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AP homework on Newton’s Laws WS #1 - #1 Ch.5 4th ed. - Newton's Laws The Problem 1. A force, F, applied to an object of mass m1 produces an acceleration of 3.0 m/s². The same force applied to a second object of mass m2 produces an acceleration of 1.0 m/s². A. What is the value of the ratio m1/m2? F Draw the M1 diagram M2 F WS #1 - #1 Ch.5 4th ed. - Newton's Laws The Solution a1 3.0 m a2 1.0 m s² s² m1 a2 1.0 m s² .33 m2 a1 3.0 m s² If m1 and m2 are combined, find their acceleration under the action of the force F. a2 m1 m2 m 1 1 a12 m2 m2 a12 a2 1 m1 1.0 m s² 1 .33 m2 a12 .75 m s² WS #2 - #26 Ch.5 4th ed - Newton's Laws 2. Find the tension in each cord for the systems described below. (Neglect the mass of the cords.) 40.°= 1 = 50.° 60.° 2 T T2 1 T1 90.° T3 T2 T3 5.0 kg 10 kg (a) (b) WS #2 - #26 Ch.5 4th ed - Newton's Laws Free body Diagrams T1 T2 T 1 = 40.° = 50.° 1= 60.° 2 = 0.0° T2 1 2 T3 T3 WS #2 - #26 Ch.5 4th ed - Newton's Laws The Working Equations T3 mg ˆ j ˆ T1 T1 cos1 i T1 sin1 ˆ j ˆ T2 T2 cos 2 i T2 sin 2 ˆ j F x ma x T1 cos1 T2 cos 2 0 T1 cos1 T2 cos 2 cos1 T2 T1 cos 2 WS #2 - #26 Ch.5 4th ed - Newton's Laws The Working Equations F y ma y T1 sin 1 T2 sin 2 T3 0 cos1 T1 sin1 T1 sin 2 T3 cos 2 cos1 sin 2 T1 sin 1 T3 cos 2 sin1 cos 2 cos1 sin 2 T1 T3 cos 2 WS #2 - #26 Ch.5 4th ed - Newton's Laws The Working Equations T3 cos 2 T1 sin1 cos 2 cos1 sin 2 T3 cos 2 T1 sin 1 2 From Symmetry T3 cos 1 T2 sin 2 1 WS #2 - #26 Ch.5 4th ed - Newton's Laws The Solution Part a 1 40 2 50 m 5.0 kg T3 mg 5.0 kg 9.80 m s² 49 N T3 cos 2 49 N cos 50 T1 31.5 N sin 1 2 sin 40 50 T3 cos1 49 N cos 40 T2 37.5 N sin 2 1 sin 50 40 WS #2 - #26 Ch.5 4th ed - Newton's Laws The Solution Part b 1 60 2 0 m 10. kg T3 mg 10 kg 9.80 m s² 98 N T3 cos 2 98 N cos 0 T1 110 N sin 1 2 sin 60 0 T3 cos1 49 N cos 60 T2 57 N sin 2 1 sin 0 60 WS #3 - #28 Ch.5 4th ed - Newton's Laws The Problem 3. A 225 N weight is tied to the middle of a strong rope, and two people pull at opposite ends of the rope in an attempt to lift the weight. a. What is the magnitude F of the force that each person must apply in order to suspend the weight as shown in the Figure (P5.28)? b. Can they pull in such a way as to make the rope horizontal? Explain. T1 T2 10.°= 1 2= 10.° T3 225 N WS #3 - #28 Ch.5 4th ed - Newton's Laws The Diagram T1 T2 10.°= 1 2= 10.° T3 225 N WS #3 - #28 Ch.5 4th ed - Newton's Laws The Solution - The equations come from problem 26 1 10 2 10 T3 225 N T3 cos 2 225 N cos10 T1 sin 1 2 sin 10 10 T1 650 N T3 cos1 225 N cos10 T2 sin 2 1 sin 10 10 T2 650 N WS #4 - #31 Ch.5 4th ed - Newton's Laws The Problem 4. A bag of cement hangs from three wires as shown on page 134 figure p5.31. Two of the wires make angels of q1 and q2 with the horizontal. If the system is in equilibrium. w cos 2 Show that T2 sin 1 2 See problem 26 WS #4 - #31 Ch.5 4th ed - Newton's Laws The Problem 4. Given that w = 325 N, q1 = 10°,and q2 = 25°, find the tensions T1, T2, and T3 in the wires. 1 10 2 25 T3 325 N T3 cos 2 325 N cos 25 T1 sin 1 2 sin 10 25 T1 510 N T3 cos1 325 N cos10 T2 sin 2 1 sin 10 25 T2 560 N WS #5 - #17 Ch.5 3th ed - Newton's Laws The Problem 5. Two forces F1 = (3i - 5j) N and F2 = (2i + j) N act on a 1.50 kg body. 1. What are the magnitude and direction of the acceleration? 2. If the body is initially resting at the origin, what are the coordinates of its position at t = 4.00 s? WS #5 - #17 Ch.5 3th ed - Newton's Laws The Solution ˆ F1 3i 5 ˆ N j ˆ j F2 2i ˆ N m 1.50kg F ma a F ˆ j 3i 5 ˆ N 2i ˆ N ˆ j ˆ 5i 4 ˆ N j m 1.50kg 1.50kg a 3.33i 2.67 ˆ ˆ j m s2 a 4.27 m s2 at -38.7 WS #5 - #17 Ch.5 3th ed - Newton's Laws The Problem and The Solution B. If the body is initially resting at the origin, what are the coordinates of its position at t = 4.00 s? a 3.33i 2.67 ˆ ˆ j m s2 vi 0 so 0 t 4.00s s so vi t at 1 2 2 s 0 0 t 1 2 ˆ 3.33i 2.67 ˆ j m s2 4.00s 2 s 26.6i 21.4 ˆ m ˆ j WS #6 - #18 Ch.5 4th ed - Newton's Laws The Problem and Diagrams 6. Two forces F1 and F2 act on a 5.0 kg mass. If F1 = 20. N and F2 = 15 N, Find the acceleration in both of the following figures. WS #6 - #18 Ch.5 4th ed - Newton's Laws The Solution to diagram 1 m 5.0kg ˆ F1 20.iN F2 15 ˆN j a F F 1 F2 m m a ˆ 20.i 15 ˆ N j 5.0kg a 4.0i 3.0 ˆ ˆ j m s2 WS #6 - #18 Ch.5 4th ed - Newton's Laws The Solution to diagram 2 m 5.0kg ˆ F1 20.iN F2 15cos 60 i 15sin 60 ˆ N ˆ j a F F 1 F2 m m a 20.iˆ 15cos 60iˆ 15sin 60 ˆj N 5.0kg a 5.5i 2.6 ˆ ˆ j m s2 WS #7 - #67 Ch.5 3rd ed - Newton's Laws The Problem 7. The largest caliber antiaircraft gun operated by the Luftwaffe during World War II was the 12.8-cm Flak 40. This weapon fired a 25.8 kg shell with a muzzle velocity of 880 m/s. What propulsive force was necessary to attain the muzzle velocity within the 6.0 m barrel? (Assume constant acceleration and neglect the earth’s gravitational effect.) (Physics for Scientists and Engineers by Serway, 3rd Edition, Ch 5# 67) WS #7 - #67 Ch.5 3rd ed - Newton's Laws The Diagram 6.0 m 25.8 kg F=? V = 880 m/s WS #7 - #67 Ch.5 3rd ed - Newton's Laws The Solution d 6.0 m v f 880 m/s m 25.8 kg vo 0 F ? F ma v 2 vi2 2ad f v 2 vi2 a f 2d v 2 vi2 880 m/s 0 2 F m 25.8 kg f 2d 2 6.0 m F 1.7 106 N WS #8 - #43 Ch.5 4th ed - Newton's Laws The Problem 8. A 72.0-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 0.800s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.50 s and comes to rest. What does the spring scale register: a. before the elevator starts to move? b. during the first 0.80 s.? c. while the elevator is traveling at constant speed? d. during the time it is slowing down? WS #8 - #43 Ch.5 4th ed - Newton's Laws The Diagram Elevator Normal 72 kg Force of the scale is your weight 1.2m/s Force of Gravity Scale WS #8 - #43 Ch.5 4th ed - Newton's Laws The Solution a. What does the spring scale register: before the elevator starts to move? m 72.0kg g 9.80 m s 2 ay 0 F y ma y N Fg 0 N Fg mg 72.0kg 9.8 0 m s 2 706 N WS #8 - #43 Ch.5 4th ed - Newton's Laws The Solution b. What does the spring scale register: during the first 0.80 s.? m 72.0kg g 9.80 m s2 v f 1.20 m s t 0.80 s vi 0 v f v i at v f vi 1.20 m 0 a s 1.50 m s2 t 0.80s Fy ma y N Fg ma y N Fg ma y N mg ma y m g a y 72.0kg 9.80 m s2 1.50 m s2 814N WS #8 - #43 Ch.5 4th ed - Newton's Laws The Solution c. What does the spring scale register: while the elevator is traveling at constant speed m 72.0kg g 9.80 m s2 ay 0 F y ma y N Fg 0 N Fg mg 72.0kg 9.80 m s2 706N WS #8 - #43 Ch.5 4th ed - Newton's Laws The Solution d. What does the spring scale register: during the time it is slowing down? m 72.0kg g 9.80 m s 2 vf 0 t 1.50s v i 1.2 0 m s v f v i at v f vi 0 1.20 m a s 0.800 m s 2 t 1.50s F y ma y N Fg ma y N Fg ma y N mg ma y m g a y 72.0kg 9.80 m s 2 0.8 00 m s 2 648 N WS #9 - #45 Ch.5 4th ed - Newton's Laws The Problem and Solution 9. A net horizontal force F = A + Bt3 acts on a 3.50-kg object, where A = 8.60 N and B = 2.50 N/s3. What is the horizontal speed of this object 3.00 s after it starts from rest? F = A + Bt 3 A 8.60N B 2.50 N s3 m 3.50kg t 3.00s vi 0 vf ? F ma A + Bt 3 ma A + Bt 3 a m 3.00 3.00 3.00 3.00 A + Bt 3 A Bt 3 v f a dt dt dt dt 0 0 m 0 m 0 m 2.50 3.00s A B 0 3.00 4 4 A Bt 4 8.60N N vf t 3.00s s3 0 m 4m 0 3.50kg 4 3.50kg m 4m v f 7.37 m 14.5 m 21.8 m s s s WS #10 - #48 Ch.5 4th ed - Newton's Laws The Problem 10. A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion. After it is in motion, a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find the coefficients of static and kinetic friction from this information. WS #10 - #48 Ch.5 4th ed - Newton's Laws The Solution m 25kg Fs 75N Fk 60.N s ? k ? N Fy ma y 0 f F Fy N mg 0 N mg 25kg F x ma x 0 F x F f 0 f F mg fs F 75N s s .306 N mg 25kg 9.80 m s2 fk Fk 60N k .245 N m g 25kg 9.80 s2 m WS #11 - #72 Ch.5 4th ed - Newton's Laws The Problem 11. In the figure, the man and the platform together weigh 750 N. Determine how hard the man would have to pull to hold himself off the ground. (Or is it impossible? If so, explain why.) F y ma 0 T T T T T 750N 0 T 750N 750N T down on the man’s arms WS #12 - #53 Ch.5 4th ed – Newton’s Law The Problem 11. A boy drags his 60.0 N sled at constant speed up a 15° hill. He does so by pulling with a 25 N force on a rope attached to the sled. If the rope is inclined at 35° to the horizontal 1. what is the coefficient of kinetic friction between the sled and the snow? 2. At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope? (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53) WS #12 - #53 Ch.5 4th ed – Newton’s Law The Diagram WS #12 - #53 Ch.5 4th ed – Newton’s Law The Solution mg 60.N F 25N =20° 15 F y ma y 0 N F sin mg cos 0 N mg cos F sin F x ma x 0 F cos mg sin F f 0 F cos mg sin N 0 WS #12 - #53 Ch.5 4th ed – Newton’s Law The Solution F cos mg sin mg cos F sin 0 mg cos F sin mg sin F cos mg cos F sin F cos mg sin F cos m g sin mg cos F sin 25N cos 20° 60.N sin 15 .16 60.N cos 15 25N sin 20° WS #12 - #53 Ch.5 4th ed – Newton’s Law The solution Part b 1. At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope? (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53) F y ma y 0 N m g cos 0 N m g cos WS #12 - #53 Ch.5 4th ed – Newton’s Law The solution Part b 1. At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope? (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53) F x ma x mg sin F f ma x mg sin N ma x mg sin m g cos ma x a x g sin g cos g sin cos a x 9.80 m s 2 sin 15 .16 cos 15 a x 1.0 m s 2 WS #13 - #39 Ch.5 4th ed - Newton's Laws The Problem 13. A small bug is placed between two blocks of masses m1 and m2 (m1 > m2) on a frictionless table. A horizontal forcec F, can be applied to either m1, as in the first figure below or m2 as in the second figure below. For which of these two cases does the bug have a greater chance to surviving? Show! (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 39) F m1 m1 F m2 m2 WS #13 - #39 Ch.5 4th ed – Newton’s Laws The Solution a F F m m 1 m2 When force is on m1 When force is on m1 m1 exerts a force F2 on m2 m2 exerts a force F1 on m1 F2 m2 a F1 m1a m2 F m1 F F2 F1 m1 m2 m1 m2 if m2 m1 F2 F1 WS #14 - #50 Ch.5 4th ed – Newton’s Laws The Problem 14. What force must be exerted on Block A in order for Block B not to fall. The coefficient of static friction between blocks A and B is 0.55, and the horizontal surface is frictionless. (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 50) WS #14 - #50 Ch.5 4th ed – Newton’s Laws The Diagram Block B Block A 10.0 kg 100. kg WS #14 - #50 Ch.5 4th ed – Newton’s Laws The Free Body Diagrams WS #14 - #50 Ch.5 4th ed – Newton’s Laws The Solution – 10 kg mass F y 0 F x mb a F f mb g 0 N mb a F f mb g mb g mb a N mb g g N mb g a WS #14 - #50 Ch.5 4th ed – Newton’s Laws The Solution – both mass together g 9.80 m s 2 .55 ma 100 kg m b 10.0 kg g a F x a m g F ma mb 9.80 m s 2 F 100 kg 10.0 kg .5 5 F 1960 N WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Problem 15. Three baggage carts of masses m1, m2, and m3 are towed by a tractor of mass M along an airport apron. The wheels of the tractor exert a total frictional force F on the ground as shown. In the following, express your answers in terms of F, M, m1, m2, m3 and g. a. What are the magnitude and direction of the horizontal force exerted on the tractor by the ground? b. What is the smallest value of the coefficient of static friction that will prevent the wheels from slipping? Assume that each of the two drive wheels on the tractor bears 1/3 of the tractor’s weight. c. What is the acceleration, a, of the system (tractor plus baggage carts)? d. What are the tensions T1, T2, and T3 in the connecting cables? e. What is the net force on the cart of mass m2? (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 71) WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Solution Part a a. What are the magnitude and direction of the horizontal force exerted on the tractor by the ground? a. From Newton’s Third Law since the force of the tractor on the ground is F, the magnitude of the force of the ground on the tractor must also be F but in the opposite direction. Therefore the force exerted on the tractor is F to the right. WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Solution Part b b. What is the smallest value of the coefficient of static friction that will prevent the wheels from slipping? Assume that each of the two drive wheels on the tractor bears 1/3 of the tractor’s weight. N F y ma y 0 N 2 mg 0 3 N 2 mg F 3 F f friction N 2/3 mg F F 2 N 3 mg 3F 2m g WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Solution Part c c. What is the acceleration, a, of the system (tractor plus baggage carts)? a F external m F a M m1 m2 m3 WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Solution Part d d. What are the tensions T1, T2, and T3 in the connecting cables? There is one force on m1 T2 is accelerating two masses m1 and m2 F 1 m1a F m 1 m2 a F T1 m1 M m1 m2 m3 T2 m2 m1 F M m1 m2 m3 m1 F T1 M m1 m2 m3 WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Solution Part d T3 is accelerating three masses m1 and m2 and m3 F m 1 m2 m3 a T3 m1 m2 m3 F M m1 m2 m3 WS #15 - #71 Ch.5 4th ed – Newton’s Laws The Solution Part e e. What is the net force on the cart of mass m2? The net force is F F 2 m2a F F2 m2 M m m m 1 2 3 m2F F2 M m m m 1 2 3 WS #16 - #37 Ch.5 4th ed – Newton’s Laws The Problem 1. In the system on the next slide, a horizontal force Fx acts on a 8.00 kg mass. a. For what values of Fx does the 2.00 kg mass accelerate upward? b. For what values of Fx is the tension in the cord zero? c. Plot the acceleration of the 8.00 kg mass versus Fx. Include values of Fx from -100 N to +100 N. (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 37) WS #16 - #37 Ch.5 4th ed – Newton’s Laws The Diagram WS #16 - #37 Ch.5 4th ed – Newton’s Laws The Diagram y x x y WS #16 - #37 Ch.5 4th ed – Newton’s Laws The Free Body Diagram y y x x WS #16 - #37 Ch.5 4th ed – Newton’s Laws Find F such that a > 0 m 2.0 kg 4m 8.0 kg F x ma x F mg m 4m a a 0 if F mg 0 F mg F 2.0 kg 9.80 m s2 19.6 N WS #16 - #37 Ch.5 4th ed – Newton’s Laws Find F such that T = 0 For m For 4m F ma F ma F T 4ma T mg m a F 0 4ma 0 mg ma F 4m g 4 2.0 kg 9.80 m s2 a g F 78.4 N WS #16 - #37 Ch.5 4th ed – Newton’s Laws Part C – The Plot WS #17 - #47 Ch.5 4th ed – Newton’s Laws The Problem 17. Two blocks of masses m and 4m are connected by a light string passing over a fixed pulley as shown at the right. The 4m mass is on a smooth frictionless surface and is pushed toward the left by a constant horizontal force F. a. What value of F will give the system an acceleration of 0.400 g b. Find the tension in the string when F has the value determined in (A). c. What happens when F>4.00 mg? (Physics for Scientists and Engineers by Serway, 3rd Edition, Ch 5# 47) WS #17 - #47 Ch.5 4th ed – Newton’s Laws The Diagram y x x y WS #17 - #47 Ch.5 4th ed – Newton’s Laws The Solution (A) What value of F will give the system an acceleration of 0.400 g a system F m F mg 0.400 g 4m m 0.400 g 4m m F mg F 0.400 g 5m mg 0.400 g 5m mg 2 m g mg F mg WS #17 - #47 Ch.5 4th ed – Newton’s Laws The Solution (b) Find the tension in the string when F has the value determined in (A). a F 4m y m F T 0.400 g 4m x F T 0.400 g 4m T 4m 0.400 g F T 4m 0.400 g mg T .600mg WS #17 - #47 Ch.5 4th ed – Newton’s Laws The Solution (c) What happens when F>4.00 mg? a F 4m m y F T a 4m F T 4ma x T 4ma F T 4ma 4mg 4m a g but amax g T 0 The String goes slack WS #18 - #46 Ch.5 4th ed – Newton’s Laws The Problem 18. Mass m1 on a smooth horizontal table is connected to mass m2 through a very light pulley P1 and a light fixed pulley P2 as shown. a. If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relationship between these accelerations? b. Express the tensions in the strings, and c. the accelerations a1 and a2 in terms of the masses m1, m2, and g. (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 46) WS #18 - #46 Ch.5 4th ed – Newton’s Laws The Diagram T’ P1 T P2 m1 T’ T m2 m2g WS #18 - #46 Ch.5 4th ed – Newton’s Laws If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relationship between these acceleration A) Find a1 in terms of a 2 . 2d1 2d 2 a1 = 2 a2 = 2 But when m 2 moves a distance d 2 , t t Pulley P1 moves a distance d 2 , which means that the string between m1 and P1 gets a distance d 2 shorter. This means that m1 moves 2d 2 . 2d1 2 2d 2 2d a1 = 2 = = 2 2 2 = 2a 2 t t2 t a1 = 2a 2 WS #18 - #46 Ch.5 4th ed – Newton’s Laws Express the (B) tensions in the strings, For m1 F 1 = m 1 a1 T1 = m1a1 T1 a1 = m1 For P1 F p1 = m p1 ap1 = 0 T2 - T1 - T1 = 0 T2 T1 = 2 WS #18 - #46 Ch.5 4th ed – Newton’s Laws Express the (B) tensions in the strings, For m 2 F 2 = m 2a2 T1 a1 m1 m 2 g - T2 = m 2 a 2 = m 2 = m2 2 2 T2 T1 m 2 T2 m 2 g - T2 = m 2 = m2 2 = 2m1 2m1 4m1 m 2 T2 m 2 g - T2 = 4m1 4m1m 2 g - 4m1 T2 = m 2 T2 4m1m 2 g T2 = m 2 + 4m1 WS #18 - #46 Ch.5 4th ed – Newton’s Laws Express the (B) tensions in the strings, 4m1m 2 g T2 m 2 + 4m1 2m1m 2 g T1 = = = 2 2 m 2 + 4m1 WS #18 - #46 Ch.5 4th ed – Newton’s Laws (c) the accelerations a1 and a2 in terms of the masses m1, m2, and g 2m1m 2 g a1 = T1 = m 2 + 4m1 = 2m 2 g m1 m1 m 2 + 4m1 2m 2 g a1 m 2 + 4m1 m2g a2 = = = 2 2 m 2 + 4m1 WS #19 - #85 Ch.5 4th ed - Newtons Laws The Problem 19. The three blocks to the right are connected by mass- less strings that pass over massless frictionless pulleys. The acceleration of the system is 2.35 m/s2 to the left and the surfaces are rough. Mass 1 is 10.0 kg, mass 2 is 5.00 kg, and mass 3 is 3.00 kg. Find: a. the tensions in the strings and b. the coefficient of kinetic friction between blocks and surfaces. (Assume the same m for both blocks) (Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 85) 5.00 kg T1 T2 3.00 kg 10.0 kg WS #19 - #85 Ch.5 4th ed - Newtons Laws The Diagram WS #19 - #85 Ch.5 4th ed - Newtons Laws The Free body Diagram They all acceleration as a unit. The tensions are internal. WS #19 - #85 Ch.5 4th ed - Newtons Laws (b) the coefficient of kinetic friction between blocks and surfaces. m1 10.0kg m2 5.00kg m3 3.00kg a 2.35 m s2 25.0 For body 2 For body 3 F y ma y 0 F y ma y 0 N 2 m2 g 0 N 3 m3 g cos 0 N 2 m2 g N 3 m3 g cos WS #19 - #85 Ch.5 4th ed - Newtons Laws (b) the coefficient of kinetic friction between blocks and surfaces. F x ma x F f 3 m3 g sin F f 2 m1 g m1 m2 m3 a N 3 m3 g sin N 2 m1 g m1 m2 m3 a N 2 N 3 m1 m2 m3 a m1 g m3 g sin m1 m2 m3 a m1 g m3 g sin N2 N3 m1 m2 m3 a m1 g m3 g sin m3 g cos m2 g WS #19 - #85 Ch.5 4th ed - Newtons Laws (b) the coefficient of kinetic friction between blocks and surfaces. m1 m2 m3 a m1 g m3 g sin m3 g cos m2 g 10.0 5.00 3.00 2.35 10.0 9.80 3.00 9.80 sin 25.0 kg m s 2 9.8 0 m s 2 3.00kg cos 25.0 5.0 0kg .57 2 WS #19 - #85 Ch.5 4th ed - Newtons Laws (a) the tensions in the strings Find T1 using body 1 F m a 1 T1 m1 g m1a T1 m1a m1 g m1 a g T1 10.0kg 2.35 m s 2 9.80 m s 2 T1 74.5N WS #19 - #85 Ch.5 4th ed - Newtons Laws (a) the tensions in the strings Find T2 using body 3 F m a 3 m3 g sin F f 3 T2 m3 a T2 m3 a m3 g sin m3 g cos T2 m3 a g sin cos T2 3.00kg 2.35 m s 2 9.80 m s 2 sin 25.0 .572 cos 25.0 T2 34.7N WS #20 - #65 Ch.5 4th ed - Newtons Laws 20. A mass M is held in place by an applied force FA and a pulley system as shown at the right. The pulleys are mass-less and frictionless. Find the tension in each section of rope, T1, T2, T3, T4, and T5 and the applied force FA. WS #20 - #65 Ch.5 4th ed - Newtons Laws Starting solutions Mass M y x On mass M F m Ma 0 T5 Mg 0 T5 Mg WS #20 - #65 Ch.5 4th ed - Newtons Laws The Solution - T2,T3 On Pulley P2 Pulley 2 T2 T3 T5 0 y T2 and T3 are the same string T2 T3 T3 T3 T5 2T3 T5 0 x T5 Mg T3 2 2 Mg T3 T2 2 WS #20 - #65 Ch.5 4th ed - Newtons Laws The Solution – T1,T4 On Pulley P1 T1 T2 T3 T4 0 Pulley 1 T1 , T2 and T3 are the same string T1 T2 T3 y Mg T1 2 x Mg Mg Mg T4 0 2 2 2 3 Mg T4 2 Mg F T1 2 Newton’s Law 1998 AP Question - Solution From the 1998 Physics C Mech AP Test 1. Block 1 of mass ml is placed on block 2 of mass m2, which m 1 is then placed on a table. A string connecting block 2 to a m 2 hanging mass M passes over a pulley attached to one end of the table, as shown to the right. The mass and friction M of the pulley are negligible. The coefficients of friction between blocks 1 and 2 and between block 2 and the tabletop are nonzero and are given in the following table. Newton’s Law - Solution From the 1998 Physics C Mech AP Test Coefficient Between Coefficient Between Blocks 1 and 2 Block 2 and the Tabletop Static s1 s2 Kinetic k1 k2 Express your answers in terms of the masses, coefficients of friction, and g, the acceleration due to gravity. Newton’s Law - Solution From the 1998 Physics C Mech AP Test (a) Suppose that the value of M is small enough that m1 the blocks remain at rest when released. For each m2 of the following forces, determine the magnitude of the force and draw a vector on the block provided to indicate the direction of the force if it M is nonzero. i. The normal force N1 exerted on block 1 by block 2 F Forces on m1 N1 N1 y =0 N1 -m1g = 0 m1g m 1 N 1 = m 1g Newton’s Law - Solution From the 1998 Physics C Mech AP Test ii. The friction force f1 exerted m 1 on block 1 by block 2 m 2 F x = max = m1 (0) = 0 M f1 = 0 m 1 This is true since there is no motion and friction only opposes motion, there is no horizontal force for the friction to oppose. Newton’s Law - Solution From the 1998 Physics C Mech AP Test iii. The force T exerted on block m1 2 by the string m2 M Forces on m2 N2 F = ma x = 0 f2 T m 2 T x Mg - T = 0 N1 m2g T = Mg Newton’s Law - Solution From the 1998 Physics C Mech AP Test m 1 iv. The normal force N2 exerted m 2 on block 2 by the tabletop Forces on m2 N2 F y =0 M f2 T N 2 - N1 - m 2 g = 0 N1 m2g N 2 = N1 + m 2 g N 2 = m1g + m 2 g N2 N 2 = g(m1 + m 2 ) m 2 Newton’s Law - Solution From the 1998 Physics C Mech AP Test m 1 m 2 v. The friction force f2 exerted on block 2 by the tabletop Forces on F x =0 M m2 N2 T - f2 = 0 f2 T N1 m2g f 2 = T = Mg f2 m 2 Newton’s Law - Solution From the 1998 Physics C Mech AP Test m 1 (b) Determine the largest value of M for which the blocks m 2 can remain at rest. From (a) v. above M f 2max = Mg f 2max s2 N 2 M max = = g g but from (a) iv. above N 2 = g(m1 + m 2 ) s2 g(m1 + m 2 ) M max = = s2 (m1 + m 2 ) g Newton’s Law - Solution m From the 1998 Physics C Mech AP Test 1 m 2 (c) Now suppose that M is large enough that the hanging block Forces descends when the blocks are Forces on on M released. Assume that blocks 1 m N2 2 M and 2 are moving as a unit (no slippage). Determine the f2 T magnitude a of their N1 T Mg acceleration. m2g F 2x a2 x M Mg f 2 a2 x m1 m2 M M 2k m1 m2 a2 x g Mg 2 k g m1 m2 m1 m2 M a2 x m1 m2 M Newton’s Law - Solution From the 1998 Physics C Mech AP Test (d) Now suppose that M is large enough that as the hanging block descends, block 1 is slipping on block 2. Determine each of the following. m 1 i. The magnitude a1 of the m 2 acceleration of block 1 Forces on m1 N1 Fx m1a x M f1 f1 m1 g 1k ax m1g m1 m1 a x g 1k Newton’s Law - Solution From the 1998 Physics C Mech AP Test (d) Now suppose that M is large enough that as the hanging block descends, block 1 is slipping on block 2. Determine each of the following. m 1 m 2 ii. The magnitude a2 of the acceleration of block 2 Forces Forces M on on m2 N2 M N 2 m2 g N 1 f2 T N 2 m2 g m1 g f1 T Mg N 2 g m2 m1 N1 m2g Newton’s Law - Solution From the 1998 Physics C Mech AP Test (d) Now suppose that M is large enough that as the hanging block descends, block 1 is slipping on block 2. Determine each of the following. m 1 m 2 ii. The magnitude a2 of the Forces Forces acceleration of block 2 on on M m2 N2 M N 2 g m2 m1 f2 T f1 N1 T Mg F x ax m m2g ax F x Mg f1 f 2 m m2 M Mg m1 g 1k 2 k g m1 m2 Mg m1 g 1k 2 k N 2 ax ax m2 M m2 M g M m11k 2 k m1 m2 ax m2 M