Chapter 6 Force and Motion II I Drag forces and terminal speed II Uniform circular motion III Non Uniform circular motion I Drag force and terminal speed Fluid anything that can flow Example gas liquid Drag force D Appears when there is a relative velocity between a fluid and a body Opposes the relative motion of a body in a fluid Points in the direction in which the fluid flows Assumptions Fluid air Body is blunt baseball Fast relative motion turbulent air 1 1 D C Av 2 2 6 3 C drag coefficient 0 4 1 air density mass volume A effective body s cross sectional area area perpendicular to v Terminal speed vt Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero Fg D 1 D Fg ma if a 0 C Av 2 Fg 0 2 vt 2 Fg C A 6 4 II Uniform circular motion Centripetal acceleration a v2 r 6 5 v a ctes but direction changes during motion A centripetal force accelerates a body by changing the direction of the body s velocity without changing its speed Centripetal force F m v2 R 6 6 a F are directed toward the center of curvature of the particle s path 2 III Non Uniform circular motion A particle moves with varying speed in a circular path The acceleration has two components Radial ar v2 R Tangential at dv dt at causes the change in the speed of the particle a ar2 at2 d v v2 r a at ar dt r r t F F F In uniform circular motion v constant at 0 a ar 49 A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table What speed keeps the cylinder at rest N For M T Mg ac 0 T For m T m mg v2 v2 Mg m v r r Mgr m T Mg 33E Calculate the drag force on a missile 53cm in diameter cruising with a speed of 250m s at low altitude where the density of air is 1 2kg m3 Assume C 0 75 1 2 D C Av2 0 5 0 75 1 2kg m3 0 53m 2 2 250m s 6 2kN 2 32 The terminal speed of a ski diver is 160 km h in the spread eagle position and 310 km h in the nosedive position Assuming that the diver s drag coefficient C does not change from one point to another find the ratio of the effective cross sectional area A in the slower position to that of the faster position 2Fg 2Fg C AE A 160km h A vt D E 3 7 C A 310km h AD 2Fg AE C AD 3 11P A worker wishes to pile a cone of sand onto a circular area in his yard The radius of the circle is R and no sand is to spill into the surrounding area If s is the static coefficient of friction between each layer of sand along the slope and the sand beneath it along which it might slip show that the greatest volume of sand that can be stored in this manner is s R3 3 The volume of a cone is Ah 3 where A is the base area and h is the cone s height To pile the most sand without extending the radius sand is added to make the height h as great as possible Eventually the sides become so steep that sand at the surface begins to slip Goal find the greatest height greatest slope for which the sand does not slide Cross section of sand s cone Static friction grain does not move N F gy mg cos y h If grain does not slide f F gx mg sin N f Fgy Fgx mg sin f s max s N s mg cos s tan Fgx mg The surface of the cone has the greatest slope and the height of the cone is maximum if h s tan R x Vcone R h R s A h R 2 R s s R 3 3 3 3 21 Block B weighs 711N The coefficient of static friction between the block and the table is 0 25 assume that the cord between B and the knot is horizontal Find the maximum weight of block A for which the system will be stationary N System stationary f s max s N Block B N m B g T2 f Knot T1 T 2 x T 2 cos 30 T 2 177 75 N 205 25 N cos 30 T1 T3 T3 T1 T1 f s max 0 T1 0 25 711 N 177 75 N FgB T 2 y T 2 sin 30 T3 FgA Block A T3 m A g T 2 sin 30 0 5 205 25 N 102 62 N 23P Two blocks of weights 3 6N and 7 2N are connected by a massless string and slide down a 30 inclined plane The coefficient of kinetic friction between the lighter block and the plane is 0 10 that between the heavier block and the plane is 0 20 Assuming that the lighter block leads find a the magnitude of the acceleration of the blocks and b the tension in the string c Describe the motion if instead the heavier block leads Block A NA T FgxA Block B NB fkA FgxB fkB T t en em v Mo NB fk A NA T A FgyA Light block A leads FgyB T fk B B FgB FgA 4 Light block A leads Block A N A F gyA m A g cos 30 3 12 N f kA kA N A 0 1 3 12 N 0 312 N F gxA f kA T m A a 3 6 N sin 30 0 312 N T 0 37 a 1 49 T 0 37 a a 3 49 m s 2 Block B N B F gyB m B g cos 30 6 23 N T 0 2 N f kB kB N B 0 2 6 23 N 1 25 N F gxB T f kB m B a 7 2 N sin 30 T 1 25 N 0 73 a 2 35 T 0 73 a W AW B T WA WB kB kA cos 0 2 N Heavy block B leads Reversing the blocks is equivalent to switching the labels This would give T kA kB 0 impossible The above set of equations is not valid in this circumstance aA aB The blocks move independently from each other 74 A block weighing 22N is held against a vertical wall by a horizontal force F of magnitude 60N The coefficient of static friction between the wall and the block is 0 55 and the coefficient of kinetic friction between them is 0 38 A second P acting parallel to the wall is applied to the block For the following magnitudes and directions of P determine whether the block moves the …