1. Lecture 5Purdue University, Physics 2201 Lecture 05 Forces and Motion beyond 1 D Textbook Sections 3.7, 4.1 PHYSICS 220

2. Lecture 5Purdue University, Physics 2202 UNIMPORTABLE: #00E9BF56,2.00,2.00 #09B63F80,6.00,6.00 #0E133429,2.00,2.00 #225BEB92,4.00,4.00 #80330CBF,6.00,6.00 #803AF64C,6.00,6.00 #809B0813,4.00,4.00 #80E6096F,6.00,6.00 #810C8D00,2.00,2.00 #81709E6F,6.00,6.00 #817EE11E,6.00,6.00 #8279AE55,6.00,6.00 #831033A0,4.00,4.00 #8341E92B,4.00,4.00 #834C955A,6.00,6.00 #83CA7831,6.00,6.00

3. Lecture 5Purdue University, Physics 2203 Exercise Fred throws a ball 30 m/s vertically upward. How long does it take to hit the ground 2 meters below where he released it? y = y 0 + v y0 t + 1/2 gt 2 v y = v y0 + gt v y 2 = v y0 2 + 2g(y-y 0 ) y = y 0 + v y0 t - 0.5 g t 2 y-y 0 - v y0 t + 0.5g t 2 = 0 -2 – 30 t + 0.5*9.8 t 2 = 0 t = 6.19 s or -.06 s

4. Lecture 5Purdue University, Physics 2204 Exercise Fred throws a ball 30 m/s vertically upward. What is the maximum height the ball reaches? How long does it take to reach this height? v 2 -v o 2 = 2 a  y  y = (v 2 -v o 2 )/ (2 a) = -30 2 / (2 * (-9.8)) = 46 m v = v 0 + a t t = (v-v 0 ) / a = (0 – 30 m/s )/ (-9.8 m/s 2 ) = 3.1 seconds

5. Lecture 4Purdue University, Physics 2205 Contact Force: Spring Force exerted by a spring is directly proportional to the amount by which it is stretched or compressed. F spring = k x always trying to restore its original length Example: When a 5 kg mass is suspended from a spring, the spring stretches 8 cm. Determine the spring constant. F spring - F gravity = 0 F spring = F gravity k x = m g y x k = m g / x = (5 kg) x (9.8 m/s 2 ) /(0.08 m) = 612 N/m F spring F gravity

6. Lecture 4Purdue University, Physics 2206 Contact Force: Tension Tension – A force transmitted by a rope, cord, cable or the like which transmits a force from one end to an object attached at the other end Ideal string (or cord, rope, etc.): –Always maintains constant tension everywhere. –Has a zero mass. –Tension is parallel to the string Pulley changes the direction of the force associated with the tension in the rope.

7. Lecture 4Purdue University, Physics 2207 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1) T - m 1 g = m 1 a 1 2) T - m 2 g = -m 2 a 1 using a 1 = -a 2 2) T = m 2 g -m 2 a 1 1) m 2 g -m 2 a 1 - m 1 g = m 1 a 1 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) 1 T m1gm1g 2 T m2gm2g Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | y x

8. Lecture 4Purdue University, Physics 2208 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1 T m1gm1g 2 T m2gm2g y x a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) a = 2.45 m/s 2  x = v 0 t + ½ a t 2  x = ½ a t 2 t = sqrt(2  x/a) t = 0.81 seconds

9. Lecture 4Purdue University, Physics 2209 iClicker Two boxes are connected by a string over a frictionless pulley. In equilibrium, box 2 is lower than box 1. Compare the weight of the two boxes. A) They have the same weight B) Box 1 is heavier C) Box 2 is heavier 1 2  F = 0 1) T - m 1 g = 0 2) T – m 2 g = 0 => m 1 = m 2 1 T m1gm1g 2 T m2gm2g

10. Lecture 4Purdue University, Physics 22010 Tension Example Determine the force exerted by the hand to suspend the 45 kg mass as shown in the picture. y x  F = 0 T + T – W = 0 2 T = W T = m g / 2 = (45 kg x (9.8 m/s 2 )/ 2 = 220 N T T W 1) 220 N2) 440 N3) 660 N 4) 880 N5) 1100 N Remember the magnitude of the tension is the same everywhere along the rope!

11. Lecture 4Purdue University, Physics 22011 Question What is the force on the ceiling? y x  F = 0 F c -T - T – T = 0 F c = 3 T F c = 3 x 220 N = 660 N FcFc T A) 220 NB) 440 NC) 660 N D) 880 NE) 1100 N

12. Lecture 4Purdue University, Physics 22012 Question What does the scale read? A) 225 N B) 550 NC) 1100 N 30 The sum of the forces here will be 0. Thus, the force tension will equal the force gravity. Since the force of gravity is 550N, the force of tension (which is measured by the scale) will also be 550N.

13. Lecture 4Purdue University, Physics 22013 Two blocks one sliding one hanging A block of mass m 1 =3kg rests on a frictionless horizontal surface. A second block of mass m 2 =2kg hangs from an ideal cord of negligible mass who runs over an ideal pulley. Block 2 starts from rest 0.8 meters above the floor, how long does it take to hit the floor? a a a 1 = (m 2 )g / (m 1 +m 2 ) a = 3.9 m/s 2  y = v 0 t + ½ a t 2 = ½ a t 2 t = sqrt(2  y/a) t = 0.64 seconds

14. Lecture 4Purdue University, Physics 22014 Friction Magnitude of frictional force is proportional to the normal force. F kinetic =  k N  k coefficient of kinetic friction F static   s N  s coefficient of static friction Be Careful! Static friction , can be any value up to  s N Direction always opposes motion F FfFf

15. Lecture 5Purdue University, Physics 22015 Inclined Plane N = m g cos  m g sin  - F f = 0 F f =  N m g sin  -  m g cos  = 0 N FfFf mg  Special case 1: Start with  at zero and slowly increase . Just before it slides  s m g cos  = m g sin  tan  =  s Special case 2: Object is sliding down at constant velocity, that is a = 0  k m g cos  = m g sin  tan  =  k 

16. Lecture 5Purdue University, Physics 22016 iClicker What is the normal force of ramp on block? A) N > mg B) N = mg C) N < mg T N W In “y” direction:  F = ma N – mg cos  = 0 N = mg cos  y x  W  W sin  W cos 

17. Lecture 5Purdue University, Physics 22017 Force at Angle Example A person is pushing a 15 kg block across a floor with  k = 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block? y x Normal Weight Pushing x- direction:  F x = 0 F push cos  – F friction = 0 F push cos  –  F Normal = 0 F Normal = F push cos  /  y- direction:  F y = 0 F Normal –F weight – F Push sin  = 0 F Normal –mg – F Push sin  = 0 Combine: ( F push cos  /  –mg – F Push sin  = 0 F push ( cos  /  - sin  ) = mg F push = m g / ( cos  /  – sin  )  Friction  F push = 80 N