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Newton's law of motion Newton’s second law ( law of inertia ) Every body remains in a state of rest or uniform velocity unless acted on by a resultant / net / unbalanced force. i.e. F = 0 Dv = 0 Inertia is the tendency of a body to maintain its state of rest or of constant velocity. Mass is a measure of inertia. A large mass requires a large force to produce a particular acceleration. 1 Newton’s second law ( law of dynamics ) The rate of change of momentum of a particle is directly proportional to the resultant force acting on it. i.e. F ∝ d(mv)/dt or F = k d(mv)/dt If the mass is constant, F = kmd(v)/dt = k ma i.e. The acceleration of an object is directly proportional to the unbalanced force acting on it. 1 N is defined as the force which gives a mass of 1 kg an acceleration of 1 ms-2 . ∴ k = 1 and F = ma For F = 0 a = 0 if m ≠ 0 Dv = 0 v is constant. i.e. First law is a special case of second law. 2 Sand drops vertically with 2 kgs-1 on to a conveyor belt moving horizontally with 0.1 ms-1. Calculate the external force required to keep belt moving with 0.1 ms-1. By F = d(mv)/dt = v(dm/dt) (v is constant) External force required = 0.1 x 2 = 0.2 N v = 0.1 ms-1 F 3 A jet of water emerges from a hose of cross-section area 5 x 10-3 m2 with a speed of 0.3 ms-1 and strikes a wall at right angles. Calculate the force exerted on the water by the wall assuming the water in brought to rest and does not rebound. (Density of water = 1000 kgm-3) 0.3 m s -1 5 x 10-3 m2 water 4 0.3 m In 1 s 5 x 10-3 m2 water In 1 second: Volume of water striking the wall = 5 x 10-3 x 0.3 = 0.0015 m3 Mass of water striking the wall = 1000 x 0.0015 = 1.5 kg (density = mass / volume) The force exerted on the water = rate of change of momentum of the water = (mv – mu) / t = (0 – 1.5 x 0.3) = -0.45 N Negative sign means the force is opposite in direction to the speed of water hitting the wall. 5 Mass The mass m of an object is a measure of its inertia. Inertia is the resistance of an object to change its state of rest or uniform motion. An object with the greater mass is more difficult to start or stop moving. 6 Weight The weight W of an object is the gravitational force acting on it towards the centre of the earth. Weight is a force and W = mg. The weight of a body is proportional to its mass. Thus, we can find the mass of a body m by comparing its weight with that of a standard masse m0 on a beam balance (天秤). In the same place, W = mg and W0 = m0g. If the beam balance is in equilibrium, W = W0 ⇒ m = m0. 7 Mass Weight varies with position on the earth’s surface independent of the presence decreases if the object becomes or absence of the earth far away from the earth scalar vector constant 8 Newton’s third law ( Action and Reaction ) If body A exerts a force on body B, then B will exert an equal and opposite force on A. i.e. FAB = FBA Forces never occur alone but always in pairs. Action and reaction act on different bodies. 9 A worker of mass 60 kg standing on a lift pulls down a rope with a force. The mass of the lift is 20 kg and the acceleration of the lift is 0.5 ms-2. (a) Draw free body diagrams for the man and the lift. Free body diagrams 10 (a) Free body diagrams T R T R: reaction from the lift R’: from acting on the lift by the man T: tension Mg: weight of the man mg: weight of the lift Mg mg Any action and reaction pair? R’ 11 (b) Find the tension of the rope and normal reaction on the man. R = R’ (Action & Reaction) T For the man, R T R + T – Mg = Ma R + T – 600 = 60 x 0.5 R + T = 630 --- (1) For the lift, T – mg – R’ = ma T – 200 – R = 20 x 0.5 T – R = 210 --- (2) Mg (1) + (2): mg 2T = 840 ⇒ T = 420 N (1) – (2): R’ 2R = 420 ⇒ R = 210 N 12 Example 4 A long trolley of mass M with a flat top is placed on a smooth horizontal bench. A block of mass m is projected horizontally on the top of the trolley with speed u and the coefficient of kinetic friction is m. u (a) (b) Draw free body diagrams for the block and the trolley. (i) the deceleration of the block, the acceleration of the trolley and the distance traveled by the block before it stops moving on the trolley. (ii) the final common velocity of the system. 13 Example 4 (a) Draw free body diagrams for the block and the trolley. u R f f’ mg R’ f: frictional force acting on the block f’: frictional force acting on the trolley mg: weight of the block Mg: weight of the trolley R: reaction on the block R’: force acting on the trolley from the block S S: reaction on the trolley from the floor Any action and reaction pair? Mg 14 (b) (i) Find the deceleration of the block and the acceleration of the trolley before the block stop moving. R f S f’ mg R’ For the block, f = mR and R = mg f = mmg Deceleration = f/m = mg Mg For the trolley, f’ = f (action and reaction) Acceleration = f’/M = mmg/M 15 Find the distance traveled by the block before it stops moving on the trolley. For the block, Initial velocity = u Final velocity = 0 (relative to the trolley) Acceleration = -mg By 2as = v2 – u2 2(-mg)s = 02 – u2 s = u2/(2mg) 16 (b) (ii) the final common velocity of the system. u Common velocity v At rest By conservation of momentum, mu + 0 = (m + M)v v = mu/(m + M) 17