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Transcript
Kinematics in One Dimension MECHANICS comes in two parts: kinematics: motion (displacement, time, velocity) x, t, v, a dynamics: motion and forces x, t, v, a, p, F v v0 a t a const v v0 v t 0 0 2 2 1 x x0 v0 t 2 a t v v 2a x x0 2 2 0 Dynamics: Laws of Motion Recall the two parts of MECHANICS - kinematics dynamics Introduce the motivation for motion as a derived concept called FORCE. F = ma Have you seen a force lately? A push or a pull; motion not required What is the effect of a force acting on a free body? F g m accelerated motion Newton’s First Law of Motion Every body continues in its state of rest or uniform speed in a straight line unless acted on by a nonzero net force. Preceding defines INERTIA Preceding defines MASS (m W) Best when observed in the absence of friction. Newton’s Second Law of Motion The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object. a= F m F = m a dp (F= ) dt Newton’s Second Law of Motion F = m a F x m ax F y m ay Units for FORCE kg m/s = N (Newton) 2 g cm/s 2 = dyne (dyne) 2 slug ft/s = lb (pound) Weight, and a Mystery in a Box. You have received a gift box, of mass 10.0 kg, with a mystery object inside. You place it on a table. What is the weight of the box and the normal force on it (from where)? If a force of 40.0 N is added to the box, what is the normal force? If a force of 40.0 N is subtracted from the box, what is the normal force? Weight, and a Mystery in a Box. W = mg = 10.0 kg 9.80 m/s2 = 98.0 N FN = mg Weight, and a Mystery in a Box FN - W - 40.0 N = 0 FN = mg + 40.0 N = 98.0 N + 40.0 N = 138 N Weight, and a Mystery in a Box FN - W + 40.0 N = 0 FN = mg - 40.0 N = 98.0 N - 40.0 N = 58.0 N Force, and the Mystery in a Box The box is now pulled across a frictionless (!) table with a force of 40.0 N at an angle of 30.0. What is its acceleration? What is the normal force exerted by the table? Checking the normal force to see if the box can fly... FN W FPy 0 FN mg FPy 98.0 N 40.0 N sin(30.0) 78.0 N Slip, sliding away… m ax FPx 0 FPx ax m 40.0 N cos(30.0) 10.0 kg 3.46 m/s 2 Still going down the slippery slope... The mystery box slides freely down a ramp 6.0 m long inclined at 9.5. How long does it take the box to reach the bottom? Would this change if the box’s mass were doubled? F = m a Y direction Forces FN - Wy = 0 X direction Forces Wx = m a ma Wx mg sin( ) a g sin( ) 9.80 m/s 2 sin(9.5) 1.62 m/s 2 x 12 at 2 FN = Wy mg cos( ) 98.0 N cos(9.5) 97.7 N 2x t a 2(6.00 m) 1.62 m/s 2 2.7 s One 3.5-kg paint bucket is hanging by a cord from another 3.5-kg paint bucket which is also hanging by a cord. What is the tension in the cords? If the two buckets are pulled upward with an acceleration of 1.60 m/s2, what is the new tension in the cords? T1 - W1 - W2 = 0 T1 = mg + mg = 2 ( 3.5 kg 9.80 m/s2 ) = 68.6 N T2 - W2 = 0 T2 = mg = 34.3 N T1 - W1 - W2 = (m + m) a T1 = 2 mg + 2m a = 2 ( 34.3 N ) + 2 ( 3.5 kg) 1.60 m/s2 = 79.8 N T2 - W2 = m a T2 = mg + m a = 39.9 N The Bane of Galileo: Friction Friction is everywhere! There is little wonder why it played such a prominent part in MECHANICS for the ancients (e.g., Aristotle). Galileo recognized friction as separate from motion, so the equations of kinematics could be discovered (using geometry). Friction - treated as a force (though not a vector) - always opposite to the motion - sometimes related to the motion, sometimes not so related Friction Static friction is always greater than moving (kinetic) friction. • Calculating frictional force, Ffr • The force of friction, Ffr or f, is equal to the normal force times the coefficient of friction. fs ≤µs·N (Static) fk = µk·N (kinetic) • µ is like the % of the force pushing the two substances together which results in friction. Kinetic friction Ffr = k FN Static friction Ffr s FN where s k • Static friction and kinetic friction plotted for a pull versus time. • In lab you will plot Frictional force vs the Mass and the slope will be mµ. A 10.0-kg box rests on a surface and experiences a force of 40.0 N at an angle of 30. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is 0.30. Will the box move? If it moves, how fast will it move? FN - mg + FPy = 0 FPx ? s FN FPx - Ffr = m a FN - mg + FPy = 0 FN = mg FPy 98.0 N 40 N sin(30) 78.0 N FPx FP cos( ) 40.0 N cos(30) 34.6 N Fsf s FN 0.40 78.0 N 31.2 N Since FPx Fsf , the box can move! FPx - Ffr = m a FPx Ffr a m FP cos( ) s FN m 40.0 N cos(30) 0.30 78.0 N 10.0 kg 1.1 m/s 2 A block is placed at the top of an inclined plane. The incline is 30.0 and 9.3 m long, with a coefficient of kinetic friction of 0.17. What is the acceleration of the block? What speed does the block have when it reaches the bottom? FN - Wy = 0 FN = mg cos() Wx - Ffr = m a m a = mg sin() - s FN FN = mg cos() ma = mg sin() - k FN Wx Ffr a m mg sin( ) k mg cos( ) m g sin( ) k cos( ) 9.80 m/s 2 sin(30) 0.17 cos(30) 3.46 m/s 2 How fast does it go? v 2 v02 2ax v 2ax 2 3.46 m/s 2 9.3 m 8.0 m/s Summary: When solving problems: 1) Draw a diagram with the forces. 2) Resolve the components into x and y components. 3) Remember Newton’s Second Law 4) Write Newton’s 2nd law for the x and y components. Next time: Gravity and Centripetal force