1. Vertical Suspension Bowling Ball PIPE ROPE Lift Vertical Up Gravity Vertical Down There are Two Forces acting on the Ball There are no Horizontal Forces F h = 0 There is no Vertical Acceleration, consequently F lift + F gravity = 0

2. Horizontal Movement There are Four Forces acting on the Box of Bowling Balls Accelerating to the Right Gravity The Force of Gravity on an Object is commonly known as Weight. Normal Force The Force the supporting surface exerts on the object is known as the Normal Force. Pull Friction

3. PullFriction Gravity Normal Force In order for the Box to Accelerate, the pull force must exceed the frictional force, causing a net horizontal force. F f + F p = F h There is no Vertical Acceleration, consequently F g + F n = 0 We’ll come back to friction in a bit…

4. Friction Gravity Normal Force Movement on an Inclined Plane There are Three Forces acting on this Box of Bowling Balls Axis of Motion Box sliding down a rake

5. Friction Gravity Normal Force Axis of Motion Box sliding down a rake Tilting the Frame of Reference

6. Gravity Axis of Motion Vertical Down Horizontal To the Left Angle of the Incline Off the Horizontal (  Box sliding down a rake Breaking the Force down into Vertical and Horizontal Components Trigonometric Funtions F v = -(F g * cos  F h = -(F g * sin  F g = Weight “-” indicates direction

7. Box sliding down a rake Friction Gravity Normal Force Axis of Motion FgvFgv FghFgh

8. Box sliding down a rake Friction Gravity Normal Force Axis of Motion FgvFgv FghFgh FghFgh Friction FgvFgv Normal Force The Box is Accelerating Horizontally. Consequently, there is a Net Horizontal Force. F g h + F f = F h There is no Vertical Acceleration, so… F g v + F n = 0

9. Assigning Values to the Terms Gravity = Weight Normal Force = Weight * cos  How to Calculate Friction. What we know so far… What’s Next… When Gravity is the only other Vertical Force

10. Three Kinds of Friction Static Kinetic Rolling Pulling Force If F f < F p, then Acceleration If F f = F p, then Constant Velocity Once moving, if F f > F p, then Deceleration

11. How to Find the Frictional Force F f = F n *   is the  Coefficient of Friction Coefficients of Static and Kinetic Friction are unitless numbers that quantify how well two surfaces will slide against one another. These Coefficients are specific to the material combination NOT any individual material. IE. Rubber on Asphalt:  s =.6,  k =.4

12.  s  k Rubber on Asphalt.6.4 Steel on Ice.1.05 Steel on Steel (dry).6.5 Steel on Steel (greased).1.05 Rope on Wood.5.3 Teflon on Steel.04.04 Teflon on Teflon.04.04 Shoes on Ice.1.05 Rubber Soled Shoes on Wood.9.7 Leather Soled Shoes on Wood.3.2 Climbing Boots on Rock1.0.8

13.  s  k On UHMW Lauan w/grain.25.23 Lauan against grain.31.27 Poplar.25.22 Oak.27.23 Steel.21.18 On MDF UHMW.26.22 MDF.45.36 Pine.40.28 On Glazed Lauan UHMW.29.22  Values for Theatre, w/ Uncertainty

14. Say the Box and its contents weigh 50 lbs. The base of the box is UHMW and the surface the box is sitting on is MDF. F f = F n *  s Find F n F n = F g * cos 0 F n = 50 lbs * 1 = 50 lbs  s of UHMW on MDF is.26 So… F f = 50 lbs *.26 F f = 13 lbs  k of UHMW on MDF is.22 So to keep the box moving… F f = 50 lbs *.22 F f = 11 lbs

15. A Note on Surface Area Except in extreme circumstances, the Surface Area of the materials does NOT affect the Friction between the materials. More surface to bond, Less Pressure between surfaces. Less surface to bond, More Pressure between surfaces. Advantage of less surface is cancelled by the increased pressure.

16. Rolling Friction v. Static and Kinetic Friction Similar to Static and Kinetic Friction: F f = F n *  r Different from Static and Kinetic Friction: Rolling Friction is a property of an individual caster. The wheel and bearing materials, diameter of the wheel, and type of construction define the COF for any given caster. As long as the wheel doesn’t slip on the surface, the surface material is negligible in determining the  r. More Casters does not dramatically effect overall Friction Expressed as a Coefficient:  r

17. Where to find Coefficients of Friction for your application Online Manufacturer’s Specs Measure it Yourself COF are affected by minor variances in materials and manufacturing, dust, dirt, impurities, age and condition of surfaces.

18. 940 lbs  s for UHMW on glazed lauan =.29 Define the Forces and Draw the FBD Gravity Normal Force Drive Friction Gravity = Weight = -940 lbs Normal = Gravity = 940 lbs Friction s = Normal *  s 940 lbs *.29 = -272.6 lbs Drive s = ma + Friction s 32 ft over 4 s, accelerated in the wings over 4 s Friction k = Normal *  k 940 lbs *.27 = -253.8 lbs  k for UHMW on glazed lauan =.27 m = Fg/g = 940 lbs/32.2 ft/s 2 = 29.2 slugs UHMW on Glazed Lauan Drive k = ma + Friction k 29.2 slugs * 2 ft/s2 + 272.6 lbs = 331.0 lbs 29.2 slugs * 0 ft/s2 + 253.8 lbs = 253.8 lbs

19. 940 lbs Describe the motion Max Velocity = d/t = 32 ft/s / 4 s = 8 ft/s 32 ft over 4 s, accelerated in the wings over 4 s UHMW on Glazed Lauan Acceleration = v1-v0 / t = 8 ft/s – 0 ft/s / 4 s = 2 ft/s 2 Deceleration = v1-v0 / t = 0 ft/s – 8 ft/s / 4 s = -2 ft/s 2 Velocity in ft/s Time in s

20. 940 lbs  r for a generic wagon on decent casters =.05 What if it were on wheels? Gravity Normal Force Drive Friction Gravity = Weight = -940 lbs Normal = Gravity = 940 lbs Friction s = Normal *  r 940 lbs *.05 = -47 lbs Drive 0 = ma + Friction = 105.4 lbs 32 ft over 4 s, accelerated in the wings over 4 s Drive 1 = ma + Friction = 47 lbs On Wheels

21. HP = F * ft / s 550 Weight = 940 lbs UHMW on glazed lauan =.29 Define the Forces Gravity Normal Force Pull Friction Gravity = Weight = 940 lbs Normal = Gravity = 940 lbs Friction = Normal *  s 940 lbs *.29 = 272.6 lbs Pull = Friction = 272.6 lbs 32 ft over 4 s, accelerated in the wings over 4 s Calculate the HP HP = 141 lbs * 32 ft / 4 s 550 HP = 272.6 lbs * 8 ft/s 550 *.08 HP = 2.05 lb ft/s

22. 32 ft over 4 s 2.05 HP will generate 141 lbs of Pull Force at this velocity Static Friction = 141 lbs Pull Force > Static Friction to Accelerate Object Weight = 940 lbs UHMW on Waxed Wood,  s =.15 How much HP do we need to accelerate the box?

23. 32 ft over 4 s F n = m * a Weight = 940 lbs UHMW on Waxed Wood,  s =.15 Force = Mass * Acceleration (F f + F n ) * v 550 HP = (F f + lbs/32 * a) * v 550 HP = (141 lbs + 940 lbs/32 ft/sec 2 * 8 ft/s 2 ) * 8 ft/s 550 HP = a = 8 ft/sec 2 HP = 2.48 lbft/s

24. 32 ft over 4 s HP by friction HP by acceleration Weight = 940 lbs UHMW on Waxed Wood,  s =.15 2.04 2.48 2.72 3.30 Design Factor: Do not use more than 75% of name plate rating to overcome friction or if you will accelerate for more than 5 sec.