Chapter 4. Forces and Newton s Laws of Motion. continued
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1 Chapter 4 Forces and Newton s Laws of Motion continued
2 Exam Arrive ON TIME by 10:20 or you will not take the exam. Only Section 1 students. Sit ONLY in your assigned seat. Latest assignments will be on the website 1 Sheet (both sides) of information prepared by YOU not commercially.
3 Quiz 4 1. C&J page 111 (middle), Check Your Understanding #21
4 Quiz 4 1. C&J page 111 (middle), Check Your Understanding #21 2. An passenger, mass m = 70 kg, is on a scale in an airplane that hits a strong downdraft causing a downward acceleration with magntitude 3.8 m/s 2. What apparent weight does the scale read? a) 0N b) 270N c) 970 N d) 420N e) 690N
5 Quiz 4 1. C&J page 111 (middle), Check Your Understanding #21 2. An passenger, mass m = 70 kg, is on a scale in an airplane that hits a strong downdraft causing a downward acceleration with magntitude 3.8 m/s 2. What apparent weight does the scale read? a) 0N b) 270N c) 970 N d) 420N e) 690N 3. On a frictionless surface, two boxes with the masses shown are pushed together by a force with magnitude F. What is the force that the smaller mass block exerts on the larger mass block? a) F / 3 b) F / 5 c) 2F / 3 d) 3F / 2 e) 2F / 5 F 203m kg 20 kg 2m
6 4. A block of mass 5 kg is initally at rest and pressed against a wall with a force of F. The coefficient of static friction is 0.5, and the coefficient of kinetic friction is less than this. What is the minimum force magnitude to keep the block at rest. a) 25 N b) 50 N c) 75 Ν d) 100 N e) No amount of force will keep it at rest. F
7 4. A block of mass 5 kg is initally at rest and pressed against a wall with a force of F. The coefficient of static friction is 0.5, and the coefficient of kinetic friction is less than this. What is the minimum force magnitude to keep the block at rest. a) 25 N b) 50 N c) 75 Ν d) 100 N e) No amount of force will keep it at rest. 5. I. A person pulls with a force magnitude F on a string attached to a wall. II. A person at each end of the same string pulls on it with a force F. Consider the two statements above. Which statement below is true? a) The tension at the center of the string is zero b) The string tension in II. is half as big as the tension in I. c) The string tension in II. is the same as the tension in I. d) The string tension in II. is twice as big as the tension in I. e) The string tension in II. is four times as big as the tension in I. F
8 Quiz 4 1. C&J page 111 (middle), Check Your Understanding #21 2. An passenger, mass m = 70 kg, is on a scale in an airplane that hits a strong downdraft causing a downward acceleration with magntitude 3.8 m/s 2. What apparent weight does the scale read? a) 0N Weight = mg = (70 kg)(9.8 m/s 2 ) = 690 N b) 270N Acceleration is 3.8 m/s 2, net force = ma = 270 N c) 970 N Net force = mg + F d) 420N, Scale reads F. e) 690N 270 N = 690 N + F F = ( ) N = 420N 3. On a frictionless surface, two boxes with the masses shown are pushed together by a force with magnitude F. What is the force that the smaller mass block exerts on the larger mass block? a) F / 3 b) F / 5 c) 2F / 3 d) 3F / 2 e) 2F / 5 a = F 5m ; net force of 2m on 3m F 2m = 2ma = 2 5 F F 203m kg d) Balanced forces 20 kg 2m
9 4. A block of mass 5 kg is initally at rest and pressed against a wall with a force of F. The coefficient of static friction is 0.5, and the coefficient of kinetic friction is less than this. What is the a) 25 N b) 50 N c) 75 Ν minimum force magnitude to keep the block at rest. d) 100 N F = F f max = µf = µf; slides if, mg > f max mg = µf; F = mg µ = 5kg(9.8 m/s2 ) 0.5 e) No amount of force will keep it at rest. 5. I. A person pulls with a force magnitude F on a string attached to a wall. II. A person at each end of the same string pulls on it with a force F. a) The tension at the center of the string is zero = 100 N Consider the two statements above. Which statement below is true? b) The string tension in II. is half as big as the tension in I. c) The string tension in II. is the same as the tension in I. d) The string tension in II. is twice as big as the tension in I. e) The string tension in II. is four times as big as the tension in I. F mg f F
10 4.11 Equilibrium Application of Newton s Laws of Motion Pulleys (massless, frictionless) 1. Tension magnitdue is the same at every location on the rope. 2. The same tension acts on object at each end of each section of rope Cases with mass at rest (force vectors labeled with magnitude) # = number of rope sections supporting the weight of the mass Block & tackle T T T T T T n-rope sections T m T T = W 2 T T T T = W 4 T = W n F 1 = W W = mg Reverses force F 2 = W 2 m W F 4 = W 4 m W F n = W n M W = Mg
11 4.11 Equilibrium Application of Newton s Laws of Motion Equilibrium requires net force = zero, in every direction (x and y) 1) T = W = mg 2) Net force vector = 0 3) Use x direction along leg 4) y is perpendicular to x T 1 = T 2 = T = mg (rope and pulleys insure this) x : + T cos35 + T cos35 F = 0 F = 2mg cos35 = 2(2.2 kg)(9.8 m/s 2 )(.82) = 35 N
12 4.11 Equilibrium Application of Newton s Laws of Motion x :T 1 sin10 T 2 sin80 = 0 y :T 1 cos10 mg T 2 cos80 = 0 from x, solve for T 1 = T 2 sin80 / sin10 W = mg = 3150N from y, solve for T 2 T 2 = mg cos10 sin80 / sin10 cos80 = 580N, T 1 = 3300N
13 4.12 Nonequilibrium Application of Newton s Laws of Motion When an object is accelerating, it is not in equilibrium.
14 4.12 Nonequilibrium Application of Newton s Laws of Motion The acceleration is only along the x axis : y :T 1 = T 2 = T to make a y = 0 a x = m/s 2 a y = 0 x : 2T cos30 + D R = ma x T = ( ma x D + R) ( 2cos30 ) = N
15 Units, Scalars, Vectors Vector decomposition and vector addition. Vector A points directly downward. What are the two components of the vector A along the axes rotated by 35 from the (x,y) axes. + y 35 + x A
16 Units, Scalars, Vectors Vector decomposition and vector addition. Vector A points directly downward. What are the two components of the vector A along the axes rotated by 35 from the (x,y) axes. + y 35 + x A Acos35 Asin35
17 Units, Scalars, Vectors Vector decomposition and vector addition. Vector A points directly downward. What are the two components of the vector A along the axes rotated by 35 from the (x,y) axes. + y 35 Asin35 A Acos35 + x x component : Asin35 y component : Acos35
18 Units, Scalars, Vectors Vector addition and vector decomposition. Vector A points along +y and vector B points along +x. What is the magnitude and direction of vector C = A + B? A C θ A C = A 2 + B 2 ; θ = tan 1 B B
19 Velocity, Acceleration, Displacement, initial values at t = 0 1D motion equations for constant acceleration t = time relative to the start of the clock (t = 0) x = displacement over the time t v 0 = velocity at time t = 0 v = final velocity after time t or displacement x a = constant acceleration (typical units: m/s 2 )
20 1D motion equations for constant acceleration Bird runs north at a speed of 13.0 m/s, and slows down to 10.6 m/s in 4.0 seconds. What is the direction of the bird s acceleration? What is the bird s velocity after an additional 2.0 seconds? v 0 = m/s, north v = m/s t = 0 t = 4 s
21 2D motion equations for constant acceleration Bird runs north at a speed of 13.0 m/s, and slows down to 10.6 m/s in 4.0 seconds. What is the direction of the bird s acceleration? What is the bird s velocity after an additional 2.0 seconds? v 0 = m/s, north v = m/s t = 0 t = 4 s a = v v 0 = t t = (4 + 2) s = 6 s ( ) m/s 4.0 s v = v 0 + at = 13.0 m/s + ( 0.60 m/s 2 ) 6 s = ( )m/s = 9.4 m/s = 0.60 m/s 2 (points south) ( )
22 2D motion equations for constant acceleration x = 1 2 x-direction v x = v ox + a x t ( v + v ) t ox x y = 1 2 y-direction v y = v oy + a y t ( v + v ) t oy y v 2 x = v 2 ox + 2a x x v 2 y = v 2 oy + 2a y y x = v ox t a x t 2 y = v oy t a y t 2
23 2D motion equations for constant acceleration An projectle is fired at an angle of 45 with respect to the horizontal at a velocity of 100 m/s. There is a 100 m deep cliff, 100 m from the point of release. What is the range of the projectile? v 0 y x = m/s m 100 m v 0x = v 0 y = 70.7 m/s y = 0 v y y = 100m
24 2D motion equations for constant acceleration An projectle is fired at an angle of 45 with respect to the horizontal at a velocity of 100 m/s. There is a 100 m deep cliff, 100 m from the point of release. What is the range of the projectile? y direction v y 2 = v 2 0 y 2gy = (70.7)2 m 2 /s 2 2(9.8 m/s 2 )( 100 m) v y = 83.4 m/s t = y 0.5(v 0 y + v y ) = v 0 y 100 m x = m/s m v 0x = v 0 y = 70.7 m/s 100 m 0.5( )m/s = 15.7s y = 0 v y y = 100
25 2D motion equations for constant acceleration y direction v y 2 = v 2 0 y 2gy = (70.7)2 m 2 /s 2 2(9.8 m/s 2 )( 100 m) v y = 83.4 m/s t = An projectle is fired at an angle of 45 with respect to the horizontal at a velocity of 100 m/s. There is a 100 m deep cliff, 100 m from the point of release. What is the range of the projectile? y 0.5(v 0 y + v y ) = v 0 y 100 m x = m/s m v 0x = v 0 y = 70.7 m/s 100 m 0.5( )m/s = 15.7s x direction y = 0 ( )( 15.7s) x = v 0x t = 70.7 m/s = 1110 m v y y = 100
26 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction A force of magnitude, F, acts on three identical blocks. Rank the normal force on the three blocks. F F F F A F B F C
27 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction A force of magnitude, F, acts on three identical blocks of mass m. Rank the normal force on the three blocks. F doesn't affect normal force F Adds upward force component to mg F F Adds downward force component to mg F A F B F C F C > F A > F B
28 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction There is a graviational force, F, between two masses, m 1 and m 2, at a separation distance of R is F. If the distance between the masses is increased by a factor of 2, what is the effect on the gravitational force? F = G m 1 m 2 R 2 ; new r = 2R F = G m m 1 2 = G m m 1 2 r 2 (2R) 2 = G m m 1 2 4R = G m m 1 2 R 2 = 1 4 F
29 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction A mass m rests on a inclined plane with angle. If the coefficient of friction is 0.5, at what angle will the mass just begin to slide? θ F mg sinθ + y f m θ W = mg mg cosθ W = mg θ
30 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction A mass m rests on a inclined plane with angle. If the coefficient of friction is 0.5, at what angle will the mass just begin to slide? θ F + y Will slide if mg sinθ > f max m θ W = mg mg sinθ f mg cosθ W = mg θ
31 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction A mass m rests on a inclined plane with angle. If the coefficient of friction is 0.5, at what angle will the mass just begin to slide? θ F + y mg sinθ W = mg θ f Will slide if mg sinθ > f max mg cosθ m θ W = mg f max = µf = µmg cosθ mg sinθ > µmg cosθ µ = sinθ cosθ = tanθ θ = tan 1 µ = tan = 26.5
32 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction Three masses shown are stacked. What is the normal force of the 5m mass on the 3m mass? 20 m kg 203m kg 20 5mkg
33 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction Three masses shown are stacked. What is the normal force of the 5m mass on the 3m mass? F = 3mg + mg = 4mg How does the 5m mass know that there is a 1m mass on top? 20 m kg 203m kg 20 5mkg F = 4mg a) The 5m mass can see the 1m mass on the top. b) The 1m mass pushes on the 3m mass, and the 3m mass has a weight of 3mg. c) The 5m mass pushes up on the 3m mass and the 3m mass pushes up on 1m. d) The 1m mass and the 3m mass are glued together to make a 4m mass. e) All 3 masses must be glued together.
34 Newton s laws of motion Forces: gravity, tension, compression, normal, static and kinetic friction Three masses shown are stacked. What is the normal force of the 5m mass on the 3m mass? F = 3mg + mg = 4mg F 1 = 1mg W = 3mg 20 m kg 203m kg How does the 5m mass know that there is a 1m mass on top? 20 5mkg F = 4mg a) The 5m mass can see the 1m mass on the top. b) The 1m mass pushes on the 3m mass, and the 3m mass has a weight of 3mg. c) The 5m mass pushes up on the 3m mass and the 3m mass pushes up on 1m. d) The 1m mass and the 3m mass are glued together to make a 4m mass. e) All 3 masses must be glued together.
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