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Transcript
Chapters 5, 6 Force and Laws of Motion Newtonian mechanics Sir Isaac Newton (1643 – 1727) • Describes motion and interaction of objects • Applicable for speeds much slower than the speed of light • Applicable on scales much greater than the atomic scale • Applicable for inertial reference frames – frames that don’t accelerate themselves Force • What is a force? • Colloquial understanding of a force – a push or a pull • Forces can have different nature • Forces are vectors • Several forces can act on a single object at a time – they will add as vectors Force superposition • Forces applied to the same object are adding as vectors – superposition • The net force – a vector sum of all the forces applied to the same object Newton’s First Law • If the net force on the body is zero, the body’s acceleration is zero Fnet 0 a 0 Newton’s Second Law • If the net force on the body is not zero, the body’s acceleration is not zero Fnet 0 a 0 • Acceleration of the body is directly proportional to the net force on the body • The coefficient of proportionality is equal to the mass (the amount of substance) of the object ma Fnet Fnet a m Newton’s Second Law • SI unit of force kg*m/s2 = N (Newton) • Newton’s Second Law can be applied to all the components separately • To solve problems with Newton’s Second Law we need to consider a free-body diagram • If the system consists of more than one body, only external forces acting on the system have to be considered • Forces acting between the bodies of the system are internal and are not considered Chapter 5 Problem 14 Three forces acting on an object are given by F1 = (– 2.00^i + 2.00^j) N, F2 = (5.00^i – 3.00^j) N, and F3 = (–45.0^i) N. The object experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What is the mass of the object? Newton’s Third Law • When two bodies interact with each other, they exert forces on each other • The forces that interacting bodies exert on each other, are equal in magnitude and opposite in direction F12 F21 Forces of different origins • Gravitational force • Normal force • Tension force • Frictional force (friction) • Drag force • Spring force Gravity force (a bit of Ch. 13) • Any two (or more) massive bodies attract each other • Gravitational force (Newton's law of gravitation) m1m2 F G 2 rˆ r • Gravitational constant G = 6.67*10 –11 N*m2/kg2 = 6.67*10 –11 m3/(kg*s2) – universal constant Gravity force at the surface of the Earth mEarthmCrate ˆ m1m2 FCrate G 2 rˆ G j 2 r REarth GmEarth mCrate ˆj g mCrate ˆj FCrate 2 REarth g = 9.8 m/s2 Gravity force at the surface of the Earth • The apple is attracted by the Earth • According to the Newton’s Third Law, the Earth should be attracted by the apple with the force of the same magnitude mEarthmApple m1m2 ˆj FEarth G 2 rˆ G 2 r REarth aEarth G mEarthm Apple 2 Earth R mEarth m Apple GmEarth mApple ˆ ˆj ˆj g j R2 m mEarth Earth Earth Weight • Weight (W) of a body is a force that the body exerts on a support as a result of gravity pull from the Earth • Weight at the surface of the Earth: W = mg • While the mass of a body is a constant, the weight may change under different circumstances Tension force • A weightless cord (string, rope, etc.) attached to the object can pull the object • The force of the pull is tension ( T ) • The tension is pointing away from the body Free-body diagrams Normal force • When the body presses against the surface (support), the surface deforms and pushes on the body with a normal force (n) that is perpendicular to the surface • The nature of the normal force – reaction of the molecules and atoms to the deformation of material Normal force • The normal force is not always equal to the gravitational force of the object Free-body diagrams Free-body diagrams Chapter 5 Problem 28 Two objects are connected by a light string that passes over a frictionless pulley as shown. Draw free-body diagrams of both objects. Assuming the incline is frictionless, m1 = 2.00 kg, m2 = 6.00 kg, and θ = 55.0° find (a) the accelerations of the objects, (b) the tension in the string. Frictional force • Friction ( f ) - resistance to the sliding attempt • Direction of friction – opposite to the direction of attempted sliding (along the surface) • The origin of friction – bonding between the sliding surfaces (microscopic cold-welding) Static friction and kinetic friction • Moving an object: static friction vs. kinetic Friction coefficient • Experiments show that friction is related to the magnitude of the normal force • Coefficient of static friction μs f s ,max s n • Coefficient of kinetic friction μk f k k n • Values of the friction coefficients depend on the combination of surfaces in contact and their conditions (experimentally determined) Free-body diagrams Free-body diagrams Chapter 5 Problem 42 Three objects are connected on a table. The rough table has a coefficient of kinetic friction of 0.350. The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg, as shown, and the pulleys are frictionless. Draw a free-body diagram for each object. (a) Determine the acceleration of each object and their directions. (b) Determine the tensions in the two cords. Drag force • Fluid – a substance that can flow (gases, liquids) • If there is a relative motion between a fluid and a body in this fluid, the body experiences a resistance (drag) • Drag force (R) R = ½DρAv2 • D - drag coefficient; ρ – fluid density; A – effective cross-sectional area of the body (area of a crosssection taken perpendicular to the velocity); v - speed Terminal velocity • When objects falls in air, the drag force points upward (resistance to motion) • According to the Newton’s Second Law ma = mg – R = mg – ½DρAv2 • As v grows, a decreases. At some point acceleration becomes zero, and the speed value riches maximum value – terminal speed ½DρAvt2 = mg Terminal velocity • Solving ½DρAvt2 = mg we obtain vt 2mg DA vt = 300 km/h vt = 10 km/h Drag force proportional to speed • In dense fluids (liquids) a resistance force can be proportional to speed R bv • b depends on the property of the medium, and on the shape and dimensions of the object • The negative sign indicates that the force is in the opposite direction to motion Spring force (a bit of Ch. 7) • Spring in the relaxed state • Spring force (restoring force) acts to restore the relaxed state from a deformed state Hooke’s law • For relatively small deformations Fs kd Robert Hooke (1635 – 1703) • Spring force is proportional to the deformation and opposite in direction • k – spring constant • Spring force is a variable force • Hooke’s law can be applied not to springs only, but to all elastic materials and objects Centripetal force • For an object in a uniform circular motion, the centripetal acceleration is 2 v ac R • According to the Newton’s Second Law, a force must cause this acceleration – centripetal force mv Fc mac R 2 • A centripetal force accelerates a body by changing the direction of the body’s velocity without changing the speed Centripetal force • Centripetal forces may have different origins • Gravitation can be a centripetal force • Tension can be a centripetal force • Etc. Centripetal force • Centripetal forces may have different origins • Gravitation can be a centripetal force • Tension can be a centripetal force • Etc. Free-body diagram Chapter 6 Problem 14 A roller-coaster car has a mass of 500 kg when fully loaded with passengers. (a) If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the car at this point? (b) What is the in maximum speed the vehicle can have at point B and still remain on the track? Answers to the even-numbered problems Chapter 5 Problem 2 (a) 1/3; (b) 0.750 m/s2 Answers to the even-numbered problems Chapter 5 Problem 6 (a) 534 N down; (b) 54.5 kg Answers to the even-numbered problems Chapter 5 Problem 18 (b) 1.03 N; (c) 0.805 N to the right Answers to the even-numbered problems Chapter 5 Problem 36 0.306; 0.245 Answers to the even-numbered problems Chapter 6 Problem 2 215 N horizontally inward Answers to the even-numbered problems Chapter 6 Problem 4 (a) 1.65 km/s; (b) 6.84 × 103 s Answers to the even-numbered problems Chapter 6 Problem 12 (a) 1.33 m/s2; (b) 1.79 m/s2 forward and 48.0° inward Answers to the even-numbered problems Chapter 6 Problem 26 (a) 6.27 m/s2; (b) 784 N up; (c) 283 N up Answers to the even-numbered problems Chapter 6 Problem 40 8.88 N