2. Physics 1501: Lecture 9, Pg 1 Physics 1501: Lecture 9 l Announcements çHomework #3 : due next Monday çBut HW 04 will be due the following Friday (same week). çMidterm 1: Monday Oct. 3 l Topics çReview Friction çCircular motion and Newton’s Laws

3. Physics 1501: Lecture 9, Pg 2 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

4. Physics 1501: Lecture 9, Pg 3 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

5. Physics 1501: Lecture 9, Pg 4 Lecture 9, ACT 1 Friction and Motion A box of mass m 1 = 1 kg is being pulled by a horizontal string having tension T = 30 N. The box 1 is on top of a second box of mass m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are  s =3.5 and  k = 0.5. This second box can slide on an ice rink (frictionless). çThe acceleration of box 1 is (a)Greater than (b) Equal to (c) Smaller than the acceleration of box 2 ? m2m2m2m2 T m1m1m1m1  friction coefficients  s =3.5 and  k =0.5 slides without friction a2a2 a1a1

6. Physics 1501: Lecture 9, Pg 5 Lecture 9, ACT 1 Solution l First, we need to know if box 1 will slide on box 2 çFBD of box 1 çBox 1 does not slide if m1m1 N1N1 m1gm1g T f f MAX =  s m 1 g = 3.5  1 kg  10 m/s 2 = 35 N T - f = 0  T = f l The maximum value for the static friction force is  f MAX =  s N 1 =  s m 1 g çBox 1 does not slide if f = T ≤ f MAX çHere f = T = 30 N ≤ f MAX = 35 N  No sliding Both boxes have the same acceleration ANSWER: (b) a 1 =a 2 =T/(m 1 +m 2 )T=Ma 

7. Physics 1501: Lecture 9, Pg 6 Lecture 9, ACT 1 More details l Let us find how FBD and Newton’s laws lead to our intuitive result m1m1 N1N1 m1gm1g T f l FBD of box 1 l Action/reaction pairs (Newton’s 3 rd law) m1m1 N1N1 m1gm1g T f -f -N 1 -T -m 1 g

8. Physics 1501: Lecture 9, Pg 7 Lecture 9, ACT 1 More details l FBD of box 2 l Action/reaction pairs (Newton’s 3 rd law) m2m2 m2gm2g N2N2 -f -N 1 m2m2 m2gm2g -f f -m 2 g N1N1 -N 1 N2N2 -N 2

9. Physics 1501: Lecture 9, Pg 8 Lecture 9, ACT 1 More details l Box 1 l Box 2 T + f = m 1 a 1 N 1 + m 1 g = 0 i j i : j :  T - f = m 1 a 1 N 1 = m 1 g i : j : -f = m 2 a 2 N 2 - N 1 + m 2 g = 0 m1m1 N1N1 m1gm1g T f a1a1 m2m2 m2gm2g N2N2 -f -N 1 a2a2 f = m 2 a 2 N 2 = (m 1 + m 2 )g  a 1 = a 2 = a (no sliding) T = f+m 1 a 1 = m 2 a 2 +m 1 a 1 = (m 2 +m 1 )a a= 30 N / (1 kg + 2 kg) = 10 m/s 2

10. Physics 1501: Lecture 9, Pg 9 Lecture 9, ACT 2 Friction and Motion l Consider the same boxes as ACT 1, but now the box 1 is being pulled by a horizontal string having tension T = 40 N. Now, the acceleration of box 1 is (a)Greater than (b) Equal to (c) Smaller than the acceleration of box 2 ? m2m2m2m2 T m1m1m1m1  friction coefficients  s =3.5 and  k =0.5 slides without friction a2a2 a1a1

11. Physics 1501: Lecture 9, Pg 10 Lecture 9, ACT 2 Solution l First, we need to know if box 1 will slide on box 2 çFBD of box 1 çBox 1 does not slide if m1m1 N1N1 m1gm1g T f f MAX =  s m 1 g = 3.5  1 kg  10 m/s 2 = 35 N T - f = 0  T = f l The maximum value for the static friction force is  f MAX =  s N 1 =  s m 1 g çBox 1 does not slide if f = T ≤ f MAX çHere T = 40 N > f MAX = 35 N  It will slide Both boxes have different accelerations ANSWER: (b) l The maximum value for the static friction force is çANSWER: (a)

12. Physics 1501: Lecture 9, Pg 11 Lecture 9, ACT 2 Solution l The force of friction is therefore due to the kinetic coefficient of friction  k l The FBD and force pairs are the same as before, except that we now have for f f =  k | N 1 | =  k m 1 g m1m1 N1N1 m1gm1g T f l So only the i component will be modified.

13. Physics 1501: Lecture 9, Pg 12 Lecture 9, ACT 2 More details l Box 1 l Box 2 i j m1m1 N1N1 m1gm1g T f a1a1 m2m2 m2gm2g N2N2 -f -N 1 a2a2 T + f = m 1 a 1 T - f = m 1 a 1   a 1 =(T -  k m 1 g) /m 1 a 1 =(40 N - 0.5  1kg  10 m/s 2 ) /1kg = 35 m/s 2   -f = m 2 a 2 f = m 2 a 2 a 2 =  k m 1 g / m 2  a 2 = 0.5  1kg  10 m/s 2 /2 kg = 2.5 m/s 2 

14. Physics 1501: Lecture 9, Pg 13 Newton’s Laws and Circular Motion v R Centripedal Acceleration a C = v 2 /R What is Centripedal Force ? F C = ma C = mv 2 /R aCaC Animation

15. Physics 1501: Lecture 9, Pg 14 Example Problem I am feeling very energized while I shower. So I swing a soap on a rope around in a horizontal circle over my head. Eventually the soap on a rope breaks, the soap scatters about the shower and I slip and fall after stepping on the soap. To decide whether to sue ACME SOAP I think about how fast I was swinging the soap (frequency) and if the rope should have survived. From the manufacturers web site I find a few details such as the mass of the soap is 0.1 kg (before use), the length of the rope is 0.1 m and the rope will break with a force of 40 N. (assume F BS is large versus the weight of the soap)

16. Physics 1501: Lecture 9, Pg 15 Example Problem Step 1 In this case the picture was given. I need to find the frequency of the soap’s motion that caused the rope to break. I will use Newton’s Second Law and uniform circular motion.

17. Physics 1501: Lecture 9, Pg 16 Example Problem Step 2 Diagram. I will solve for the frequecy f. { I know M=0.1 kg, R=0.1m, and F BS = 40N } I will use F = ma (Newton’s Second Law), a=v 2 /r=  2 r (circular motion) 2  f =  v T

18. Physics 1501: Lecture 9, Pg 17 Example Problem Step 3 – Solve Symbolically 1  F = T = ma 2. a =  2 R  T = m  2 R 3.  = 2  f  T = 4  2 f 2 mR In general When it breaks

19. Physics 1501: Lecture 9, Pg 18 Example Problem Step 4 – Numbers

20. Physics 1501: Lecture 9, Pg 19 Example Problem Step 5 – Analyze A.The units worked out to give s -1, which is correct for a frequency. B.It seems that the suit is in trouble, if not necessarily dead. Being able to twirl your soap safely 10 rev/s is pretty good. I might have been excessively boisterous.

21. Physics 1501: Lecture 9, Pg 20 Lecture 9, ACT 3 Circular Motion Forces l How fast can the race car go ? (How fast can it round a corner with this radius of curvature ?) m car = 1500 kg  S = 0.5 for tire/road R = 80 m R A) 10 m/s B) 20 m/s C) 75 m/s D) 750 m/s

22. Physics 1501: Lecture 9, Pg 21 Banked Corners In the previous ACT, we drew the following free body diagram for a race car going around a curve on a flat track. N mgmg FfFf What differs on a banked curve ?

23. Physics 1501: Lecture 9, Pg 22 Banked Corners Free Body Diagram for a banked curve. N mgmg FfFf For small banking angles, you can assume that F f is parallel to ma. This is equivalent to the small angle approximation sin  = tan . Can you show that ? mama

24. Physics 1501: Lecture 9, Pg 23 Nonuniform Circular Motion Earlier we saw that for an object moving in a circle with nonuniform speed then a = a r + a t. arar atat What are F r and F t ?

25. Physics 1501: Lecture 9, Pg 24 My match box car is going to do a loop the loop. What must be its minimum speed at the top so that it can make the loop successfully ?? Example Exercise 1

26. Physics 1501: Lecture 9, Pg 25 Example Exercise 1 mgmg Radial : F r = N + mg cos  mv 2 /R Tangential : F t = mg sin  Solve the first for v. mgmg  N   

27. Physics 1501: Lecture 9, Pg 26 Example Exercise 1 mgmg N  To stay on the track there must be a non-zero normal force. Why ? The limiting condition is where N = 0. And at the top,  = 0.