As observed from the frame of reference of the sidewalk:

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1 Section 3.1: Inertial and Non-inertial Frames of Reference Tutorial 1 Practice, pae a) When the car is movin with constant velocity, I see the ball lie still on the floor. I would see the same situation when the car is at rest. b) To an observer on the sidewalk, the ball appears to be movin with a constant velocity of 14 m/s [E]. c) If the car accelerates forward, I see the ball roll backward on the floor. d) As observed from the frame of reference of the car: As observed from the frame of reference of the sidewalk: The car s frame of reference is non-inertial. I observe a fictitious force the force pushin the ball backward, west) in the car s frame of reference.. a) Given: m =.0 = 0.00 k; θ = 3.5 Required: a Analysis: Look at the situation from an Earth inertial) frame of reference. The horizontal component of the tension balances the acceleration and the vertical component of the tension balances the ravitational force. Express the components of the tension in terms of the horizontal and vertical applied forces. Then calculate the manitude of the acceleration. Solution: Vertical component of force:!f y = 0 cos" # m = 0 = m cos" Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-1

2 Horizontal component of force:!f x = ma sin" = ma # m & cos" ' sin" = ma # sin" & cos" ' = a tan" = a a = 9.8 m/s tan3.5 a = 6. m/s Statement: The manitude of the boat s acceleration is 6. m/s. I did not need to know the mass of the ball to make the calculation because that value was cancelled out to obtain the forces. b) Given: m =.0 = 0.00 k; θ = 3.5 Required: Analysis: From part a), = m cos!. Solution: = m cos! 0.00 k = 9.8 m/s ) cos3.5 = 0.6 N Statement: The manitude of the tension in the strin is 0.6 N. I need to know the mass to make this calculation because the force is the mass multiplied by the acceleration. 3. a) When the subway is movin at a constant velocity, there is no tension in the strap. b) Given: m = 14 k; θ = 35 ; a = 1.4 m/s Required: Analysis: Look at the situation from an Earth inertial) frame of reference. The horizontal component of the tension balances the acceleration, so express the x-components of the tension in terms of the horizontal applied forces. The vertical forces of ravity and the normal force balance each other.!f x = ma sin" = ma = ma sin" Solution: = ma sin! 14 k) 1.4 m/s ) = tan35 = 34 N Statement: The tension on the strap durin acceleration is 34 N. Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-

3 4. Given: µ s = 0.4 Required: maximum a Analysis: The force due to the acceleration of the train, F = ma, must be less than the force of static friction, F s = µ s, where = m = m. So the maximum acceleration occurs when F = F s. F = F s ma =! s ma =! s m) a =! s Solution: a =! s 9.8 m/s = 0.4 a = 4.1 m/s Statement: The maximum acceleration of the train before the passener beins to slip alon the floor is 4.1 m/s. Tutorial Practice, pae a) Given: m = 55 k; a! =.9 m/s [up] Required: Analysis: Use up as positive and solve for the normal force when + + m) = ma. + +!m) = ma = ma + m Solution: = ma + m.9 m/s = 55 k + 55 k 9.8 m/s = 7.0!10 N Statement: The student s apparent weiht is N. b) Given: m = 55 k; a! =.9 m/s [down] Required: Analysis: Use down as positive and solve for the normal force when + m) = ma. + m) = ma = ma m Solution: = m! ma 9.8 m/s = 55 k! 55 k.9 m/s = 3.8 "10 N Statement: The student s apparent weiht is N. Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-3

4 . a) Given: = 9.5 k; m =.5 k; = 70.0 N Required: a! Analysis: Use up as positive and solve for the acceleration when + + m) = ma. + +!m) = ma a = F! m N m = F! m N + ) Solution: a = F! m N + ) = 9.8 m/s 70.0 k " m! 9.5 k +.5 k s 9.5 k +.5 k =!3.967 m/s two extra diits carried) a =!4.0 m/s Statement: The acceleration of the elevator is 4.0 m/s [down]. b) Given: m =.5 k;! a = m/s [down] Required:! Analysis: Use down as positive and solve for the normal force when + m) = ma. Solution:! + m) = ma = m! ma 9.8 m/s =.5 k = 15 N!.5 k m/s Statement: The force on the smaller box is 15 N [up]. 3. a) Given: = 4. k; m =.6 k; a = 0 Required: A ; B Analysis: Since this is an inertial frame of reference, the tension on rope B balances the force of ravity on block. The tension on rope A balances the force of ravity on block 1 and the tension on rope B. F = m Solution: Determine the force on rope B: B = m 9.8 m/s ) =.6 k = 5.48 N two extra diits carried) B = 5 N Determine the force on rope A: A = + B = 4. k 9.8 m/s ) N A = 67 N Statement: The tension on the rope A is 67 N and the tension on the rope B is 5 N. Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-4

5 b) Given: = 4. k; m =.6 k a! = 1. m/s [up] Required: A ; B Analysis: This is now a non-inertial frame of reference, so instead of just ravity, the acceleration is a), or + a, where down is positive. Solution: Determine the force on rope B: B = m + a 9.8 m/s + 1. m/s ) =.6 k = 8.6 N one extra diit carried) B = 9 N Determine the force on rope A: A = + a + B 9.8 m/s + 1. m/s ) N = 4. k A = 75 N Statement: The tension on the rope A is 75 N and the tension on the rope B is 9 N. 4. Given:! a = 0.98 m/s [down]; m = 61 k Required: Analysis: Use down as positive and solve for the normal force when + m) = ma. Solution:! + m) = ma = m! ma 9.8 m/s = 61 k! 61 k 0.98 m/s = 5.4 "10 N Statement: The passener s apparent weiht is m/s. Section 3.1 Questions, pae a) The ball would appear to move straiht up and down because I am movin with the same velocity as the other train. From my viewpoint, the other train and passener are standin still, and the ball is not affected by the train s motion. b) If the trains moved in opposite directions, the ball would appear to have horizontal motion, so I would see the path as a parabola.. Given: a = 1.5 m/s Required: θ Analysis: The horizontal component of the tension balances the acceleration, and the vertical component of the tension balances the ravitational force. Express the tanent ratio of the anle in terms of the applied force and the ravitational force, then solve for the anle. tan! = F # ma &! = tan "1 m ' # a &! = tan "1 ' Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-5

6 # a & Solution:! = tan "1 ' # m & 1.5 = tan "1 s 9.8 m s '! = 8.7 Statement: The strin makes an 8.7 anle with the vertical. 3. Given: v f = 55 km/h; t = 10.0 s Required: θ Analysis: Determine the acceleration usin v f = a t or a = v f. The horizontal component of the!t tension balances the acceleration and the vertical component of the tension balances the ravitational force. Express the tanent ratio of the anle in terms of the applied force and the ravitational force, then solve for the anle. Solution: Determine the acceleration of the plane: a = v f!t 55 km = h! 1000 m 10.0 s 1 km! 1 h 60 min! 1 min 60 s = m/s two extra diits carried) a = 7.08 m/s Determine the anle the strin makes with the vertical: tan! = F # ma &! = tan "1 m ' # a & = tan "1 ' # m & = tan "1 s 9.8 m s '! = 35.9 Statement: The strin makes an 35.9 anle with the vertical. 4. Given: θ = 16 Required: a Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-6

7 Analysis: Look at the situation from an Earth inertial) frame of reference. The horizontal component of the tension balances the acceleration and the vertical component of the tension must balance the ravitational force since the cork ball does not move. Express the components of the tension in terms of the horizontal and vertical applied forces. Then calculate the manitude of the acceleration. Vertical component of force:!f y = 0 cos" # m = 0 = m cos" Horizontal component of force:!f x = ma sin" = ma # m & cos" ' sin" = ma # sin" & cos" ' = a tan" = a Solution: a = tan! = 9.8 m/s )tan16 a =.8 m/s Statement: The manitude of the car s acceleration is.8 m/s. 5. Given: v f = 6.0 m/s; t = 10.0 s; m = 64 k Required: Analysis: Determine the upward acceleration usin v f = a t or a = v f. Use up as positive and!t solve for the normal force when + + m) = ma. + +!m = ma = ma + m Solution: Determine the upward acceleration: a = v f!t = 6.0 m/s 10.0 s a = 0.60 m/s Determine the apparent weiht: = ma + m 0.6 m/s = 64 k + 64 k 9.8 m/s = 6.7!10 N Statement: The passener s apparent weiht is N. Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-7

8 6. Given: = 55 N; m = 5 k Required: a Analysis: Use up as positive and solve for the acceleration when + + m) = ma. + +!m = ma a =! m m Solution: a =! m m 9.8 m/s 55 k " m! 5 k = s 5 k a =!4.9 m/s Statement: The acceleration of the ride is 4.9 m/s [down]. 7. a) At rest: Acceleratin downhill: When the car is at rest, the dice are alined with the vertical with respect to level round), and make an anle of 17 with respect to the normal perpendicular to the roof). The only forces actin on the dice are ravity and tension. When the car is acceleratin, a horizontal fictitious force pointin to the rear of the car deflects the dice so that they are alined with the normal, and make an anle of 17 with respect to the vertical, as viewed from the level round. b) Given: θ = 17 Required: a Analysis: Look at the situation from an inertial frame of reference on the same anle as the hill. The vertical component of the ravitational force F balances the tension and the horizontal component of the ravitational force F balances the force applied by the acceleration. Express the horizontal component of F in terms of the applied forces. Then calculate the manitude of the acceleration.!f x = 0 F sin" # ma = 0 m sin" # ma = 0 sin" # a = 0 a = sin" Solution: a = sin! = 9.8 m/s sin17 a =.9 m/s Statement: The manitude of the car s acceleration is.9 m/s. Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-8

9 8. a) Given: = 1.8 k; m = 1. k; m = 1. k; = 70.0 N Required: Analysis: Since mass 1 does not slide, the acceleration due to the horizontal force must balance the acceleration due to the tension, which equals m. Use F = ma to determine the acceleration. = + m 3 = m + m 3 = m m + m 3 ) 1 Solution: = m m + m 3 ) 1 1. k) 9.8 m/s 1.8 k 1.8 k +1. k k) = 39 N Statement: The applied force is 39 N. b) Given: = 1. k; m =.8 k; θ = 5 Required: Analysis: Since mass 1 does not slide, the applied force on mass 1, the ravitational force, and the normal force must balance each other: ΣF = 0. Use F = ma to determine the acceleration. Since the applied force is entirely horizontal and the ravitational force is entirely vertical, use the tanent ratio to relate them. Determine the applied acceleration: F = ma = )a a = Determine the horizontal acceleration of mass 1: tan! = m a 1 = a a = tan! Determine the applied force: = tan! = tan! ) Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion 3.1-9

10 Solution: = tan! 9.8 m/s ) tan5 = 17 N 1. k +.8 k) Statement: The applied force is 17 N. Copyriht 01 Nelson Education Ltd. Chapter 3: Uniform Circular Motion