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Transcript
Dynamics Review Statics: The study of objects at equilibrium [Acceleration = Zero]. Rotational Equilibrium: 0 net Torque 0 Translational Equilibrium: a 0 Fnet Forces 0 F 0, and F 0, and F 0 x y z You should have known (had in your head) all of the above! 841000211 5/11/2017 10:39:00 AM Page 1 of 15 Dynamics: The study of the connection between force and motion. “The study of why objects move.” Newton’s Second Law: Fnet = ma Actually, defined in terms of acceleration … Fnet a m “acceleration is … proportional to net force acting on an object, AND inversely proportional to the mass of that object.” Dynamics Example Problems: 841000211 5/11/2017 10:39:00 AM Page 2 of 15 Example 4-3: What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h within a distance of 55 m? m ____________ x ____________ vinitial _________ v final __________ a ____________ t _____________ F ma ________ 841000211 5/11/2017 10:39:00 AM Page 3 of 15 m 1500 kg x 55 m vinitial 28 m / s v final 0 m / s a 7.1 m / s / s t who cares! F 1.1x10 4 N 841000211 5/11/2017 10:39:00 AM Page 4 of 15 Example 4-6: Action-Reaction Force Pairs Newton’s Third Law: Whenever one object exerts a force on a second object, the second object exerts a force in the opposite direction on the first. Note: Two different objects. Equal but opposite. 841000211 5/11/2017 10:39:00 AM Page 5 of 15 Fg = mg = (10.0 kg)(9.80 m/s/s) = 98.0 N Determine FN in each case below: 841000211 5/11/2017 10:39:00 AM Page 6 of 15 Example 4-7: Let applied force to the crate be 100.0 N. Determine the acceleration of the box. 841000211 5/11/2017 10:39:00 AM Page 7 of 15 Example 4-19: Friction and Angled Forces: A 10.0 kg box is pulled along a horizontal surface by a force of 40.0 N applied at an angle of 30.0° [always assumed to be above the horizontal]. The coefficient of kinetic friction of 0.30. Calculate the acceleration of the box. Sketch the problem, showing all forces acting on the box. [Hint: There are four forces acting on the box.] Notes: The coefficient of friction. " mu" F friction FNormal F f FN F f ,kinetic k FN F f ,static s FN s k F f , s F f ,k 841000211 5/11/2017 10:39:00 AM Page 8 of 15 841000211 5/11/2017 10:39:00 AM Page 9 of 15 A 10.0 kg box is pulled along a horizontal surface by a force of 40.0 N applied at an angle of 30.0°. The coefficient of kinetic friction of 0.30. Calculate the acceleration of the box. 841000211 5/11/2017 10:39:00 AM Page 10 of 15 Inclined Planes: Gravitational acceleration [g] is straight down, thus, weight [FG] acts straight down! (Gravity pulls toward the center of the Earth.) Normal Force is perpendicular to the surface! Thus, weight and normal force are NOT in the same direction! The component of FG perpendicular to the plane is parallel to FN. 841000211 5/11/2017 10:39:00 AM Page 11 of 15 FG must be split into components. A suitable coordinate system must be chosen so as to determine FNet and a: o Parallel to the plane and Perpendicular to the plane! (Text shows x and y … I don’t like that convention … their x is parallel, y is perpendicular!) Fnet , F Fnet ,// F// ma FN Fg , ma// Ff Fg ,// o NOT Horizontal and Vertical! (You could work it this way, but it would be a lot less fun.) 841000211 5/11/2017 10:39:00 AM Page 12 of 15 Fnet , F ma FN Fg , Fapplied,, if any a 0 ma 0 Fg , Fg cos mg cos ma FN Fg , 0 FN mg cos Fapplied,, if any Fnet ,// F// ma// F f Fg ,// Fapplied,//,if any Fg ,// mg sin ma// k FN mg sin Fapplied,//,if any 841000211 5/11/2017 10:39:00 AM Page 13 of 15 P(4): 52, 53: Inclined Planes. 52. The carton shown in Fig. 4–55 lies on a plane tilted at an angle 22 .0º to the horizontal, with k 0.12. (a) Determine the acceleration of the carton as it slides down the plane. (b) If the carton starts from rest 9.30 m up the plane from its base, what will be the carton’s speed when it reaches the bottom of the incline? 53. A carton is given an initial speed of 3.0 m s up the 22.0º plane shown in Fig. 4–55. (a) How far up the plane will it go? (b) How much time elapses before it returns to its starting point? Ignore friction. Rework, not ignoring friction. 841000211 5/11/2017 10:39:00 AM Page 14 of 15 52. (a) Consider the free-body diagram for the carton on the surface. There is no motion in the y direction and thus no acceleration in the y direction. Write Newton’s 2nd law for both directions. Fx FN mg cos 0 FN mg cos F x FN Ffr y mg sin Ffr ma x mg ma mg sin k FN mg sin k mg cos a g sin k cos 9.80 m s 2 sin 22.0 o 0.12 cos 22.0o 2.58 2.6 m s 2 (b) Now use Eq. 2-11c, with an initial velocity of 0, to find the final velocity. v 2 v02 2a x x0 v 2a x x0 2 2.58 m s 2 9.30 m 6.9 m s 53. (a) Consider the free-body diagram for the carton on the frictionless surface. There is no acceleration in the y direction. Write Newton’s 2nd law for the x direction. Fx mg sin ma a g sin Use Eq. 2-11c with v0 3.0 m s and v 0 m s to find the distance that it slides before stopping. v 2 v02 2a x x0 x x0 v 2 v02 2a 0 3.0 m s FN y x mg 2 2 9.8 m s 2 sin 22.0o 1.2 m The negative sign means that the block is displaced up the plane, which is the negative direction. (b) The time for a round trip can be found from Eq. 2-11a. The free-body diagram (and thus the acceleration) is the same whether the block is rising or falling. For the entire trip, v0 3.0 m s and v 3.0 m s . v v0 at t 841000211 v v0 a 3.0 m s 3.0 m s 9.8 m s sin 22 2 o 5/11/2017 10:39:00 AM 1.6 s Page 15 of 15