Tutorial 1 Practice, page (a) An FBD of the top book during its acceleration is shown below.

Transcription

1 ection.4: Forces of Friction Mini Investiation: Liht from Friction, pae 86 A. When we crushed the mints, the candy briefly ave off liht. B. The crystals of suar in the mints rubbed toether to create the friction that produces the liht. Tutorial Practice, pae 89. (a) An FBD of the top book durin its acceleration is shown below. (b) The force of static friction causes the top book to accelerate horizontally.. Given: a.7 m/s Required: µ Analysis: F µ FN ; F m a olution: Equation for y-components of the force: F y " m m Equation for x-components of the force: F x m a F m a µ FN m a µ m m a a µ.7 m/s 9.8 m/s µ 0.8 tatement: The smallest coefficient of static friction between dinner plates that will prevent slippae is Given: F T 8 N [forward 9 up]; µ 0.45 ; µ 0.45; F Required: m Analysis: F m a Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-

2 olution: Equation for y-components of the force: F y + sin" # m m # sin" (Equation ) Equation for x-components of the force: F x m a cos" # F m a cos" # µ FN m a (Equation ) ubstitute Equation () into Equation () and solve for m: cos " µ FN cos " µ ( m " F T sin ) (cos + µ sin) µ m m (cos + µ sin) µ (8 N)(cos sin9 ) 0.45(9.8 m/s ) m 6.9 k tatement: The smallest possible mass for the box is 6.9 k. 4. Given: θ 6.0 ; v i m/s [down slope]; v f 0 m/s ; µ 0.4 Required: Δ d Analysis: F m a ; vf vi + aδ d olution: Equation for y-components of the force: F y " m cos# m cos# Equation for x-components of the force: F x m a x m sin" # F m a x m sin" # µ m cos" m a x a x (sin" # µ cos") (9.8 m/s )(sin6.0 # 0.4cos6.0 ) a x #0.340 m/s (two extra diits carried) Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-

3 olve for the distance travelled: v f v i + a x d d v " v f i a x (0 m/s) " ( m/s) ("0.340 m/s ) d.#0 m tatement: The sled will slide for. 0 m before comin to rest. 5. Given: m 39 k ; direction of rope [forward up] ; µ 0.3; F Required: Analysis: F olution: Equation for y-components of the force: F y + y " m m " sin# Equation for x-components of the force: F x x " F cos# " µ FN cos# " µ (m " sin#) (cos# + µ sin#) µ m µ m cos# + µ sin# (0.3)(39 k)(9.8 m/s ) cos + 0.3sin 87 N tatement: A tension of 87 N in the rope is needed to keep the box movin at a constant velocity. 6. (a) Given: m 4 k; m 4 k; µ 0.3; F a.8 0 N [forward 5 up] Required: a Analysis: F y ; F x m a Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-3

4 olution: Equation for y-components of the force: F y + y " m m " sin# F µ FN (4 k)(9.8 m/s ) " (80 N)sin5 59. N (two extra diits carried) 0.5(59. N) F N (two extra diits carried) Equation for x-components of the force: F x m a x " F " F T m a (Equation ) Equation for y-components of the force: F y " m m (4 k)(9.8 m/s ) 37. N (two extra diits carried) F µ FN 0.5(37. N) F 34.3 N Equation for x-components of the force: F x m a " F m a (Equation ) Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-4

5 Add equations () and () to eliminate the tension of the rope. olve for a. ( F ) + ( x F ) m a + m a x F F (m + m )a a F F x m + m (80 N)cos N 34.3 N 38 k.797 m/s (two extra diits carried) a.8 m/s tatement: The acceleration of the boxes is.8 m/s. (b) Given: m 4 k;; µ 0.3; F a.8 0 N [forward 5 up] Required: Analysis: F m a olution: F m a µ FN + m a 0.5(37.) + (4 k)(.797 m/s ) 59 N tatement: The tension in the rope is 59 N. 7. Given: µ 0.0; µ 0.5 ; m 00.0 k Required: a Analysis: F ; F x m a x olution: For the y-components of the force: F y " m m (00.0 k)(9.8 m/s ) 980 N For the x-components of the force: F x " F F µ s FN 0.5(980 N) 45 N Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-5

6 Once the refrierator is movin, an applied for 45 N produces acceleration, a: F x m a " F m a a " µ FN m 45 N " 0.0(980 N) 00.0 k a 0.49 m/s tatement: The acceleration when you apply minimum force needed to move the refrierator is 0.49 m/s. 8. Given: µ 0.5; m 0 k Required: Analysis: F µ FN ; F olution: For the y-components of the force: F y " m m (0 k)(9.8 m/s ) 078 N (two extra diits carried) For the x-components of the force: F x " F F µ s FN 0.5(078 N).7 #0 N tatement: The minimum force required to just set the stae prop into motion is.7 0 N. Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-6

7 ection.4 Questions, pae 90. Given: vi 0 m/s; vf 0 m/s; Δ d 40 m Required: coefficient of friction between the tires and the road, µ Analysis: v v + aδ d; F x m a x. Choose forward as positive. f i olution: Find the acceleration a : v f v i + ad a " v i d (0 m/s) " (40 m) a "5.0 m/s Calculate µ : F x m a " F m a "µ m m a a µ " 5.0 m/s 9.8 m/s µ 0.5 tatement: The coefficient of friction between the tires and the road is Given: v i 50.0 m/s ; Δ t 0.0 s ; µ Required: v f Analysis: F x m a. Choose forward as positive. olution: Find the acceleration a: F x m a " F m a "µ m m a a "µ "0.030(9.8 m/s ) a "0.94 m/s (one extra diit carried) Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-7

8 Calculate v f : v f v i + at 50.0 m/s + ("0.94 m/s )(0.0 s) v f 47 m/s tatement: The speed of the puck after 0.0 s is 47 m/s. 3. (a) Given: m.0 0 k; F a 3.50 N Required: µ Analysis: F ; F µ FN olution: For the y-components of the force: F y " m m For the x-components of the force: F x " F F µ FN F a F µ a m 350 N (00 k )(9.8 m/s ) µ 0.8 tatement: The coefficient of static friction between the floor and the sofa is 0.8. (b) Given: m.0 0 k; F a 3.50 N ; v i 0 m/s ; v f.0 m/s; Δ t 5.0 s Required: µ Analysis: F m a olution: Find the acceleration, a : a v t.0 m/s 5.0 s a 0.40 m/s Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-8

9 Usin the x-components of the force: F x m a " F m a F " m a µ FN F a " m a F µ a " m a m 350 N " (00 k )(0.40 m/s ) (00 k )(9.8 m/s ) µ 0.4 tatement: The coefficient of kinetic friction between the sofa and the floor is (a) Given: µ 0.9 Required: Analysis: F olution: For the y-components of the force: F y " m cos# m cos# For the x-components of the force: F x m sin" # F m sin" µ FN m sin" µ m cos" tan" µ " tan # (0.9) " 6 tatement: The crate just beins to slip when the anle of inclination,θ, is 6. (b) Given: µ 0.6; θ 6.7 Required: a Analysis: F x m a Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-9

10 olution: F x m a m sin" # F m a m sin" # µ m cos" m a a (sin" # µ cos") (9.8 m/s )(sin6.7 # 0.6cos6.7 ) a 0.8 m/s tatement: The crate accelerates at 0.8 m/s [down the incline] when the coefficient of kinetic friction is (a) Two situations in which friction is helpful for an object movin on a horizontal surface are when runnin, so you can push yourself forward, and when walkin on a snowy field so you can use traction from the snow to move yourself forward. (b) Two situations in which it would be ideal if there were no friction when an object moves across a horizontal surface are: shootin the puck across the ice when playin hockey; and when tryin to move a sled over a snowy sidewalk. 6. (a) Given: m 45 k ; m k; µ 0.45; µ 0.35 Required: F µ F N Analysis: To determine if this system is in static equilibrium, you need to determine the manitude of the static friction, F, for mass m and compare it to the tension in the strin, F, for mass m. Find F for mass m. F µ FN µ m 0.45(45 k)(9.8 m/s ) F 00 N Find F for mass m. F m ( k)(9.8 m/s ) F 0 N tatement: Yes, the system is in static equilibrium. The tension in the strin for mass m is not sufficient to break the force of static friction for mass m, so there is no acceleration. (b) Given: F 0 N Required: F T Analysis: As lon as the two masses remain at rest, the tension in the strin is equal to the force of ravity on mass m, F T F m. Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-0

11 olution: F T F 0 N tatement: The tension in the strin is 0 N. (c) Given: m 45 k; m 3 k; µ 0.35 Required: a Analysis: F m a olution: Find F T for mass m. F x m a " F m a " µ m m a Find for mass m. F y m a m " F T m a olve for a : µ m ( ) + m F ( T ) m a + m a (m µ m ) (m + m ) a a (m µ m ) m + m (3 k 0.35(45 k))(9.8 m/s ) 77 k a. m/s tatement: The acceleration of the system is. m/s, eliminatin the strin tension. 7. (a) Given: θ 4 Required: µ Analysis: F olution: For the y-components of the force: F y " m cos# m cos# Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-

12 For the x-components of the force: F x m sin" # F m sin" µ FN m sin" µ m cos" tan" µ µ tan4 µ 0.90 tatement: The coefficient of static friction is (b) Given: θ 35 Required: µ Analysis: F x olution: F x m sin" # F m sin" µ m cos" µ tan" tan35 µ 0.70 tatement: The coefficient of kinetic friction is Given: m 8.0 k; m k ; θ 6 ; θ 39 ; µ 0. Required: a Analysis: F m a olution: For the y-components of the force (mass m ): F y " m cos# m cos# For the y-components of the force (mass m ): F y " m cos# m cos# Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-

13 For the x-components of the force (mass m ): F x m a " m sin# " F m a " m sin# " µ m cos# m a " 49.7 N m a (two extra diits carried) For the x-components of the force (mass m ): F x m a m sin" # F # m a m sin" # µ m cos" # m a 54.8 N # m a (two extra diits carried) Add the final equations for mass m and mass m to eliminate the strin tension. ( 49.7 N) + (54.8 N ) (m + m ) a a 54.8 N 49.7 N 0 k a 0.8 m/s tatement: The acceleration of the two masses (as a system) is 0.8 m/s [clockwise]. Copyriht 0 Nelson Education Ltd. Chapter : Dynamics.4-3