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Physic 231 Lecture 7 • • • Newton’s N ’ 3d Law: L – “When a body exerts a force on another, the second body exerts an equal oppositely directed force on the first body.” Frictional forces: – kinetic friction: f k = μ k N – static friction f s < μs N Examples. Example F1 andd F2, actt on the • Two T forces, f th 5.00 5 00 kg k block bl k shown h in i the th drawing. d i The magnitudes of the forces are F1=60 N and F2 = 25 N. What is the horizontal acceleration (magnitude and direction) of the block? F1 600 Normal Normal force cancels the y components of F1 and W F1,y + Normal y − W = 0 F2 W F1 60 N F2 25 N m 5 kg θ1 60o a ? ( ) F1,x = 60 cos 60o N = 30N Fnet = F1,x − F2 = 30N − 25N = 5N a = Fnet / m = 5N / 5kg = 1m / s 2 Block accelerates to the right. conceptual question • Consider C id a person standing t di iin an elevator l t th thatt iis accelerating l ti upward. The upward normal force N exerted by the elevator floor on the person is – a)) llarger th than – b) identical to – c) smaller than the downward weight W of the person. Example: with balance scale marked in Newtons, not the usual scale, but a quite reasonable one. • A 100 kkg man stands t d on a bbathroom th scale l iin an elevator. l t St Starting ti from f rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.80 s. It travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.5 1 5 s and comes to rest. rest What does the scale register (a) before the elevator starts to move? (b) during the first 0.8 s? (c) while the elevator is traveling at constant speed? (d) during the Scale reading g =Normal force negative acceleration? N ? m 100 kg vf 1.2 m/s Δt 0.8 s N − mg = ma : scale = N (in Newtons) a) : a=0; N = mg : scale = N = 980 Newtons b) : N − mg = ma : scale l =N a = Δv / Δt = (1.2m / s) / 0.8s = 1.5m / s 2 N = m(g + a);scale = m(g + a) scale = (100kg)(9.8m / s 2 + 1.5m / s 2 ) = 1130Newtons N vi 1.2 m/s vf 0 Δt 1.5 s c) : N = mg : scale = N = 980Newtons d) : a = (v f − vi ) / Δt = −1.2m / s / (1.5s) = −0.8m / s 2 N = m(g + a);scale = m(g + a) = 900Newtons No Acceleration: Static Equilibrium • All objects bj t are att restt andd remain i so. The Th nett force f on any object bj t mustt vanish. I.e. on an object: Fi = 0 i • Example: Three ropes are arranged so as to support a 4 kg mass as shown below. Determine the tension in each rope. ( ) Fi,x = 0 = − T1 + T2 cos 600 T2 T1 A T3 4 kg i 0 = − T1 + T2 / 2; T1 = T2 / 2 60o ( ) Fi,y = 0 = T2 sin 600 − T3 i T3 ( ) T2 = T3 / sin 600 = 1.16T3 4 kg Solve by considering forces at point A mg T3 = mg = 39.1N;T 39 1N;T2 = 45.3N 45 3N T1 = 22.6N Atwood’s machine • • Consider the Atwood machine to the right. The massless string passes over a massless and frictionless pulley. It is under tension T, which we define to be the magnitude of the tension force. By this definition it is a positive number. Choosing up to be positive , what is the net force on mass 1? – a) T-m1g – b) T+m1g – c) m1g-T – d) none of the above m1 m2 Atwood’s machine • Choosing g up p to be ppositive , what is the net force on mass 2? – a) m2g-T – b) T+m2g – c) T-m2g – d) none of the above m1 m2 Atwood’s machine • From Newton’s 2nd law: T − m1 g = m1a1 T = m1a1 + m1 g • Also T − m2 g = m2a2 m1 T = m2a2 + m2 g • If m2 exceeds m1, m2 g goes down and m1 goes up. If m1 exceeds m2, m1 goes down and m2 goes up. In either case Δy2 = − Δy1 v2 = − v1 a2 = −a1 m2 • Putting it together: m1a1 + m1 g = m2a2 + m2 g = −m2a1 + m2 g m1a1 + m2a1 = m2 g − m1 g a1 f ( m2 − m1 )g ⇐ net force = pulling m up 1 total mass m1 + m2 and m down. 2 Reading Quiz 4. An action/reaction pair of forces A. B B. C. D. point in the same direction. act on the same object. object are always long-range forces. act on two different objects. Slide 4-13 Newton’s Third Law • Wh When a body b d exerts a force f on another, h the h secondd body b d exerts an equal oppositely directed force on the first body. – Note: the two forces act on different bodies body 1 body 2 • Force on body 2 due to body 1: F21 • Force on body 1 due to body 2: F12 • 3d law: l F21=- F12 Example • Two skaters, T k t an 82 kg k man andd a 48 kg k woman, are standing t di on ice. i Neglect any friction between the skate blades and the ice. The woman pushes the man with a force of 45 N due east. Determine the accelerations (magnitude and direction) of the man and the woman. woman x components (West is positive) −45N = 0.55m / s 2 east a man = 82kg 45N a woman = = 0.94m / s 2 west 48kg East quiz • Two skaters, T k t an 100 kg k man andd a 50 kg k woman, are standing t di on ice. i Neglect any friction between the skate blades and the ice. By pushing the man, the woman is accelerated at 2 m/s2 in the direction of due west What is the corresponding acceleration of the man? west. – a) 4 m/s2 due east – b) 1 m/s2 due east – c) 2 m/s2 due east – d) 1.5 m/s2 due east Fwoman _ on _ man = − Fman _ on _ woman m man a man = − m woman a woman East choosing west to be negative (just to be contrary) m wo 50 woman a a man,x = − a woman,x = − ( −2m / s 2 ) = 1m / s 2 east m man 100 Note: the direction would still be east even if you chose west to be positive Example F = 60 N • N12 N 21 m1=10 kg m2= 5 kg A bl blockk with ith mass 5 kg k andd a secondd block bl k with ith mass 10 kg k are supported by a frictionless surface. A force of 60 N is applied to the 10 kg mass. What is the force of the 5 kg block on the 10 kg block? x components: t body 2: N 21,x = m 2a x ; N 21,x = − N12,x from Newtons 3d law body 1: Fx + N12,x = m1a x ; Fx = − N12,x + m1a x Fx = N 21,x + m1a x = m 2 a x + m1a x = ( m 2 + m1 ) a x a x = Fx / ( m 2 + m1 ) N12,x = − N 21,x = − m 2 a x = − m 2 Fx / ( m 2 + m1 ) N12,x = −5kg 60N / (15kg) = −20N N12,x = 20N to the left If objects move together, the acceleration is governed by the total mass Friction • • Friction F i ti impedes i d the th motion ti off one object bj t along l the th surfaces f off another. th It occurs because the surfaces of the two objects temporarily stick together via “microwelds”. The frictional force can be larger if the two surfaces are at rest with respect to each other. other Experimentally we have two cases: N v fk – kinetic friction: f k = μk N – static friction f s < μs N fs N F • The coefficient of static friction generally exceeds that for kinetic friction: μ s > μk • Frictional F i ti l forces f always l oppose the th motion ti off one surface f with ith respectt to the other. Example with static friction • Consider C id the th figure fi below, b l with ith M1=105 105 kkg and d M2=44.1 44 1 kkg. Wh Whatt is i the minimum static coefficient of friction necessary to keep the block from slipping. T = M 2g If M1 doesn’t move T = f s ≤ μs N = μs M1g Putting it together M 2 g ≤ μs M1g M2 ≤ μs M1 If this isn’t true, M1 will slip