FrictionEx. 10kg block. FN = weight = mg = 98N. Static coef. = 0.50; Kinetic coef. = 0.30. 0

  • Block at rest. Draw a Force Diagram for the block.0

  • Three boxes are pushed by force F with v > 0 along a horizontal surface with mk = 0.291.0

  • Derive the Angle of Repose relation:Maximum angle block remains at rest:

  • Atwood with Friction. m1=1kg m2=2kg. Kinetic friction = 0.5.

  • Drag Forces

  • Motion Along a Curved PathForce required turn and to change speedCoordinates usually used are F/B (tangential) and L/R (radial)Sum forces tangential = mass x tangential accelerationSum forces L/R (radial centripetal) = mass x centripetal acceleration/

  • What is the fastest speed the car can go without sliding? Assume the car has m = 1200kg and ms = 0.92. 0

  • A block loops the loop. Which force diagram is correct for when it passed point D?

  • Center of Mass Definition

  • Center of Mass Acceleration

  • Center of Mass when Net External Force is ZeroZero Net Force implies center of mass acceleration is also zero. So if CM originally at rest, it remains at rest. If CM moving, its velocity remains same.Example: two people standing on ice push off one another/

  • Ex. Center of Mass. A 100kg person walks 6 feet forward in a 50kg canoe. How far did he move relative to the shore?

  • SummaryFriction depends on Surface Composition and Normal ForceDrag Force vary with speedForce required to move along curved path even at constant speedCenter of Mass stays same when only internal forces operate

  • Can you stop in time?Buggy rolls. You slide.

  • Given m = 75kg, M = 20kg, D = 3.5m, vo =1.1m/s. What frictional coefficient is needed?Insert values, determine ax.

  • Diagramming Refresher:

  • Accelerating with F2WD.Stopping with 4W Disc-Brakes

  • A 3kg box at rest on level surface with ms = 0.55. What is the largest F acting 60 below horizontal for which the box remains at rest?(60 is close to maximum angle)

  • vpcvpgRelative vs. Absolute Velocity

  • Which force diagram applies to the object at Point B?

  • Which force diagram applies to the object at Point C?

  • Assume mass = 1.2kg and radius = 45cm. If speed at Point D is 3.6m/s, what is the size of the normal force acting at Point D?-cen+cen

  • Q. Assume mass = 1.2kg and radius = 45cm. If speed at Point B is 5.1m/s, what is the size of the normal force acting at Point B?

  • Given: T = 50N, q = 30, r = 1m Find: mg and v.

  • The speed is now 6.5m/s and r = 1.0m. Angle, tension, mass?For example, if m = 1.0kg,then T = 42.3N.

  • Q. The speed of a mass on a string of length L is 6.5m/s. The radius r = 2.0m. Find angle, tension, mass, and L.For example, if m = 1.0kg,then T = 23.9N.

  • Practice Q: What is F such that 0.5kg block stays at rest if all surfaces are frictionless?

  • Banked Turn:

  • Banked Turn:

  • Modified Atwood Machine with friction.Let m1 = 2kg, m2 = 3kg, q = 30, sliding friction coeff. 0.44Objects are in CW motion.

  • Q. Recalculate last problem with m1 = 6kg m2 = 1kg. (All else remaining the same)

  • Practice Q: Find the variable relationships.

  • Figures

  • Q. Assume the car has m = 1200kg and ms = 0.92.How large is the frictional force if v = 15m/s?0

  • 5-3Drag Forces0

  • Drag ForcesCan be approximated as,Fdrag = bvnwhere b and n are constants0

  • Example: Air drag, n = 2.If b = 25N/(m/s)2, at what speed would the object be resisted by 10N?At what speed would the same object be resisted by 30N?0

  • Drag force grows quickly with v:Terminal Velocity:Reached when drag force equals weight force0

    **