FRICTIONAL FORCES. Direction of frictional forces... (not always obvious)... CHAPTER 5 APPLICATIONS OF NEWTON S LAWS

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1 RICTIONAL ORCES CHAPTER 5 APPLICATIONS O NEWTON S LAWS rictional forces Static friction Kinetic friction Centripetal force Centripetal acceleration Loop-the-loop Drag force Terminal velocity Direction of frictional forces... (not always obvious)... M C GM GC frictional force frictional force of the ground on the man of the ground on the crate Here s the easy way to remember... Usually, the frictional force is opposite to the direction of the motion without friction. crate moves to right. GC without friction foot would slip to the left. GM

2 MG GM CM MC M C f 1 f 2 CG GC Note that GM and GC are frictional forces. Consider the man and crate as a system. The forces MC and CM are internal to the system; also CM = MC, and so they do not contribute to any motion. GM GC The remaining forces acting on the system are GM and GC. Thus, if GM > GC the system, i.e., the man and the crate, moves to the right. So, it is the frictional forces f 1 and f 2 that determine the outcome f f frictional force [1] Block initially at rest: The normal force N is always present when an object rests on a surface. You can think of it as a reaction force of the surface to the force of the object on the surface. Note y y = N m g = 0, i.e., N = m g. The frictional force that has to be overcome is: f s µ s N, normal force applied force weight w (= m g ) where µ s is called the coefficient of static friction. If > f s the block will move. Therefore, a minimum applied force ( = µ s N ) is required to start moving an object over a surface. If f s the net force on the block is zero, so the frictional force in this case is f =. N orces acting ON the crate

3 v f k [2] Block moving: When the object is moving, the frictional force is: f k = µ k N, where µ k is the coefficient of kinetic friction. Thus, the net force on the object is then f k f k frictional force normal force applied force weight w (= m g ) orces acting ON the crate towards the right ( = ma ). Note, until the applied force > f s ( = µ s N ), the net force is zero, i.e., = f. N Here is some experimental data: orce sensor kg A B frictional force f weight w (= m g ) At A ( breakout ): = f s = µ s N = µ s mg N normal force applied force Time (s) No motion Motion at constant velocity (static) (kinetic) C frictional force (f) f s = µ s N f k = µ k N i.e., µ s = mg = (5 N) ( m/s 2 ) = rom B to C (net force = 0): = µ k N = µ k mg f = at rest in motion Applied force () i.e., µ k = mg = (3.7 N) ( m/s 2 ) =

4 Question 5.1: Two forces, 1 and 2, act on different objects on a surface, as shown. If the objects are all moving with constant velocity, in which case is the coefficient of kinetic friction between the object and surface greatest? 2 = 3 N 6 kg 1 = 12 N 2 = 4 N 4 kg 1 = 8 N A B 2 = 2 N 3 kg 1 = 8 N 2 = 2 N 2 kg 1 = 8 N C D

5 2 = 3 N 6 kg A 1 = 12 N 2 = 4 N 4 kg B 1 = 8 N (a) θ 2 = 2 N 1 = 8 N 2 = 2 N 1 = 8 N 3 kg 2 kg C D 2 f 1 Draw the BD. The net force on each block is 1 2 f = 0 since they are moving with constant velocity. f = 1 2 = µ N, i.e., µ = ( 1 2 ). N µ A = 9 N 6g = 1.5 N g ; µ B = 4 N 4g = 1 N g, smallest µ C = 6 N 3g = 2 N g ; µ D = 6 N 2g = 3N g. greatest (b) θ Question 5.2: Your daughter is sitting on a sled and asks you to propel her across a flat, horizontal surface. You have a choice; (a) you can push on her shoulders, or (b) you can pull her using a rope attached to the sled. Which would require less force from you to get the sled moving, case (a), case (b) or is the force the same in both cases?

6 N a (a) θ f mg θ θ N b (b) θ f mg θ Question 5.3: The coefficient of friction between the tires of a car and the road on a particular day is What is the steepest slope of the road on which the car is parked if Look at the BD s of the girl/sled system. The minimum force required to start the sled moving in either case is: the car is not to slide down the hill? Note, the brakes are fully on so the wheels are locked. x = min cosθ = µ s N. The angle θ is the same in both cases, so min N. a y = Na sin θ = mg, i.e., Na = mg + sin θ. b y = Nb + sin θ = mg, i.e., Nb = mg sin θ. N a > N b. So, min (a) > min (b), i.e., a smaller force is required to start the sled moving in (b) than in (a).

7 f s N y When the car is just about to move: x = 0 and y = 0. rom chapter 4, we have: forces along x: mg sin θ f s = 0 forces along y: N mg cosθ = 0 But f s = µ s N. θ mg sin θ µ s N = 0. Substituting for N from (ii) we get: mg sin θ µ s mg cosθ = 0. tan θ = µ s. w = m g So, at this angle θ the car is just about to slide (i), (ii). If µ s = 0.70, then θ = tan 1 (0.70) = 35 ". Note: it is independent of the mass of the car and has nothing to do with the strength of the brakes x

8 Wheels on surfaces: Two possibilities: rolling wheel - no slipping - STATIC RICTION locked wheel - skidding - KINETIC RICTION Since µ k < µ s then f k < f s and so there is less frictional force in a skid you travel further Look at the web-site anti-lock (abs) brakes

9 Question 5.4: The driver of a 1200 kg car moving at 15.0 m/s is forced to slam on the brakes. The car skids to a halt after traveling a distance of 25.5 m. (a) What is the coefficient of kinetic friction between the road and the tires? (b) How long did it take the car to stop? (c) If the car had been fitted with an anti-skid control system (ABS) and the coefficient of static friction between the road and the tires was µ = 0.65, how far would the car have traveled before it came to a complete stop? v (a) ind the acceleration. We have 0 = v 2 + 2a(x x ), i.e., a = v 2 2l = (15.0 m/s)2 2(25.5 m) = 4.41 m/s2. The only force in the x-direction is the kinetic frictional force, that is the net force, so, by Newton s 2nd Law: f k = ma = µ k N = µ k mg µ k = a g = ( 4.41 m/s2 ) 9.81 m/s 2 = Note that the result is independent of the mass of the car (b) We have v = v + at, i.e., t = v v a l v = 0 x x f k 15.0 m/s = 2 = 3.40 s m/s N mg

10 (c) In part (a) we found a = v 2 2l. Also f k = ma = µ s N = µ s mg, i.e., a = µ s g = v 2 2l. l = v 2 2µ s g = (15.0 m/s) 2 2(0.65)(9.81 m/s 2 = 17.6 m. ) This distance is 7.9 m less, i.e., 31% less, than the result in part (b). The moral is when coming to a stop, try not to skid Question 5.5: In the figure m 1 = 4.0 kg and the coefficient of static friction between the block and the inclined surface is ind the range of the possible values of m 2 for which the blocks are stationary.

11 There are two scenarios to consider, (a) when to move up the slope, and (b) when down the slope. m 1 m 1 is about is about to move (b) N Now: (a) N 1 x = T + f s + ( m 1 gsin30 ) = 0 (4) But rom (1): 1 x = T + ( f s ) + ( m 1 gsin30 ) = 0 1 y = N + ( m 1 gcos30 ) = 0 2 y = T + ( m 2 g) = 0 (1)... (2) (3) N = m 1 gcos30 = (4.0 kg)(9.81 m/s 2 )(0.866) = 34.0 N. f s = µ s N = (0.40)(34.0 N) = 13.6 N. T = f s + m 1 gsin30 rom (4): So, from (3): 1 y = N + ( m 1 gcos30 ) = 0 2 y = T + ( m 2 g) = 0 But... (2) unchanged (3) unchanged N = m 1 gcos30 = (4.0 kg)(9.81 m/s 2 )(0.866) = 34.0 N. f s = µ s N = (0.40)(34.0 N) = 13.6 N. T = f s + m 1 gsin30 = ( 13.6 N) + (4.0 kg)(9.81 m/s 2 )(0.50) = 6.0 N. m 2 = T g = 6.0 N = 0.61 kg m/s = (13.6 N) + (4.0 kg)(9.81 m/s 2 )(0.50) = 33.2 N. So, from (3): m 2 = T g = 33.2 N = 3.39 kg m/s Therefore, m 1 remains stationary providing 0.61 kg m kg.

12 Question 5.6: To prevent a box from sliding down an incline, physics student Isaac pushes on the box parallel to the incline with just enough force to hold stationary. If the mass of the box is 6.00 kg, the slope of the incline is 30 and the coefficient of static friction between the box and the incline is µ s = 0.15, what is the minimum force Isaac has to apply? N f s x y mg Draw the BD for the box on the incline. The x-component of the net force is + f s + ( mgsinθ) = 0, and the y-component is N + ( mgcosθ) = 0. But f s = µ s N = µ s mgcosθ. = mgsinθ µ s mgcosθ = mg(sinθ µ s cosθ) = (6.00 kg)(9.81 m/s 2 )(0.50 (0.15)(0.866)) = 21.8 N.

13 AB GA = f A BA GB = f B A B Question 5.7: Two people (A and B) are tugging at each other. By Newton s 3rd law the force that A applies on B is equal to the force that B applies on A. How come one of them can win this tug-of-war? A B Identify the forces acting on each person. AB and BA are an A/R pair. AB = BA. Note that the (frictional) forces are one part of A/R pairs also, i.e., GA = AG and GB = BG Since AB and BA cancel each other the result depends on the frictional forces GA and GB. If GA > GB then A wins If GB > GA then B wins If f = µ s N for both people then the person with greater mass should win Note that if there s no friction (e.g., on ice) then GA = GB = 0 so no one wins

14 Centripetal force acceleration with magnitude In chapter 3 we determined that an object traveling with a constant speed v in a circle of radius r experiences a centripetal (radial) a r = v2 r. The corresponding radial force acting on the object is r = ma r = mv2 r which is called the centripetal force. In the absence of this force, the object would continue in a straight line., Question 5.8: A stone, of mass 0.75 kg, is attached to a string and whirled in a horizontal circle of radius of 35 cm, like a conical pendulum. If the string makes an angle of 30 with the vertical, find (a) the tension in the string, (b) the speed of the stone, and (c) the time for one revolution. However, there is a difference between circular motion is a horizontal plane and in a vertical plane.

15 (a) In the x-direction Tsin30 = ma r = mv2 r And in the y-direction rom (2) (b) rom (1) But (c) Orbital time Tcos30 + ( mg) = 0 T = mg (0.75 kg)(9.81 m/s = cos a r = v2 r a r = Tsin30 m = (1) (2) 2 ) (8.50 N)(0.50) 0.75 kg = 8.50 N. = 5.67 m/s 2. v = a r r = (5.67 m/s 2 )(0.35 m) = 2πr v = 1.41 m/s. = 2π(0.35 m) 1.41 m/s = 1.56 s. Question 5.9: A curve in the road, with a radius of 70.0 m, is banked at an angle of 15. If the coefficient of friction between the road and car tires is 0.70, what is the maximum speed a car can make the corner without sliding? Hint: eliminate the normal force.

16 mg θ θ N f s But Draw the BD for the car on the banked curve. We have in the y-direction: (1) In the x-direction we have (note that the net force is the centripetal force) i.e., N sinθ + µ s N cosθ = mv2 (2) R Dividing (2) by (1), i.e., eliminating y x N cosθ mg f s sinθ = 0. f s = µ s N. N cosθ µ s N sinθ = mg N sinθ + f s cosθ = mv2 R, N sinθ + µ s cosθ cosθ µ s sinθ = v2 gr, we get Substituting the given parameters we find v 2 = (9.81 m/s )(70 m) 1 (0.70)(0.268) = (m/s)2. v = 28.6 m/s. Note that the result is independent of the mass of the car. At home, show that: (a) with no banking (i.e., θ = 0) but with static friction ( µ s = 0.70 ), then v = 21.9 m/s. (b) with θ = 15 and µ s = 0, i.e., a banked corner but zero static friction (like ice), then v = 13.6 m/s. i.e., v 2 = gr tanθ + µ s 1 µ s tanθ.

17 Question 5.10: A pilot of mass 100 kg carries out a vertical circle of radius R = 2.70 km, i.e., a loop-the-loop, at a constant speed of 225 m/s. What is the apparent weight of the pilot (a) at the top of the loop, (b) at the bottom of the loop, and (c) at some angle inside the loop? (a) At the top the two forces acting on the pilot are the N T loop. mg N T + mg = mv2 R normal force due to the seat (N T ) and the gravitational force (mg) due to the Earth. These two forces combine to produce the centripetal force holding the pilot in the This is the force of the seat on the pilot, which, by Newton s 3 rd Law is equal and opposite to the force of the pilot on the seat, i.e., the apparent weight. Therefore, the apparent weight of the pilot is, i.e., N T = mv2 R = 894 N (91 kg). mv 2 Note that if < mg, i.e., v < gr, the pilot would R fall from the seat unless restrained by a seat belt. (Same for an upside down loop on a roller coaster. mg. m v2 R g (225 m/s) 2 = (100 kg) m/s 2 m

18 (b) Again, the two forces acting on the pilot combine to N B produce the centripetal force, i.e., Therefore, the apparent weight of the pilot is equal and opposite to the normal force of the seat on the pilot, so, the apparent weight of the pilot is mg N B mg = mv2 R. mv 2 v2 + mg = m R R + g (225 m/s) 2 = (100 kg) m/s 2 m So, at the bottom of an inside loop the apparent weight is greater than the true weight. Note also that the centripetal force is constant and is equal to 1 ( 2 N B + N T ). = 2856 N (291 kg). (c) At some angle inside the loop: m " g In the radial direction we have R = mv2 R = N + mg sin θ, where v is the instantaneous speed. After re-arranging, we obtain " N O R N = m v2 R gsin θ. Note, if we put θ = 90 (top) and θ = 270 (bottom) we obtain the previous results. θ m " g cosθ m " g m " g sin θ components of the pilot s weight Can you think of an application where this situation might be important?

19 If the passengers in a roller coaster or ferris wheel are not restrained by seat belts, what is the maximum speed for an over the top maneuver on the outside of the loop? DRAG ORCES v D fluid resistance N T mg Different from ordinary friction because it depends on speed and type of flow : Using the BD we have mg N T = mv2 R, i.e., Providing v < gr, then N T > 0, so the riders remain in their seat. However, if their seats unless restrained. N T = mg mv2 R. v > gr, the riders will fly from at low speed (non-turbulent or laminar flow) v D at high speed (turbulent flow) v D D = kv D = bv 2 The constants depend on shape, size, etc. or example: v v Large b (large area) Small b (small area)

20 no drag velocity (v) terminal velocity with drag time acceleration (a) g no drag with drag time D (= bv 2 ) w (= mg) mg bv 2 = ma y. a y = mg bv2 m. displacement (y) no drag with drag time rom rest: (y y ) = 1 2 a y t2 = 1 2 g b m v2 t 2.

21 DISCUSSION PROBLEM Question 5.11: A motorboat is traveling across a lake when the motor suddenly stops. The boat slows down If a feather and a baseball are dropped from a great height - so they each reach terminal velocity - the baseball strikes the ground first. Which object experiences the greater drag force? under a drag force D = bv. (a) What are the velocity and position of the boat as a function of time after the motor stops? (b) If the boat slows from 4.0 m/s to 1.0 m/s in 10 s, how far does it travel before stopping? A: The feather. B: The baseball. C: They experience the same drag force.

22 (a) The only force acting on the boat is D. ma = m dv dt = bv, i.e., dv v = b m dt. If the velocity is v when t = 0, then integrating this equation we get Also, v v dv v If the initial position is at x = 0 when t = 0, then x approaches a limiting value, = b t m dt, i.e., ln v 0 v = b m t. v = v e bt m. v = dx dt = v e bt m, i.e., dx = v e bt m dt. t dx = v e bt m dt, i.e., x = mv e bt m t 0 0 b 0 = mv ( 1 e bt m ). b As t increases, e bt m 0, and the position of the boat x max = mv or example, when t = 10m then b, and So, the boat s velocity and position have essentially reached their final values. v = v e 10 ~ v, x = x max ( 1 e 10 ) ~ x max. (b) With v = 4.0 m/s and v = 1.0 m/s, (1.0 m/s) = (4.0 m/s)e (b/m)(10 s). ln(0.25) = b m (10 s), i.e., b m = ln(0.25) (10 s) x max = mv b = 0.14 s m/s = = 29.6 m s b. Although it takes an infinite amount of time reach x, max the boat effectively comes to a stop after a reasonable time.