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Physics 152 Fall 2004 Tuesday, Sep 14 Lecture #7 - Chapter 5.1: FRICTION If you are having any CHIP access or viewing problems…. • On CHIP homepage, click on Java advisory. • Javascript needs to be enabled… • Need Java applets enabled, but this may not be the case for Windows XP with default Internet Explorer configuration. • There should be no problems with other browsers: Netscape 7.02, Mozilla, Safari… (This mostly affects the timer display on reading quizzes.) Properties of Friction 1. fs between any two surfaces in contact is opposite the applied force that is parallel to the surface and equal in magnitude fs< µsN µs - coefficient of static friction N - Normal force 2. When the object is on the verge of slipping fs,max = µsN Properties of Friction 3. fk is opposite the applied force that is parallel to the surface and is fk = µkN µk - coefficient of kinetic friction N - Normal force For constant velocity: F = fk Properties of Friction 4. µs and µk depend on the nature of the surface. However: µk < µs µs - coefficient of static friction µk - coefficient of kinetic friction 5. fs and fk do not depend on area because weight and µ do not depend on area. (empirical finding) Reading Quiz (1) In three experiments, three different horizontal forces are applied to the same block lying on the same countertop. The force magnitudes are F1= 12 N, F2= 4 N and F3= 8 N. In each experiment, the block remains stationary inspite of the applied force. (a) Rank the forces according to the magnitude fs of the static frictional force on the block from the countertop, greatest first. (1) F2, F 3, F 1 (2) F1, F 3, F 2 (3) F3, F 2, F 1 (b) Rank the forces according to the maximum value fs,max of that force, greatest first. (1) F2, F 3, F 1 (2) F1, F 2, F 3 (3) F3, F 2, F 1 (4) all tie (2) In the figure (a), a 'Batman' thermos is sent sliding leftwards across a long plastic tray. What are the directions of kinetic frictional forces on... (a) the thermos (a) downward (b) upward (c) leftward (d) rightward (2) In the figure (a), a 'Batman' thermos is sent sliding leftwards across a long plastic tray. What are the directions of kinetic frictional forces on... b) the tray (from each other) 1. upward 2. leftward 3. rightward 4. downward (2) In the figure (a), a 'Batman' thermos is sent sliding leftwards across a long plastic tray. (c) Does the kinetic frictional forces on the thermos increase or decrease the speed of the thermos relative to the floor? 1. decrease 2. increase 2. In figure (b), the tray is now sent sliding leftward beneath the thermos. What now are the directions of the kinetic frictional forces on... (d) the thermos 1. upward 2. rightward 3. downward 4. leftward 2. In figure (b), the tray is now sent sliding leftward beneath the thermos. What now are the directions of the kinetic frictional forces on... (e) the tray (from each other) 1. rightward 2. downward 3. leftward 4. upward 2 (f) Does the kinetic frictional forces on the thermos increase or decrease the speed of the thermos relative to the floor? decrease increase (g) Do the kinetic frictional forces always slow the objects? yes no 3) If you press an apple crate against a wall so hard that the crate cannot slide down the wall, what is the direction of... (a) the static frictional force fs on the crate from the wall? 1. upward 2. leftward 3. rightward 4. downward (b) the normal force N on the crate from the wall? 1. horizontal, left of you 2. horizontal, towards you 3. horizontal, right of you 4. horizontal, towards wall If you increase your push, what happens to... (c) Fs 1. no change 2. decreases 3. increases (d) N 1. decreases 2. no change 3. increases (e) fs,max 1. no change 2. increases 3. decreases Example: Static Friction A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is µs = 0.40 . A rope is attached to the box and pulled at an angle of q = 30o above horizontal with tension T = 40 N. Does the box move? T q static friction (= 0.40 ) m *Select axes & draw FBD of box: y *Apply FNET = ma x y: N + T sin q - mg = maY = 0 N N = mg - T sin q = 80 N T x: T cos q - fs = maX fs m q The box will move if: T cos q - fs > 0 mg y: N = 80 N y x: T cos q - fs = maX x The box will move if: T cos q - fs > 0 N T cos q = 34.6 N T fMAX = N = (0.40)(80N) = 32 N fMAX = N m q So T cos q > fMAX and the box does move mg Example: Static Friction A detached block of rock with m =1.8 x 107 kg is at rest on a “bedding plane fracture”. The coefficient of static friction between the rock and the fracture µs = 0.63 . The bedding plane fracture is inclined q = 24o above horizontal. Does the block slide? y fs x q q W y N fs x q q W Fy = 0=N - Wcosq = N - mgcosq N = mg cosq = 160 x 106 Newtons y N fs x q q W Fy = 0=N - Wcosq = N - mgcosq N = mg cosq = 160 x 106 Newtons fs = µs N = 102 x 106 Newtons Wx= mg sinq = 72 x 106 Newtons Block will not slide because Wx < fs ! y N fs x q q W Alternative Method: From example 5-2 in textbook: µs = tan q max Arctan(µs) = q max q max= 32 o Will not slide because q = 24 o < q max Example: Dynamic Friction A box of mass m1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (µk = 0.51) on top of a second box having mass m2 = 3.0 kg, which in turn slides on a frictionless floor. What is the acceleration of the second box ? T m1 slides with friction (=0.51 ) a=? m2 slides without friction *First draw FBD of the top box: N1,2 N1,2 T m1 T f k= N1 = m1g fk W1=m1g W1 (Note: a1 ~ 55 m/s2) *Newton’s 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. Action-Reaction Pair from Frictional Forces: f1,2 m1 f2,1 m2 Action-Reaction Pair from Contact Forces: N1,2 m1 m2 N2,1 *Now consider the FBD of box 2: N2, g N2, g f2,1 f2,1 = km1g m2 N2,1 N2,1 =m1g W2=m2g W2 *Finally, solve F = ma in the horizontal direction: N2, g m1g = m2a a = 2.5 m/s2 f2,1 = m1g N2,1 =m1g W2=m2g