1 1. Newton’s first law of motion Any system in mechanical equilibrium remains in mechanical equilibrium unless compelled to change that state by a nonzero total force acting on the system. 1system—a particle or particles whose motion is to be studied. §5.1 Newton’s first law of motion and the concept of force 2 Mechanical equilibrium—a system with a constant velocity is in equilibrium. 2 4Inertial reference frames—the reference frames in which Newton’s first law is valid 3 The causes of acceleration of a system--The first law of motion states that zero total force exists when the system has a constant velocity; a nonzero total force on the system causes its acceleration. 00 total ≠?≠ Fa r r §5.1 Newton’s first law of motion and the concept of force 2. The concept of force 1force is a vector quantity—the vector sum of all the forces on a system is the force on the system. total F r 2 The measurement of force and the unit of force-- N (Newton) (§5.4) §5.1 Newton’s first law of motion and the concept of force 3. The fundamental forces of nature Gravitational force Electromagnetic force Strong force Weak force 3 4. Some common forces 1 Normal force of a surface. (§5.10) 2 Tensions in ropes, strings, and cables;(§5.11) 3 Static friction and kinetic friction; (§5.12~5.13) §5.1 Newton’s first law of motion and the concept of force §5.2 Newton’s second law and third law of motion §5.2 Newton’s second law of motion and Newton’s third law of motion 1. Newton’s second law of motion and its aspects A force of magnitude 1 N on the standard kilogram produces an acceleration of magnitude 1m/s 2 . 1 N=1 kg·m/s 2 P176-177 ? ? ? ? ? = = = ?= zz yy xx maF maF maF amF total, total, total, total r r ? ? ? = = = tt nn maF maF F total, total, r or 4 2. Newton’s third law of motion If a system A exerts a force on another system B, then B exerts a force of the same magnitude on A but in the opposite direction. AonBBonA FF rr ?= A third law force pair are equal magnitude and opposite direction , but act on different systems. §5.2 Newton’s second law and third law of motion §5.3 the limitation to applying Newton’s law of motion §5.3 The limitation to applying Newton’s law of motion 1. Reference frames Newton’s law can only be used in inertial reference frames (that are not being accelerated). 2. Speed limits Newtonian mechanics is a good approximation as long as the speed of the system is much less than the speed of light. 5 3. Quantum mechanics Newtonian mechanics can not describe or account for many phenomena on the atomic and nuclear scale. 4. Force propagation Force somehow take time to propagate from one place to another. Newton’s third law does not account for such propagation delays. §5.3 the limitation to applying Newton’s law of motion 5. Chaotic—nonlinear system One of the hallmarks of Newton’s laws is their ability to predict the future behavior of a system,if we know the forces that act and the initial motion. One of the particular theme of chaotic dynamics is that tiny changes in the initial conditions of a problem can be greatly amplified and can cause substantial differences in the predicted outcomes. §5.3 the limitation to applying Newton’s law of motion 6 §5.3 the limitation to applying Newton’s law of motion The trajectory of the Cassini mission to Saturn 1. Do inertial frames really exist? Inertial reference frame is an ideal model. The earth spins on its axis. The centripetal acceleration is less than 3.4×10 -2 m/s 2 . The surface of the earth is a approximate inertial reference frame. The earth moves around the sun. The centripetal acceleration is 6×10 -3 m/s 2 . §5.4 Some topics of discussion 7 §5.4 Some topics of discussion 2. Noninertial reference frames Bob Alice 0 a r N mg A ? Newton’s law is not valid in the car. A 0 a r A mB mg N 0 a r B mg m N §5.4 Some topics of discussion 8 ω r m A τ r F r n s B O r F r 0 F r ω G r Noninertial reference frames are accelerated reference frames. §5.4 Some topics of discussion §5.4 Some topics of discussion rRr ′ += r r OOOPPO OOOPPO aaa vvv ′′ ′′ += += rrr rrr t r t R t r d d d d d d ′ += r r r O O’ r r r ′ r R r P 3. Applying Newton’s second law of motion in noninertial reference frames 0 aaa rrr + ′ = 9 amamF amamamF ′ =?+ + ′ == rr r rrr r )( 0total 0total The pseudo force arise only because of the acceleration of the noninertial reference frame. Without the pseudo force term, the accelerated frame cannot properly describe the motion of the particle using Newton’s law. 0pseudo amF r r ?=dictate called pseudo force. §5.4 Some topics of discussion How to apply Newton’s laws of motion 1identifies the system; 2 draw the free-body diagrams and illustrates all forces acting on the system with their direction indicated explicitly in the diagram; 3choose a coordinate system; 4make appropriate vector sum of all the forces and describe the motion by using Newton’s second law of motion. §5.5 applications of Newton’s law 10 Example 1: A submarine is sinking in the sea. If the buoyancy is , the kinetic friction is , where A is the area of the cross section of the submarine. Find the function of the speed of the submarine with respect to the time. F r vkAf r r ?= §5.5 applications of Newton’s law c F r f r W r o y ∫∫ = ?? =?? tv t kAvFmg vm t v mkAvFmg 00 d d d d ∫∫ = ?? ? tv t kAvFmg mg-F-kAv kA m 00 d )d( t v mg-F-kAv kA m =? 0 )ln( Solution: yy mamgkAvFF =+??= total, §5.5 applications of Newton’s law 11 ? ? ? ? ? ? ? ? ? ? = = ? ?? = ? ? ? ? t m kA t m kA e kA Fmg v e Fmg kAvFmg t Fmg mg-F-kAv kA m 1 lnv m v o t constant , d d ,00 max = ? ==∞→ ↓↑↑== kA Fmg vvt t v vtvt --terminal speed §5.5 applications of Newton’s law §5.5 applications of Newton’s law Example 2: Block m 1 in the Figure has a mass of 4.20kg and block m 2 has a mass of 2.30kg. The coefficient of kinetic friction between m 2 and the horizontal plane is 0.47. The inclined plane is frictionless. Find (a) the acceleration of the blocks and (b) the tension in the string. 12 §5.5 applications of Newton’s law o rr 27 21 2 21 = == = == θ μ aaa Nf TTT xx kk 2 N r gmW r r 22 = 2 T r k f r θ 1 N r gmW r r 11 = 1 T r θ x y x y ? ? ? =? =? 0cos sin 11 11 θ θ gmN amTgm ? ? ? =? =? 0 22 22 gmN amNT k μ Exercise 1 : As shown in figure, the block B weighs 94kg and block A weighs 29 kg. Between block B and the plane the coefficient of static friction is 0.56 and the coefficient of the kinetic friction is 0.25. (a) find the acceleration if B is moving up the plane. (b) what is the acceleration if B is moving down the plane? o 42 A B §5.5 applications of Newton’s law 13 Exercise 2: in the figure, A is a 4.4 kg block and B is a 2.6 kg block. The coefficients of static and kinetic friction between A and the table are 0.18 and 0.15. (a) Determine the minimum mass of the block C that must be placed on A to keep it from sliding. (b) Block C is suddenly lifted off A. What is the acceleration of block A? A B C §5.5 applications of Newton’s law §5.5 applications of Newton’s law O l θ Exerscise 3: an ideal pendulum with a pear of mass m and an inflexible, massless thread of length l. Keep the pendulum at horizontal initially, and then drop it from rest. When the angle between the thread and the horizontal is , find the speed of the pear and the tension in the thread. θ 14 t mmamg d d cos υ α τ == υ υ α d d d d d d dcos t s ms t msmg == υα == t s ls d d ,dd υυαα ddcos =g l mmamgT n 2 sin θ υ θ ==? Solution: θ θ sin3mgT = θυ θ sin2gl= ∫∫ = θ υθ υυαα 00 ddcosg §5.5 applications of Newton’s law O l θ T r gm r α v r sd αd M θ m Example 3: A block of mass m is placed on a frictionless inclined plane of a triangle block of mass M which is on the frictionless horizontal plane , the inclined plane is at the angle θ to the horizontal. Find (a)the magnitude of the normal force of the inclined plane on the block;(b)the acceleration of m with respect to M. §5.5 applications of Newton’s law 15 θθ sincos gamgN =′= 0cos sin =?= ′ == θ θ mgNF ammgF y x x y m θ a′ r N r gmW r r = Solution: Choose triangle block as the reference frame Draw the forces diagram The results: Are these results right? §5.5 applications of Newton’s law M θ x y Choose the ground as the reference frame: )2(0cos )1(sin =??= == θ θ NMgQF MaNF y Mx M a r For the triangle block θ NN rr ?=′ W r Q r ,0≠ M a M is not inertial reference frame. §5.5 applications of Newton’s law 16 m y x Choose the ground as the reference frame: MGmMmG aaa rrr += Mm aaa rrr + ′ = M a r a ′ r m a r θ θ sin cos Mmy Mmx aa aaa ?= ? ′ = N r W r For block of mass m ( ) )4(sincos )3(cossin θθ θθ Mmyy Mmxx mamamgNF aammamgF ?==?= ? ′ === §5.5 applications of Newton’s law ( ) )4(sincos )3(cossin θθ θθ Mmyy Mmxx mamamgNF aammamgF ?==?= ? ′ === ( ) )4(sincos )3(cossin θθ θθ Mmyy Mmxx mamamgNF aammamgF ?==?= ?′=== )1(sin Mx MaNF == θ )2(0cos )1(sin =??= == θ θ NMgQF MaNF y Mx M: m: §5.5 applications of Newton’s law 17 () θ θ θ θθ θ θ 2 2 2 sin sin sin sincos sin cos mM gmM a mM mg a mM Mmg N M + + = ′ + = + = The results are y x m M a r a ′ r m a r N r W r §5.5 applications of Newton’s law ( ) θ θ θ θθ θ θ 222 sin sin ; sin sincos ; sin cos mM gmM a mM mg a mM Mmg N M + + =′ + = + = ∞→== M20 πθθ N a a M ′ mg 0 0 0 0 g θ θ cos sin 0 mg g m M m M m M §5.5 applications of Newton’s law 18 Choose horizontal plane as reference frame: )2(0cos )1(sin =??= == θ θ NMgQF MaNF y Mx For triangle block of mass M M a r θ M x y NN vr ?=′ Q r W r §5.5 applications of Newton’s law Choose M as the reference frame: )4(0sincos )3(cossin =+?= ′=+= θθ θθ My Mx mamgNF ammamgF For block of mass m y x a ′ r m M amF r r ?= 0 N r W r §5.5 applications of Newton’s law 19 A B 1 m 3 m 2 m §5.5 applications of Newton’s law Exercise 4: There are three blocks of mass m 1 , m 2 , m 3 respectively, and m 1 >m 2 +m 3 . Pulleys and strings are massless, the kinetic friction forces between the pulley and the strings are ignored. (a) find the accelerations of m 1 , m 2 and m 3 . (b) what are the tensions of the two strings. 32 aa ′ = ′ 32321 TTTTT =+= 1111 amTgm =? 333133 amTamgm ′ ?=?+ 222122 amTamgm ′ =?+ B 1 T r 2 T r 1 a r 3 T r 1 m gm r 1 1 T r 1 a r 3 m gm r 3 3 T r 13 am r 3 a ′ r + 2 m gm r 2 2 T r 12 am r 2 a ′ r §5.5 applications of Newton’s law Solution: (a) The free-body diagrams 20 §5.5 applications of Newton’s law B 1 T r 2 T r 1 a r 3 T r 133 aaa rrr + ′ = )2( 133 aaa ? ′ ?=? 3 a r because Assume is upward then From equations (1) and (2) ,we can get a 2 an a 3 . pGpG 22 aaa mm rrr + ′ = because 122 aaa rrr + ′ = )1( 122 aaa ? ′ = 2 a r If the direction of is downward then or (b) 1 m 2 m o A The string is not a inertial reference frame! §5.5 applications of Newton’s law Exercise 5:A small metal tube is sliding down along the string with the acceleration a with respect to the string as shown in Figure. The mass of the block is m 1 , and the mass of the tube is m 2 , the string is massless, the kinetic friction force between the pulley and the string is ignored. Find the accelerations of the block and the tube with respect to the ground, the tension of the string and the friction force between the string and the tube.