Physics 1501: Lecture 8, Pg 1 Physics 1501: Lecture 8 l Announcements çHomework #3 : due next Monday l Topics çReview of Newton’s Laws. çFriction çSome applications of Newton’s laws.

Physics 1501: Lecture 8, Pg 2 Review Newton’s Laws 1, 2, 3 l Isaac Newton ( ) published Principia Mathematica in In this work, he proposed three “laws” of motion: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. FFa Law 2: For any object, F NET =  F = ma FF Law 3: Forces occur in pairs: F A,B = - F B,A (For every action there is an equal and opposite reaction.)

Physics 1501: Lecture 8, Pg 3 Exercise: Inclined plane A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ?  m a

Physics 1501: Lecture 8, Pg 4 Inclined plane... l Define convenient axes parallel and perpendicular to plane: ç Acceleration a is in x direction only.  m a i j

Physics 1501: Lecture 8, Pg 5 Inclined plane... l Consider x and y components separately: i i: mg sin  = ma a = g sin  j j: N - mg cos . N = mg cos  gmggmg N mg sin  mg cos   amaama i j

Physics 1501: Lecture 8, Pg 6 Angles of an Inclined plane    ma = mg sin    mg N

Physics 1501: Lecture 8, Pg 7 Example Gravity, Normal Forces etc. Conside a women on a swing: When is the tension on the rope largest. Is it : A) greater than B) the same as C) less than the force due to gravity acting on the woman Active Figure

Physics 1501: Lecture 8, Pg 8 New Topic: Friction l What does it do? çIt opposes motion! l How do we characterize this in terms we have learned? çFriction results in a force in a direction opposite to the direction of motion! amaama F F APPLIED f f FRICTION gmggmg N i j

Physics 1501: Lecture 8, Pg 9 Friction... l Friction is caused by the “microscopic” interactions between the two surfaces:

Physics 1501: Lecture 8, Pg 10 Friction... l Force of friction acts to oppose motion: çParallel to surface. N çPerpendicular to Normal force. amaama F ffFffF gmggmg N i j

Physics 1501: Lecture 8, Pg 11 Model for Sliding Friction N l The direction of the frictional force vector is perpendicular to the normal force vector N. f N l The magnitude of the frictional force vector |f K | is proportional to the magnitude of the normal force |N |. fN g  |f K | =  K | N | ( =  K  |  mg | in the previous example) çThe “heavier” something is, the greater the friction will be...makes sense! The constant  K is called the “coefficient of kinetic friction”.

Physics 1501: Lecture 8, Pg 12 Model... l Dynamics: i :F  K N = m a j :N = mg soF  K mg = m a amaama F gmggmg N i j  K mg

Physics 1501: Lecture 8, Pg 13 Lecture 8, ACT 1 Friction and Motion A box of mass m 1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (  k =.5) on top of a second box having mass m 2 = 2 kg, which in turn slides on an ice rink (frictionless). çWhat is the acceleration of the second box ? (a) a = 0 m/s 2 (b) a = 2.5 m/s 2 (c) a = 10 m/s 2 m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.5  ) slides without friction a = ?

Physics 1501: Lecture 8, Pg 14 Static Friction... F gmggmg N i j fSfS l So far we have considered friction acting when something moves. çWe also know that it acts in un-moving “static” systems: l In these cases, the force provided by friction will depend on the forces applied on the system.

Physics 1501: Lecture 8, Pg 15 Static Friction... l Just like in the sliding case except a = 0. i :F  f S = 0 j :N = mg F gmggmg N i j fSfS While the block is static: f S  F (unlike kinetic friction)

Physics 1501: Lecture 8, Pg 16 Static Friction... F gmggmg N i j fSfS The maximum possible force that the friction between two objects can provide is f MAX =  S N, where  s is the “coefficient of static friction”.  So f S   S N.  As one increases F, f S gets bigger until f S =  S N and the object “breaks loose” and starts to move.

Physics 1501: Lecture 8, Pg 17 Static Friction... F  S is discovered by increasing F until the block starts to slide: i :F MAX  S N = 0 j :N = mg  S  F MAX / mg F F MAX gmggmg N i j  S mg

Physics 1501: Lecture 8, Pg 18 Since f =  N, the force of friction does not depend on the area of the surfaces in contact. By definition, it must be true that  S >  K for any system (think about it...). l Graph of Frictional force vs Applied force: Additional comments on Friction: fFfF FAFA f F = F A f F =  K N f F =  S N Active Figure

Physics 1501: Lecture 8, Pg 19 Lecture 8, ACT 2 Two-body dynamics A block of mass m, when placed on a rough inclined plane (  > 0) and given a brief push, keeps moving down the plane with constant speed.  If a similar block (same  ) of mass 2m were placed on the same incline and given a brief push, it would: (a) (a) stop (b) (b) accelerate (c) (c) move with constant speed m

Physics 1501: Lecture 8, Pg 20 Example with pulley l A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F. çAssume the pulleys massless and frictionless. çAssume the rope massless.  M T5T5 T4T4 T3T3 T2T2 T1T1 F l We use the 5 step method. çDraw a picture: what are we looking for ? çWhat physics idea are applicable ? Draw a diagram and list known and unknown variables. Newton’s 2 nd law : F= m a Free-body diagram for each object

Physics 1501: Lecture 8, Pg 21 Pulleys: continued l FBD for all objects  M T5T5 T4T4 T3T3 T2T2 T1T1 F T4T4 F=T 1 T2T2 T3T3 T2T2 T3T3 T5T5 M T5T5 MgMg

Physics 1501: Lecture 8, Pg 22 Pulleys: finally l Step 3: Plan the solution (what are the relevant equations)  F= m a, static (no acceleration: mass is held in place) M T5T5 MgMg T 5 =Mg T2T2 T3T3 T5T5 T 2 +T 3 =T 5 T4T4 F=T 1 T2T2 T3T3 T 1 +T 2 +T 3 =T 4

Physics 1501: Lecture 8, Pg 23 Pulleys: really finally! l Step 4: execute the plan (solve in terms of variables) çWe have (from FBD): T 5 =MgF=T 1 T 2 +T 3 =T 5 T 1 +T 2 +T 3 =T 4  M T5T5 T4T4 T3T3 T2T2 T1T1 F T 2 =T 3 T 1 =T 3 T 2 =Mg/2 T 2 +T 3 =T 5 gives T 5 =2T 2 =Mg F=T 1 =Mg/2 T 1 =T 2 =T 3 =Mg/2 and T 4 =3Mg/2 T 5 =Mgand çPulleys are massless and frictionless l Step 5: evaluate the answer (here, dimensions are OK and no numerical values)

Physics 1501: Lecture 8, Pg 24 Lecture 8, ACT 3 Newton’s Second Law I push with a force of 2 Newtons on a cart that is initially at rest on an air table with no air. I push for a second. Because there is no air, the cart stops after I finish pushing. It has traveled a certain distance. Air Track Cart F= 2N For a second shot, I push just as hard but keep pushing for 2 seconds. The distance the cart moves the second time versus the first is, A) 4 x as longB) 2 x as longC) Same D) 1/2 as longE) 1/4 x as long t 1 =1s, v 1 t 2 =2s, v 2 t o, v o = 0 Cart x1x1 x2x2

Physics 1501: Lecture 8, Pg 25 l You are going to pull two blocks (m A =4 kg and m B =6 kg) at constant acceleration (a= 2.5 m/s 2 ) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ? l (A) YES(B) CAN’T TELL(C) NO Lecture 8, Act 4 A B a= 2.5 m/s 2 rope

Physics 1501: Lecture 8, Pg 26 Example Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m 1 = 4.00 kg, m 2 = 1.00kg and m 3 = 2.00kg. a) What is the magnitude and direction of acceleration on the three blocks ? b) What is the tension on the two cords ? m1m1 T1T1 m2m2 m3m3

Physics 1501: Lecture 8, Pg 27 m1m1 T1T1 m2m2 m3m3 m1m1 m2m2 m3m3 N=-m 2 g T 23 T 12 m1gm1g m3gm3g T 23 T12T12 T 12 T 23 T 12 - m 1 g = - m 1 a T 23 - m 3 g = m 3 a  k m 2 g a a a -T 12 + T 23 +  k m 2 g = - m 2 a SOLUTION: T 12 = = 30.0 N, T 23 = 24.2 N, a = 2.31 m/s 2 left for m 2 m2gm2g