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Chapter 12 Kinetics of Particles: Newton’s Second Law 1 Kinetics (จลนศาสตร์) • เป็ นการศึกษาการเคลื่อนที่ของวัตถุ โดยพิจารณาถึงแรงที่กระทาต่อ วัตถุน้ นั • เราใช้ Kinetics ในการทานายการเคลื่อนที่ของวัตถุเมื่อมีแรงกระทา หรื อใช้ในการคานวณแรงเพื่อให้วตั ถุน้ นั เคลื่อนที่ตามที่ตอ้ งการ 2 Newton’s Second Law of Motion • ถ้าแรงลัพธ์ที่กระทากับอนุภาคมีค่าไม่เท่ากับศูนย์ อนุภาคจะเคลื่อนที่ดว้ ย ความเร่ งที่เป็ นสัดส่ วนโดยตรงกับแรงลัพธ์ และมีทิศทางเดียวกับทิศทาง ของแรงลัพธ์ โดยอัตราส่ วนของแรงกับขนาดของความเร่ งจะมีค่าคงที่ • Consider a particle subjected to constant forces, F1 F2 F3 constant mass, m a1 a2 a3 • When a particle of mass m is acted upon by a force F , the acceleration of the particle must satisfy F ma 3 Linear Momentum of a Particle หากพิจารณา กฎข้อที่สองของนิวตัน เมื่อมวลของวัตถุคงที่ F m a d v F = m dt d d L = ( m v ) = dt dt L = linear momentum of the particle • Linear Momentum Conservation Principle: If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction. 4 Systems of Units • Of the units for the four primary dimensions (force, mass, length, and time), three may be chosen arbitrarily. The fourth must be compatible with Newton’s 2nd Law. • International System of Units (SI Units): base units are the units of length (m), mass (kg), and time (second). The unit of force is derived, kg m m 1 N 1 kg 1 2 1 2 s s 5 Eqs of Motion in Rectangular Components • Newton’s second law provides F m a • Solution for particle motion is facilitated by resolving vector equation into scalar component equations, e.g., for rectangular components, Fx i Fy j Fz k ma x i a y j a z k Fx max Fy ma y Fz maz Fx mx Fy my Fz mz 6 Eqs of Motion in Tangential and Normal Components • For tangential and normal components, F t mat dv F m t dt F n man Fn m v2 7 Eqs of Motion in Radial & Transverse Components • Consider particle at r and , in polar coordinates, Fr mar mr r 2 F ma mr 2r 8 Dynamic Equilibrium • Alternate expression of Newton’s second law, F m a 0 ma inertial vector • With the inclusion of the inertial vector, the system of forces acting on the particle is equivalent to zero. The particle is in dynamic equilibrium. • Methods developed for particles in static equilibrium may be applied, e.g., coplanar forces may be represented with a closed vector polygon. • Inertia vectors are often called inertial forces as they measure the resistance that particles offer to changes in motion, i.e., changes in speed or direction. • Inertial forces may be conceptually useful but are not like the contact and gravitational forces found in statics. 9 Sample Problem 12.1 SOLUTION: • Resolve the equation of motion for the block into two rectangular component equations. A 200-N block rests on a horizontal plane. Find the magnitude of the force P required to give the block an acceleration or 10 m/s2 to the right. The coefficient of kinetic friction between the block and plane is mk 0.25. • Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns. 10 Sample Problem 12.1 SOLUTION: • Resolve the equation of motion for the block into two rectangular component equations. + Fx ma : P cos 30 0.25N 20.3910 203.9 y O x W 200 N g 9.81 m s 2 20.39 kg F mk N m 0.25 N + Fy 0 : N P sin 30 200 N 0 • Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns. N P sin 30 200 N P cos 30 0.25P sin 30 200 N 203.9 P 343.11 N 11 Sample Problem 12.3 SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. • Write the equations of motion for the blocks and pulley. • Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension. The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord. 12 Sample Problem 12.3 O x y SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. y B 12 x A a B 12 a A • Write equations of motion for blocks and pulley. + F m a : x A A T1 100 kg a A + Fy mB aB : mB g T2 mB a B 300 kg 9.81m s 2 T2 300 kg a B T2 2940N - 300 kg a B + Fy mC aC 0 : T2 2T1 0 13 Sample Problem 12.3 • Combine kinematic relationships with equations of motion to solve for accelerations and cord tension. O x y y B 12 x A a B 12 a A T1 100 kg a A T2 2940N - 300 kg a B 2940N - 300 kg 12 a A T2 2T1 0 2940 N 150 kg a A 2100 kg a A 0 a A 8.40 m s 2 a B 12 a A 4.20 m s 2 T1 100 kg a A 840 N T2 2T1 1680 N 14 Sample Problem 12.4 SOLUTION: • The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus the acceleration of the block relative to the wedge. • Write the equations of motion for the wedge and block. The 5.4 kg block B starts from rest and slides on the 13.6 kg wedge A, which is • Solve for the accelerations. supported by a horizontal surface. Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge. 15 Sample Problem 12.4 SOLUTION: • The block is constrained to slide down the wedge. Therefore, their motions are dependent. aB a A aB A • Write equations of motion for wedge and block. + F m a : x A A N1 sin 30 mA a A y 0.5 N1 mA a A + x Fx mB a x mB a A cos 30 aB A : mB g sin 30 mB a A cos 30 aB A aB A a A cos 30 g sin 30 + Fy mB a y mB a A sin 30 : N1 mB g cos 30 mB aA sin 30 16 Sample Problem 12.4 • Solve for the accelerations. 0.5N1 mAaA , N1 2mAaA N1 mB g cos 30 mB a A sin 30 2m A a A mB g cos 30 mB a A sin 30 aA aA mB g cos 30 2m A mB sin 30 5.49.81 cos 30 213.6 5.4 sin 30 a A 1.54 m s 2 aB A a A cos 30 g sin 30 aB A 1.54 cos 30 9.81 sin 30 aB A 6.24 m s 2 17 Sample Problem 12.5 SOLUTION: • Resolve the equation of motion for the bob into tangential and normal components. • Solve the component equations for the normal and tangential accelerations. The bob of a 2-m pendulum describes an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and acceleration of the bob in that position. • Solve for the velocity in terms of the normal acceleration. 18 Sample Problem 12.5 SOLUTION: • Resolve the equation of motion for the bob into tangential and normal components. • Solve the component equations for the normal and tangential accelerations. + mg sin 30 mat Ft mat : at g sin 30 + Fn man : at 4.9 m s 2 2.5mg mg cos 30 man an g 2.5 cos 30 an 16.03 m s 2 • Solve for velocity in terms of normal acceleration. an v2 v an 2 m 16.03 m s 2 v 5.66 m s 19 Sample Problem 12.6 SOLUTION: • The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface. Determine the rated speed of a highway curve of radius = 122 m banked through an angle = 18o. The rated speed of a banked highway curve is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels. • Resolve the equation of motion for the car into vertical and normal components. • Solve for the vehicle speed. 20 Sample Problem 12.6 • Resolve the equation of motion for the car into vertical and normal components. R cos W 0 Fy 0 : W R cos Fn man : R sin SOLUTION: • The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface. W an g W W v2 sin cos g • Solve for the vehicle speed. v 2 g tan 9.81122tan18 v 19.72 m / s 21 Angular Momentum of a Particle • H O r mV moment of momentum or the angular momentum of the particle about O. • H O is perpendicular to plane containing r and mV H O rmV sin i j k rm v H x y z O mr 2 mv x mv y mvz • Derivative of angular momentum with respect to time, H O r mV r mV V mV r ma rF MO • It follows from Newton’s second law that the sum of the moments about O of the forces acting on the particle is equal to the rate of change of the angular momentum of the particle about O. 22 Conservation of Angular Momentum • When only force acting on particle is directed toward or away from a fixed point O, the particle is said to be moving under a central force. • Since the line of action of the central force passes through O, M O H O 0 and r mV H O constant • Position vector and motion of particle are in a plane perpendicular to H O . • Magnitude of angular momentum, H O rmV sin constant r0 mV0 sin 0 or H O mr 2 constant HO angular momentum r 2 h m unit mass 23 Newton’s Law of Gravitation • Gravitational force exerted by the sun on a planet or by the earth on a satellite is an important example of gravitational force. • Newton’s law of universal gravitation - two particles of mass M and m attract each other with equal and opposite force directed along the line connecting the particles, Mm F G 2 r G constant of gravitatio n 66.73 10 12 m3 kg s 2 34.4 10 9 ft 4 lb s 4 • For particle of mass m on the earth’s surface, MG m ft W m 2 mg g 9.81 2 32.2 2 R s s 24 Sample Problem 12.8 SOLUTION: • Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B. A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18820 km/h from an altitude of 240 km. Determine the velocity of the satellite as it reaches it maximum altitude of 2340 km. The radius of the earth is 6345 km. 25 Sample Problem 12.8 SOLUTION: • Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B. rm v sin H O constant rA m v A rB m v B vB v A rA rB 6345 240 km 18820 km/h 6345 2340 km v B 14269 .4 km/h 26