Secton 1. Dynamcs (Newton s Laws of Moton) Two approaches: 1) Gven all the forces actng on a body, predct the subsequent (changes n) moton. 2) Gven the (changes n) moton of a body, nfer what forces act upon t. Revew of Newton s Laws: Frst Law: body at rest remans at rest, a body n moton contnues to move at constant velocty, unless acted upon by an external force. Note: Ths s only true n an nertal frame. Example 1: You are on a tran wth a ball on the floor. The tran accelerates. hat does the ball do? Example 2: You observe the world from a rotatng carousal. ll other objects are changng ther veloctes wrt you as you turn. hat forces must be appled to them to acheve ths? Second Law: force actng on a body causes an acceleraton of the body, n the drecton of the force, proportonal to the force, and nversely proportonal to the mass. Note: Ths s only true n an nertal frame. F Express as a dv =, or F = m a, or F = m m Example 1: Force parallel to velocty. hat does the body do? Example 2: Force always at rght angles to velocty. hat does the body do? e defne the lnear momentum dp F = P = mv, so that e have two ways of measurng mass: Inertal method apply a force and measure acceleraton Gravtatonal method weght the object (no moton) mnertal m grav
Later we wll see that they are the same. Thrd Law: To every acton there s an equal and opposte reacton. Example 1: Upward force on mass s N Downwards force on plnth s = mg nd N = Example 2: rocket engne generates the force F thrust and apples the force F gas to the exhaust, nd F thrust = F gas pplcatons of Newton s Laws of Moton Example 1: Inclned Ramp n 8kg cart s pulled up a frctonless slope nclned at 20. Determne the force f the cart s to move a) th unform moton, b) th an acceleraton of 0.2 m s -2 up the plane. a) Resolve forces parallel to surface, then F = 8 g sn 20 = 26.8N b) Hence F 8 g sn 20 = ma = 8 0. 2, so F = 28.4N
Example 1: The Pulley weghtless cord hangs over a frctonless pulley. mass of 1kg hangs at one end of the cord and a mass of 2kg at the other. Calculate a) the acceleraton of the masses, b) the tenson n the cord, c) the reacton (the upwards force) exerted by the pulley. nalyss: Strng must be at constant tenson T throughout. Let upwards acceleraton be postve. Let strng accelerate at a on 1kg sde Then for the 1kg mass, T mg = ma,.e. T g = a and for the 2kg mass T mg = ma.e. T 2g = 2a a) Elmnatng T, we obtan a = g/3 b) From ether equaton, T = 4/3 g c) Then the reacton s R = 2T = 8/3 g Now solve the same problem for masses m and m Equlbrum of a Sold ody The statc equlbrum of a sold body entals two dstnct condtons: 1) The net force tendng to accelerate t s zero F = 0 Condton of Translatonal Equlbrum Note: ll the forces do not have to go through the same pont. ladder leanng aganst a wall has reacton and frcton forces at each end, whch do not go through the centre or any other sngle pont. 2) The net torque tendng to rotate t s zero T = 0 Condton of Rotatonal Equlbrum Note: body may be n translatonal equlbrum whle out of rotatonal equlbrum. It may also be n rotatonal equlbrum whle out of translatonal equlbrum.
Example: Loaded ar weghtless bar rests on two supports. Several loads are hung from t F F 1m 2m 1m 1½m 1m 1½m 200N 500N 40N 100N 300N Calculate the forces F and F nalyss: 1) Translatonal equlbrum, and note all forces are n y-drecton. So forces smply sum to zero: F = 0 = 200 + 500 + 40 + 100 + 300 F F F + F = 1140N 2) Rotatonal equlbrum, so total moment about any pont s zero. Takng moments about : T = F x = 0 = 200 1 F = 3470 5.5F 3470 So F = = 630.9N 5.5 Therefore F = 1140 F 0 + 500 2 + 40 3 + 100 4.5 F = 509.1N 5.5 + 300 7
Example: Ladder aganst all N ssume wall s frctonless,.e. F = 0 N Let the weght of the ladder be 16 kg, so that = 160 N, and let t be at 60 (π/3 radans) to the ground. Fnd the forces actng at each end of the ladder Fnd the mnmum coeffcent of frcton aganst the ground. F nalyss Translaton equlbrum: Rotatonal equlbrum: Frcton: Equate forces resolved n two axes F = N = N = 160 N Take moments about ½L cos 60º = N L sn 60º N = 80 cot 60º = 46.2 N F = N µ S N µ S 46.2 / 160 0.29
Frctonal Forces Frcton s due to nteractons between the atoms of an object and those of a surface that t touches. Mcroscopc roughness plays a role too. Macroscopc roughness s treated separately. There are two knds of frctonal forces: Statc Frcton, when the surfaces are at rest Knetc or Sldng Frcton, when the surfaces are n relatve moton. In both cases, the frctonal force opposes moton between the surfaces, and ts magntude s F = µ S N F = µ K N In general, µ S µ K where µ S and µ K are the coeffcents of statc / knetc frcton respectvely. Often approxmate to µ S ~µ K =µ. Defntons: F = µ S N s the mnmum force requred to set n moton the surfaces n contact and ntally at rest, when the normal force (contact force) s N. F = µ K N s the mnmum force requred to mantan the relatve moton of the surfaces n contact, when the normal force (contact force) s N. Example: Inclned Plane wth Frcton Sldng Uphll: Resolvng forces parallel to nclned plane, and usng F = ma, a s uphll acceleraton. F pull mg sn θ µ N = ma F pull = mg sn θ + µ mg cos θ + ma Sldng Downhll: Resolvng forces parallel to nclned plane, and usng F = ma, a s downhll acceleraton F pull + mg sn θ µ N = ma F pull = mg sn θ + µ mg cos θ +ma Just balancng, F pull = 0 = a, then µ = tan θ a = 0 for unform moton.
cceleraton wth Varyng Mass: The Rocket rocket at take-off has maxmum mass (payload, structure and fuel) whch decreases durng flght as fuel s used. Ths s a characterstc example of a varyng mass problem. Let the rocket operate by ejectng exhaust at nozzle velocty v e dm at the rate (mass per unt tme). t tme t let the rocket have velocty v ( reltve to an nertal frame and mass m (. The exhaust mass dm that departs n the next tme nterval therefore has velocty v ( = v( ve n the nertal frame. The quantty dm s an nherently negatve quantty because the mass of the rocket changes by an amount dm. Durng a postve mass ( dm) of burned fuel s ejected from the rocket. Conservng momentum p = mv: t tme t, p ( = m( v( t tme t + p ( t + d = ( m + dm)( v + dv) v dm whch expands to p( t + d = mv + mdv + vdm vdm + vedm The momentum s unchanged, so mdv = vedm dv( ve dm( So acceleraton s a( = = m( Integratng ths from ntal condtons, at t = 0, v = 0, m = m 0, v( = v e log e m 0 m( Ths s a very mportant equaton for rocket desgners. It says that the nozzle velocty must be made as hgh as possble, and that the fuel must be as hgh a proporton of the mass as possble. It mples that rockets should be multstage.
ork and Power ork s done when a force acts along a dsplacement. The work s the energy requred to acheve ths. Example: body mass m falls through a heght h. The work done by gravty s Fh = mgh. Only the component of force parallel to the dsplacement s relevant. If the force s at the angle α to the dsplacement, the work s Fs cos α. In vector notaton, ths s = F. s If the path s curved, we may want the dfferental relatonshp, d = F cos α dr = F. dr Power s the rate of dong work. d dr P = = F. = F. v
Knetc and Potental Energy Consder done s = F = ma - a force actng on a partcle to accelerate t. The work = m d = F. dr = m v. dv = ½m v 2 2 ( v ) a. dr = m dv. dr = m dr. dv e defne knetc energy accordngly as ½m v. v = ½mv 2 ork done by a force acceleratng a partcle changes the knetc energy of the partcle. ork may be done by or aganst a force due to a feld. For example, a charged partcle movng n an electrc feld under the Coulomb force, or a mass movng n a gravtatonal feld. In these cases we ntroduce the concept of potental energy. The work done on the partcle by the force s ON = F. dr = de = E E The work done by the partcle s Y = ON = E E In dfferental form, d = F. dr = de Potental Energy Curves e may wrte potental energy as a functon of poston, e.g. n one dmenson, E = E(x). Then de F( x) = dx s the force actng on the partcle. Consequently, mnma and maxma of the potental energy curve are ponts where F = 0 and a statonary partcle wll reman at rest. The mnma are stable (or metastable) and the maxma are unstable.