1 Topics � Newton’s Laws � Mechanical Behaviors of Bodies in Contact � Kinetic Relationship Mechanical Behavior of Bodies in Contact � Friction � Static, dynamic and rolling � Momentum � Conservation of momentum � Impulse � Impulse and momentum relationship � Impact � Coefficient of elasticity & restitution Friction � A force acting at the interface of surfaces in contact � Opposite to the direction of motion or impending motion Fa W Fs motion impending motion � F s = μ s R � μ s : coefficient of static friction � R: normal force (reaction to the object’s weight) R 2 Types of Friction � Static friction � Kinetic friction Rolling friction � friction Static Friction � No applied force (F a ) � No friction, no motion � F a <= F s,max Fa W Fs R � F a = F s � No motion or impending � F a > F s,max � Coefficient of friction ↓ � Kinetic friction � F k < F s,max Applied force F s,max F k F s Static Friction � F s,max = μ s R � μ s = F s,max /R Nt f f Fa W F � Nature o sur aces � Interactions of molecules � Dry surfaces � Constant, regardless of contact area s R 3 Kinetic Friction � F k = μ k R � μ k = F k / R � F 375 N Fs R Example – Moving a box � Push or Pull? � Direction of force application N F NF F Ff f •Push • More difficult? •Pul •Easier? N F Ff N F Ff 5 Example – Race Car Tires � Is a wider tire better in race cars? � Increased friction? � Not really � Fs = μ R � Why wider tire? � Increased weight Other Applications � Friction good or bad? � Benefits � Increased stabilityIncreased st bilit � Performance enhancement � Acceleration & sudden change of direction � Drawbacks � Injuries on artificial turf Other Applications � Shoe design � Material � Stud types 6 Determination of Coefficient � Force platform/force sensors C ffi i t f f i ti � oe c en o r ction � Shoes on diff. surfaces Momentum � The quantity of motion that an object possesses � At an instant V m � M = mv � Unit: kg m/s � Vector Conservation of Momentum � The total momentum in a particular direction does not change unless an external force acts on the m 2 Block at rest external acts on the system in the direction of motion � M 1 + M 2 = M 1+2 frictionless m 2 m 1 v 1 Block at rest Before impact 7 Example � M 1 + M 2 = M 1+2 � Before impact � M 1 = m 1 v 1 � M =0 frictionless m 2 m 1 v 1 Block at rest 2 = 0 Before impact frictionless v f m 1 + m 2 After impact •After impact � M 1+2 = (m 1 +m 2 )v f Example � A 90 kg hockey player traveling at a velocity of 6 m/s collides head-on with an 120 kg player traveling at - 7m/s What is their m a m b . What is combined velocity after collision? v a v b Example – Cont. � Known � m a = 90 kg, V a = 6 m/s � m b = 120 kg,V b = -7 m/s � Unknown V a+b = ? m a m b � M a + M b = M a+b � m a V a + m b V b = m a+b V a+b ()()( )()( )90 6 120 7 90 120kg kg kg V m s m s a+b +−=+ 540 840 210kg kg kg V m s m s a+b ⋅− ⋅= ⋅ V kg 210 a+b m s m s =− ⋅ =− 300 143. In the direction of player B. v a v b 8 Impulse � An external force applied over a period of time � An accumulative event Impulse: I=F t F vm � I = • � T: amount of time � F: force � Unit: Ns � Vector Example 0 200 400 600 800 1000 Fo rc e (N) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Time (sec) Typical Vertical GRF - Gait F1 F2 0 200 400 600 800 1000 Fo rc e (N) Time (sec) Typical Vertical GRF - Gait Impulse = F•t Example - Impulse 400 600 800 1000 o rce (N) Typical Vertical GRF - Gait ∑ = n ii tFI � Impulse = F•t � Impulse � Area under the curve � Total amount of effort 0 200 F o Time (sec) =i 1 9 Example - Impulse � Breaking impulse � Decelerating � Negative -300 -200 -100 0 100 200 Forc e (N ) Typical A/P GRF - Gait Peak braking force Peak propulsive force � Propulsive impulse � Accelerating � Positive -300 -200 -100 0 100 200 Forc e (N ) Time (sec) Typical A/P GRF - Gait Propulsive Impulse Breaking Impulse Impulse of Horizontal GRF 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Time (sec) Relationship: Impulse & Momentum � An external impulse applied to a system causes a change of the system’s momentum � ΔI = ΔM � FΔt= mΔv � Derived from the 2 nd law � F = ma Example – Discussion! � Impulse = momentum � ΔI = ΔM � FΔt= mΔv � F and Δt � Pitching � To do what? � � Catching � To do what? � � Any other examples? 10 Comments about Impulse and Momentum � Impulse � An accumulative event � Momentum An instantaneous event � instantaneous event � Impulse of entire takeoff � Change of velocity at the takeoff Vertical GRF in jumping Example � A golf ball of 0.08 kg is struck by a golf club with an average force of 300 N for a duration of 0.01 second. What is the momentum and the velocity of the club head at the time of release? � Known: m = 0.08 kg, Δ t = 0.01s, F = 300 N � Unknown: ΔM = ? V f = ? Example – cont � Known: m = 0.08 kg, Δ t = 0.01s, F = 300 N � ΔM = ? V f = ? � Solution: � FΔt = mΔv � 1) ΔM = Ft = (300 N)(0.01 s) = 3 kg m/s � 2) Ft = m(V f -V i ) m(V F t V Ft m (300 N)(0.01s) 0.08 kg 37.5 f f m s −=⋅ = ⋅ == V i ) 11 Impact � Collision of 2 bodies over a very short period of time � Fast force application � High loading rate = F/Δt Impact Impact � Landing � Running � Jumping 0 500 1000 1500 2000 2500 0 0.1 0.2 0.3 0.4 0.5 V e rt ical G R F ( N ) -500 Time (sec) 12 Types of Impact � Direct impact � Billiard m 1 m 2 u 1 u2 Before impact After impact � Oblique impact � Basketball: bounce pass � Tennis: volley v 1 v 2 After impact incidence angle Reflection angle Coefficient of Restitution � Under the direct impact � Ratio of the difference of velocities before and after impact m 1 m 2 u 1 u2 Before impact After impact v 1 v 2 After impact impactbeforevrelative impactaftervrelative e=− 21 21 uu vv e − − =− Coefficient of Restitution � Coefficient of Elasticity � Perfectly elastic impact � e = 1 Perfectly elastic Perfectly inelastic e=1 e = 0 � Perfectly inelastic impact � e = 0 � Most impacts � 0< e < 1 e = 13 Coefficient of Restitution � Impact of ball with the ground � A special case with ground stationary � Coefficient of r h d u 1 h b v 1 d b h h e =− 1 1 u v e =− szhang Microsoft PowerPoint - Linear Kinetics.pptx