Chapter 6 Frictional Forces Two types static applies to stationary objects kinetic applies to sliding moving objects Like the normal force the Frictional Force is a contact force but acts parallel to the interface of two objects y FA Apply Newton s 2nd Law n f x F y N mg may 0 N mg mg F x FA f max If applied force is small book does not move static ax 0 then f fs FA f S Increase applied force book still does not move Increase FA more now book moves ax 0 FA f S ma x FA f S ma x f S There is some maximum static frictional force fsmax Once the applied force exceeds it the book moves Magnitudes f max S S N not vectors s is the coefficient of static friction it is a dimensionless number different for each surface object pair wood wood wood metal also depends on surface preparation s does not depend on the mass or surface area of the object Has value 0 s 1 5 If no applied vertical force f max S S mg Push down on book y N FA f x Apply Newton s 2nd Law F y FP mg N mg FP may 0 N mg FP f max S S N S mg FP What is FA needed just to start book moving F x FA f S 0 FA f max S S mg FP When an Object is Moving fsmax is exceeded so the object can move but friction force is still being applied However less force is needed to keep an object moving against friction than to get it started We define kinetic friction f k k FN k is the coefficient of kinetic friction similar to S but always less than S Table 6 1 Now let s consider incline plane problem but with friction Book is at rest m 1 00 kg 10 0 s 0 200 y fs F y FBD fs N mg y N x mg F N mgcos 0 x x mg sin f S 0 f S mg sin N mgcos o 1 0kg 9 80 cos10 0 1 00kg 9 80 sin 10 0 1 70 N 9 65 N m s2 o m s2 Book can move slide if mgsin fsmax What is fsmax f max S S N 0 200 9 65 N 1 93 N f S Book does not move What angle is needed to cause book to slide max S mgsin f mgsin S N mgsin S mgcos tan S 1 tan S o 11 3 As is increased N decreases therefore fsmax decreases Once book is moving we need to use the kinetic coefficient of friction Lets take 15 0 and k 0 150 s F x mg sin f k ma x mg sin k mg cos ma x or a x g sin k cos 9 80 sm2 sin 15 0o 0 150 cos15 0o 1 12 sm2 Example Problem A skier is pulled up a slope at a constant velocity by a tow bar The slope is inclined at 25 0 with respect to the horizontal The force applied to the skier by the tow bar is parallel to the slope The skier s mass is 55 0 kg and the coefficient of kinetic friction between the skis and the snow is 0 120 Find the magnitude of the force that the tow bar exerts on the skier Given m 55 0 kg k 0 120 25 0 Infer since velocity is constant ax 0 also ay 0 since skier remains on slope equilibrium F 0 F x 0 Fy 0 Draw FBD apply Newton s 2nd Law Fp N Fp N fk mg fk x mg x F y 0 Fx 0 N mgcos 0 N mgcos f k FP mgsin 0 FP f k mgsin k N mgsin k mgcos mgsin mg k cos sin 286 N
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