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Chapter 4

Chapter 4. Motion With a Changing Velocity. P2.27: Find the magnitude and direction of the vector with the following components: x = -5.0 cm, y = +8.0 cm F x = +120 N, F y = -60.0 N v x = -13.7 m/s, v y = -8.8 m/s a x = 2.3 m/s 2 , a y = 6.5 cm/s 2.

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Chapter 4

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  1. Chapter 4 Motion With a Changing Velocity

  2. P2.27: Find the magnitude and direction of the vector with the following components: • x = -5.0 cm, y = +8.0 cm • Fx = +120 N, Fy = -60.0 N • vx = -13.7 m/s, vy = -8.8 m/s • ax = 2.3 m/s2, ay = 6.5 cm/s2

  3. P2.56: A 3.0 kg block is at rest on a horizontal floor. If you push horizontally on the block with a force of 12.0 N, it just starts to move. • What is the coefficient of static friction? (b) A 7.0-kg block is stacked on top of the 3.0-kg block. What is the magnitude F of the force acting horizontally on the 3.0-kg block as before, that is required to make the two blocks start to move?

  4. P3.47: A 2010-kg elevator moves with an upward acceleration of 1.50 m/s2. What is the tension that supports the elevator? P3.48: A 2010-kg elevator moves with a downward acceleration of 1.50 m/s2. What is the tension that supports the elevator?

  5. Kinematic Equations for Const. Acceleration Fnet = ma. If Fnet is const, a will also be const. Uniformly accelerated motion: a = const.

  6. A car moves at a constant acceleration of magnitude 5 m/s2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at (i) t = 2s? (ii) t = 4s? (iii) t = 10s? A car moves at a constant acceleration of magnitude 5.7 m/s2. At time t = 0, the magnitude of its velocity is 18.3 m/s. What is the magnitude of its velocity at t = 2.2s?

  7. Kinematic Equations for Const. Acceleration • Consider an object on which a net force Fnet acts on it. Thus it moves with an acceleration. • As the object moves, its velocity changes. Fnet Fnet Fnet a a Time = 0 Initial position = x0 Initial velocity = v0 Time = t Final position = x Final velocity = v

  8. Kinematic Equations for Const. Acceleration Fnet = ma. If Fnet is const, a will also be const. Uniformly accelerated motion: a = const. aave = ainst Let us use initial time, t1 = 0. Final time, t2 = t, hence t = t – 0 = t Position: initial, x1 = x0, final, x2 = x Velocity, initial v1 = v0, final, v2 = v

  9. Kinematic Equations for Constant Acceleration Uniformly accelerated motion: a = constant. Time: Initial = 0, final = t Positions: Initial = x0, final =x Velocity: Initial = v0, final = v v = v0 + at x = x0 +vot + ½ at2 v2 = v02 + 2a(x-x0) Average velocity vav = (v0 + v)/2

  10. Example Problem 4.14 A train traveling at a constant speed of 22 m/s, comes to an incline with a constant slope. While going up the incline the train slows down with a constant acceleration of magnitude 1.4 m/s2. • Draw a graph of vx versus t. • What is the speed of the train after 8.0s on the incline? • How far has the train traveled up the incline after 8.0 s?

  11. A car moving south slows down with at a constant acceleration of 3.0 m/s2. At t = 0, its velocity is 26 m/s. What is its velocity at t = 3 s? • 35 m/s south • 17 m/s south • 23 m/s south • 29 m/s south • 17 m/s north

  12. A car initially traveling at 18.6 m/s begins to slow down with a uniform acceleration of 3.00 m/s2. How long will it take to come to a stop? • 55.8 s • 15.6 s • 6.20 s • 221.6 s • None of these

  13. Free Fall • Free fall: Only force of gravity acting on an object making it fall. • Effect of air resistance is assumed negligible. • Force of gravity acting on an object near the surface of the earth is F = W = mg. • Acceleration of any object in free fall: a = g = 9.8 m/s2 down (ay = -9.8 m/s2).

  14. Free Fall ay = -9.8 m/s2 ax = 0 +y +x

  15. Free Fall contd… 1. a = g, regardless of mass of object.

  16. 2. a = g, regardless of initial velocity ay = -9.8 m/s2,ax = 0 +y v0 = 0 v0 = -15 m/s v0 = +15 m/s +x

  17. 3. Free Fall: Motion is symmetric. ay = -9.8 m/s2,ax = 0 At the maximum height: • vy = 0 • Speed at equal heights will be equal. • Equal time going up and down. +y v0 = +5 m/s +x

  18. Example: Problem 4.32 A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.5 m above the ground when launched. • How high above the ground does the stone rise? • How much time elapses before the stone hits the ground?

  19. APPARENT WEIGHT N mg A physics student whose mass is 40 kg stands inside an elevator on a scale that reads his weight in Newtons. Scale Reading = Normal force the scale exerts on the student. Scale Reading = N = mg = 40 x 9.8 N = 392 N

  20. 1. Elevator at rest. What will be the scale reading? N Fnet = N – W = may At rest means ay = 0. Hence N = W, ie apparent weight = true weight = 40 x 9.8 = 392 N W = mg

  21. 2. Elevator accelerating upwards with ay = 2.0 m/s2. What will be the scale reading?

  22. N a W = mg 2. Elevator accelerating upwards with ay = 2.0 m/s2. What will be the scale reading? Fnet = N – W = may ay = + 2.0 m/s2 (positive because acceleration is upwards) . Hence, N –W = N – mg = may. N = mg + may = m(g+ay) = 40(9.8 + 2.0) = 472 N ie, apparent weight is greater than the true weight.

  23. 3. Elevator accelerating downwards with ay = 2.0 m/s2. What will be the scale reading?

  24. N a W = mg 3. Elevator accelerating downwards with ay = 2.0 m/s2. What will be the scale reading? Fnet = N – W = may ay = - 2.0 m/s2 (negative because acceleration is downwards) . Hence N –W = N – mg = -may. N = mg - may = m(g - ay) = 40(9.8 - 2.0) = 312 N ie, apparent weight is less than the true weight.

  25. A 112.0-kg person stands on a scale inside an elevator moving downward with an acceleration of 1.80 m/s2. What will be the scale reading? • 1299 N • 1,098 N • 896 N • 112 N • 0 N

  26. A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise? • 9.8 m • 3.00 m • 1.15 m • 26.1 m • 19.6 m

  27. WEIGHTLESSNESS If the elevator was going down with an acceleration ay = g = -9.8m/s2, then N = m(g-g) = 0 ie, apparent weight = 0 This is “weightlessness” or “zero gravity” Apparent weight of an object in free fall is zero while its true weight remains unchanged.

  28. Equilibrium Newton’s 2nd Law: Fnet = ma For an object in equilibrium: Fnet = 0 Static (v = 0) 0r dynamic (v = constant) eqlbm. 2-dimensions, separate the x and y components and treat the problem as two 1-dim problems. Fx = max Fy = may For equilibrium,Fx = max = 0 and Fy = may = 0

  29. 2-Dimensions y x • X and Y are INDEPENDENT! • Break 2-D problem into two 1-D problems.

  30. Equilibrium Determine the tension in the 6 m rope if it sags 0.12 m in the center when a gymnast with weight 250 N is standing on it. y x direction: Fx = max = 0 -TL cosq + TR cosq = 0 TL = TR TR TL x W TR y direction: Fy = may = 0 TL sinq + TR sinq - W = 0 2 T sinq = W T = W/(2 sinq) = 3115 N .12 m q 3 m

  31. N N fk fs F F W W Equilibrium on a Horizontal Plane Object at rest or moving with const. velocity • Fx = max = 0 and Fy = may = 0 Object at rest Sliding with constant velocity No motion until F = > fsmax Fx = F - fs = 0 or F = fs. Fy = N - W = 0 or N = W Fx = F - fk = 0 or F = fk Fy = N - W = 0 or N = W

  32. Equilibrium on an inclined Plane y fs y x fs N x W.cos  N W Wy  W.sin  W Wx y W fs x N W.sin W.cos  An object at rest on an inclined plane • Fx = max = 0 and Fy = may = 0

  33. Equilibrium on an inclined Plane y fs x N W.sin W.cos  An object at rest on an inclined plane Fx = max = 0 and Fy = may = 0 Fy = may = 0 N - Wcos = 0 or N = Wcos Fx = max = 0 fs - Wsin = 0 or fs = Wsin If angle  is increased, the object will eventually slide down the plane. Sliding will start beyond angle max At max: fsmax = W.sinmax. But fsmax = sN = s(Wcosmax) Therefore, sWcosmax = Wsin max ORs = (Wsinmax)/Wcosmax ie, s = tanmax

  34. Equilibrium on an inclined Plane y fs x N W.sin W.cos  An object at rest on an inclined plane N = Wcos fs = Wsin • If angle  is increased, the object will eventually slide down the plane. • Sliding will start beyond angle max • At max: fsmax = W.sinmax. • But fsmax = sN = s(Wcosmax) • Therefore, sWcosmax = Wsin max • ORs = (Wsinmax)/Wcosmax ie, s = tanmax

  35. A mass m being pulled uphill by a force F y F x If m = 510 kg, s= 0.42, k = 0.33,  = 15o: • Find minimum force F needed to start the mass moving up. • If the force in (a) is maintained on the mass, what will its acceleration be? N W.sin fk  W.cos (c) To move the mass with constant speed, what must the value of F be?

  36. A block is at rest on a flat board. The flat board is gently tilted. At what angle will the block start to slide? Assume the coefficient of static friction (s) between the block and the board is 0.48. • 0.48o • 61.3o • 28.7o • 25.6o • 0.00837o

  37. Position, Velocity and Acceleration • Position, Velocity and Acceleration are Vectors! • x and y directions are INDEPENDENT! y direction x direction

  38. Velocity in Two Dimensions A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the change in velocity? x-direction vix = 5 m/s vfx = 3 m/s cos(25) Dvx = 3cos(25)–5 =-2.28m/s y-direction viy = 0 m/s vfy = 3 m/s sin(25) Dvy = 3sin(25)=+1.27 m/s y x 3 m/s 5 m/s

  39. Acceleration in Two Dimensions A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the average acceleration? [Assume force of gravity is very small]. y x-direction y-direction x 3 m/s 5 m/s

  40. A wagon of mass 50 kg is being pulled by a force F of magnitude 100 N applied through the handle at 30o from the horizontal. Ignoring friction, find the magnitude of • the horizontal component of F. • the horizontal component of acceleration. • the normal force exerted onthe wagon.

  41. Projectile Motion A projectile – An object moving in 2-dimensions near the surface of the earth with only the force of gravity acting on it. Eg: golf ball, batted base ball, kicked football, soccer ball, bullet, etc. • Assume no air resistance. • Assume g = -9.8 m/s2 constant. • We are not concerned with the process that started the motion!

  42. Free Fall: 1-dimensional motion. ay = -9.8 m/s2,ax = 0 +y At the maximum height: • vy = 0 • Speeds at equal heights will be equal. • Equal time going up/down. v0 = +5 m/s +x

  43. PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2, ax = 0 +y v0 = 5 m/s v0y  +x v0x

  44. PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2 ax = 0 v0x = v0cos v0y = v0sin What will happen to the y-component of the velocity? What will happen to the x-component of the velocity?

  45. Kinematics in Two Dimensions • x = x0 + v0xt + ½ axt2 • vx = v0x +axt • vx2 = v0x2 + 2ax(x - x0) • y = y0 + v0yt + ½ ayt2 • vy = v0y +ayt • vy2 = v0y2 + 2ay(y – y0) x and y motions are independent! They share a common time t.

  46. Kinematics for Projectile Motionax = 0 ay = -g • y = y0 + v0yt - 1/2 gt2 • vy = v0y-gt • vy2 = v0y2 - 2g y • x = x0 + v0t • vx = v0x X Y

  47. PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2,ax = 0 Once the projectile is in air, the only force acting on it is gravity. Its trajectory (path of motion) is a parabola. Fnet = ma = -mg ay = -9.8 m/s2 ax = 0

  48. PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2 andax = 0 v0x = v0cos and v0y = v0sin

  49. Two balls A and B of equal mass m. Ball A is released to fall straight down from a height h. Ball B is thrown horizontally. Which ball lands first? A B h ay = -9.8 m/s2 ax = 0 Vo = 0 v0x = 0 v0y = 0 ay = -9.8 m/s2 ax = 0 Vo 0 v0x = Vo v0y = 0

  50. A • y = y0 + v0yt + ½ ayt2 • vy = v0y +ayt • vy2 = v0y2 + 2ay(y – y0) To find time t, use y = y0 + v0yt + ½ ayt2 -h = 0 + (0 . t) + ½ (-g)t2 Gives 2h = gt2 and t = (2h/g) ay = -9.8 m/s2 ax = 0 vo = 0 v0x = 0 v0y = 0 y0 = 0, y = -h

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