Forces. Isaac Newton stated 3 laws that deal with forces and describe motion. Backbone of Physics
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1 FORCES
2 Forces Isaac Newton stated 3 laws that deal with forces and describe motion. Backbone of Physics
3 Inertia Tendency of an object to remain in the same state of motion. Resists a change in motion. Inertia and mass are directly related. An object can be weightless and still have inertia.
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10 Inertia and the Seatbelt
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12 Newton s 1 st Law of Motion An object continues in its state of rest or uniform motion in a straight line unless acted upon by an unbalanced force. Unbalanced forces cause a change in motion.
13 Ground pushes up V = 0 m/s Gravity pulls down
14 Newton s Law implies that there is no fundamental difference between an object at rest and one moving with a constant velocity.
15 Ground pushes Friction up Engine Force Gravity pulls V = 10 m/s down
16 Note: There were no unbalanced forces. If the engine force increases, the car will speed up. (Accelerate) If the engine force decreases, the car will slow down. (Decelerate) Unbalanced forces cause accelerations.
17 Newton s Second Law of Motion When an unbalanced force acts on an object, the object will be accelerated. The acceleration will vary directly with the unbalanced force... Force, A
18 and will be in the same direction as the applied force. The acceleration will vary indirectly with the mass. mass A F net = ma F = Force A = F net m m = mass A = acceleration
19 A = F net m
20 Force is a vector quantity. Why? Acceleration is a vector quantity. Therefore Force must be a vector quantity.
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23 If a golf ball, a baseball and a bowling ball are thrown with a same force, which ball will move with a greater acceleration? A. Golf ball B. Baseball ball C. Bowling ball D. The three balls will have equal acceleration. Answer: A
24 Units of Force F net = ma F net = 1 kg ( 1 m ) s 2 F net = 1 kg m s 2 = 1 Newton = 1 N A 1 N net force will accelerate a 1 kg mass at the rate of 1 m/s 2.
25 Two friends Mary and Maria are trying to pull a 10-kg chair in opposite directions. If Maria applied a force of 60 N and Mary applied a force of 40 N, in which direction will the chair move and with what acceleration? 40 N 60 N
26 A. The chair will move towards Mary with an acceleration of 2 m/s 2. B. The chair will move towards Mary with an acceleration of 10 m/s 2. C. The chair will move towards Maria with an acceleration of 2 m/s 2. D. The chair will move towards Maria with an acceleration of 10 m/s 2. Answer: C
27 Jack is boating in a river applying a contact force of 30 N, in a direction opposite to the flow of water, at the same time the water is exerting a force of 30 N on the boat. In which direction will the boat move? A. The boat will move in the direction of the flow of water. B. The boat will not move at all. C. The boat will move in the direction opposite to the flow of water. Answer: B
28 A car has a mass of 1000 kg. It starts from rest and travels 150 m N in 4 s. The car undergoes constant acceleration. What unbalanced (net) force acts on the car?
29 d = t = 150 m 4 s d = V i t + 1 At 2 2 A = 2d t 2 V i = 0 m/s V f =? A =? m = 1000 kg A = 2 ( 150 m ) ( 4s ) 2 F Net =? A = m/s 2 N
30 F net = ma F net = 1000 kg ( m ) s 2 F net = kg m s 2 F net = Newtons N
31 v yy y n yy yy yy yy yy y n yy yy yy yy yy yy During what time(s) is there a net force?
32 Mass and Weight Mass depends on the amount of matter present. Newton s 2 nd Law says that m requires an F to accelerate it at the same rate as a smaller mass. More force is needed to accelerate a bowling ball than a golf ball.
33 Drop a 1 kg ball. It accelerates at 9.8 m/s 2 because of gravity. F =? F net = 1 kg ( 9.8 m ) s 2 = 9.8 N The earth exerts a force of 9.8 N on a 1 kg mass. 2 kg 19.6 N 10 kg 98 N
34 Weight refers to the gravitational force exerted by a body. (Earth) W = mg F w = mg Weight depends on a gravitational attraction. Weight, in and of itself, is not directional. Note: Book uses F g instead of F w.
35 A car weighs 2 X 10 4 N. It starts from rest and travels 200 m S in 4 s undergoing uniform acceleration. What unbalanced (net) force acts on the car?
36 d = 200 m m =? t = 4 s F Net =? V i = 0 m/s F W = 2 x 10 4 N V f =? F W = mg m = F W g A =? m = 2 x 10 4 N 9.8 m/s 2 m = 2041 kg
37 d = V i t + 1 At 2 2 A = 2d t 2 A = 2( 200 m ) ( 4 s ) 2 A = 25 m s 2 S
38 F net = ma F net = 2041 kg ( 25 m) s 2 F net = N S
39 Frictional Force Resistance to motion between two contacting objects. Caused by weak/temporary attractions between contact points.
40 F f F A F A = Applied Force Motion F f = Frictional Force Opposes the intended motion. Acts in a direction parallel to the contacting surfaces.
41 F N F W = Weight of Object F W F N = Normal Force Force holding (pushing) the surfaces together. Acts to the contacting surfaces.
42 Note: There is no motion in the vertical direction. Therefore there are no unbalanced forces in the y direction. Σ F y = O N Σ F y = O N = F N _ + F w Summation F N = F w Forces are balanced. No net force.
43 F f = µ F N µ = F f F N µ = coefficient of friction Depends on the contacting surfaces. Varies between 0 and 1 0 < µ < 1 Frictionless Rough
44 Two Types of Friction Static Friction Kinetic Friction
45 Kinetic Friction Force F f kinetic = µ K F N Force exerted on one surface by the other when one of the objects is moving.
46 Static Friction Force F f static = µ S F N Friction that keeps an object at rest and prevents it from moving.
47 Coefficients of Friction Surface µ S µ K Rubber/Dry Concrete Wood /Wood Steel/Steel
48 Ball Bearings
49 A smooth wooden box is placed on a smooth wooden table. A 20 N force is necessary to keep a 50 N block moving at a constant velocity. A. What is the coefficient of kinetic friction?
50 F f F N F A = 20 N µ K = F f F N F W = 50 N Σ F y = O N = F N _ + F w Forces are balanced. No net force. F N F N = F w = 50 N
51 Σ F X = O N = F A _ + F f F f = F A F f = 20 N µ K = 20 N µ K =.4 50 N Note: µ is a unitless number
52 B. If a 2 kg mass is placed on the block, what force must be applied to keep the block moving at a constant velocity?
53 F N 2 kg F f F A =? F W Σ F X = O N = F A _ + F f F f = F A
54 F f = µ K F N _ Σ F y = O N = F N + F w F N = F w F w = 50 N + 2 kg ( 9.8 m/s 2 ) F w = 69.6 N F N = 69.6 N F f =.4 ( 69.6 N ) F f = 27.8 N F f = F A F A = 27.8 N
55 Net Force and Acceleration F net is what is left over after all forces have been accounted for. Σ F X = O N Occurs when all forces are balanced Forces are balanced when the acceleration is 0 m/s 2.
56 What happens if the forces are not balanced and the object is accelerating? A more general starting point is needed when doing Σ of forces. Σ F = F net = ma = F? + F?... Note: If A = 0 m/s 2 the equation reverts to: Σ F = O N
57 A block has a mass of 50 kg. It rests on a surface where the coefficient of friction is 0.2. What force must be applied to accelerate the block at a rate of 10 m/s 2 E?
58 F f F N 50 kg F A =? µ =.2 F W A = 10 m/s 2 Σ F X = F net = ma = F A _ + F f F A = ma + F f Need F f
59 F f = µ F N Σ F y = O N = F N _ + F w Forces are balanced. No net force. F N = F w F W = 50 kg (9.8 m/s 2 ) F W = 490 N F N = 490 N
60 F f =.2 (490 N) F f = 98 N F A = 50 kg ( 10 m/s 2 ) + 98 N F A = 598 N E
61 A block has a mass of 50 kg. It rests on a surface where the coefficient of friction is 0.2. If a 675 N force is applied to the block, how fast will it accelerate?
62 A 1500 N box is pulled across a floor with a rope that makes an angle of 40 o with the floor. A tension of 800 N is maintained on the rope. A. What force is actually pulling the box across the floor?
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64 800 N 40 O V V 1500 N V H Cos 40 O = V H 800 N V H = 613 N
65 B. If the box is moving at a constant velocity, what is the frictional force acting on the box? 800 N F f 40 O V V 1500 N V H
66 Σ F X = O N = V H + F f F f = V H F f = 613 N
67 C. What is the coefficient of friction between the box and the floor? µ = F f F N 800 N F f F N 40 O V V 1500 N V H F W
68 Σ F y = 0 N = F N + F w + V V F N = F w - V V Sin 40 O = V V 800 N V v = 514 N
69 F N = 1500 N N F N = 986 N µ = 613 N 986 N µ =.62
70 A 150 kg box is pulled across a floor with a rope that makes an angle of 40 o with the floor. A tension of 800 N is maintained on the rope. If the coefficient of friction is 0.20, what is the acceleration of the box?
71 Vectors and Inclined Planes F N θ 2 F W cos θ θ 1 = θ 2 θ 1 F w F W sin θ
72 F N = force holding the surfaces together. F W sin θ = portion of F W being funneled down the plane. Parallel to the plane. F W cos θ = portion of F W pushing on the plane. Perpendicular to the plane.
73 When doing Σ of forces, rotate the axis system so that the X axis is parallel to the contacting surfaces. Y X
74 A truck having a mass of 1000 kg is to be PUSHED UP a 30 o incline at a constant velocity with a N force. What is the coefficient of friction between the incline and the truck?
75 F N F f F A = N θ 2 F W cos θ F w F W sin θ θ 1 θ 1 = θ 2 = 30 O
76 µ = F f F N Σ F X = O N = F A + F f + F W sin θ F f = F A + F W sin θ F w = 1000 kg ( 9.8 m/s 2 ) = 9800 N F f = N + (9800 N ) sin 30 F f = 5100 N
77 Σ F y = 0 N = F N + F W cos θ F N = F W cos θ F N = ( 9800 N ) cos 30 F N = 8487 N µ = F f F N µ = 5100 N 8487 N µ =.6
78 A rocket having a mass of 100 kg is to be accelerated at 50 m/s 2. What force is needed? Note: F N does U S A F A F w F f = 0 N not exist
79 Σ F y = F net = ma = F A + F w F A = ma + F w F A = 100 kg ( 50 m/s 2 ) kg ( 9.8 m/s 2 ) F A = 5980 N
80 A girl dribbles a 0.75 kg basketball. If a downwards force of 20 N is exerted on the ball. What is the acceleration of the basketball? F A Σ F y = F net = ma = F A + F w F f F w A = F A + F w m
81 A = 20 N + (0.75 kg)(9.8 m/s 2 ) 0.75 kg A = 36.5 m/s 2
82 A student of mass 40 kg is now standing on a bathroom scale in an elevator. What is the acceleration (magnitude and direction) of the student if the reading on the scale is 450 N?
83 Assuming up = + Σ F y = F Net = ma = R + - F w Σ F y = F Net = ma = 450 N 40 kg(9.8 m/s 2 ) Σ F y = F Net = ma = 58 N This indicates that there is a net force acting up.
84 Σ F y = F Net = ma = 58 N 58 N = ma 58 N = 40 kg (A) A = 1.45 m/s 2
85 Newton s 3 rd Law of Motion Law of Action and Reaction: Every force is accompanied by an equal and opposite force. There are no situations where there is just a single force present.
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88 X X X X X X
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91 Clone A exerts a tension of 500 N on the rope. Thus, F CloneA on rope = 500 N. Similarly, F CloneB on rope = 500 N. But the two tensions are an interaction pair, so they are equal and opposite. Thus, the tension in the rope equals the force with which each clone pulls (i.e. 500 N). According to Newton s third law, F CloneA on rope = F CloneB on rope. The net force is zero, so the rope will stay at rest as long as the net force is zero.
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94 What is the acceleration of the following system? Block 1 Motion 2 kg µ = kg Block 2
95 Methods of Solution: View each block individually. View the system as a whole.
96 View Each Block Individually F f F N F T Block 1 2 kg F W = 2 kg ( 9.8 m/s 2 ) = 19.6 N Σ F y = O N = F N _ + F w F N = F w F N = 19.6 N
97 F f = µ F N F f =.25 (19.6 N) F f = 4.9 N _ Σ F X = F net = ma 2kg = F T + F f A 2kg = F T _ + F f m
98 F T Block 2 10 kg F W = 10 kg ( 9.8 m/s 2 ) = 98 N _ Σ F y = F net = ma 10 kg = F W + F T A 10kg = F W _ + F T m
99 A 2kg = F T _ + F f m A 10kg = F W _ + F T m Since the blocks are attached together, A 2kg and A 10kg must be the same.
100 F T + _ F f = F W + _ F T m 2 kg m 10 kg F T N = 98 N + F T 2 kg 10 kg F T = 20.4 N
101 A 10kg _ = 98 N N 10 kg A 10 kg = 7.75 m/s 2
102 Solution #2: View the System as a Whole. 2 kg Motion µ = kg Look at forces acting on the system.
103 F f = 4.9 N 2 kg Motion µ = kg F W10kg = 98 N
104 Σ F xy = F net = ma = F W 12 kg _ + F f A 12kg = 98 N N 12 kg A 12kg = 7.75 m/s 2
105 A 2.0 kg mass (m A ) and a 3.0 kg mass (m B ) are attached to a lightweight cord that passes over a frictionless pulley. Find the acceleration of the smaller mass and the tension in the cord. F XY = F Net = ma= F F w B W A 5.0 kg ( A) = 27N 19.6 N F wa F wb A = m s 2 +
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107 Vectors and Suspended Objects Find weight of the cat burglar if the tension in the horizontal cable is 845 N. V V T H V H F w R Σ F X = O N = T H + V H V H = T H
108 V H = T H = 845 N V V V H 37 O tan 37 = V V V H R V V = 637 N Σ F y = O N = V V + F w F w = V V = 637 N
109 If the sign has a mass of 25.0 kg, determine the tension in the supporting chain. V H 25 O Physics is Phun V v F w = mg m F w = 25.0kg(9.8 ) = 245N 2 s F w Σ F y = O N = V V + F w F w = V V = 245 N
110 F w = V V = 245 N R V H 25 O Physics is Phun V v cos 65 = 245N 65 O R F w R = N
111 The End
112 A picture frame having a mass of 10 kg is supported by two wires from it's top two corners. The angle between the wires is 45 o. What force is exerted by each wire?
113 V H1 V H2 θ 1 θ 2 45 O V V1 F R1 F R2 V V2 10 kg F W
114 θ 1 = θ 2 = similar triangles. 180 O = 2 θ + 45 O θ = 67.5 O Σ F X = O N = V H1 + V H2 V H1 = V H2
115 Cos 67.5 = V H1 F R1 F R1 Cos 67.5 = V H1 Cos 67.5 = V H2 F R2 F R2 Cos 67.5 = V H2 Because of the symmetry: F R1 Cos 67.5 = F R2 Cos 67.5 F R1 = F R2
116 Σ F y = 0 N = F W + V V1 + V V2 F W = V V1 + V V2 98 N = V V1 + V V2
117 Sin 67.5 = V V1 F R1 F R1 Sin 67.5 = V V1 Sin 67.5 = V V2 F R2 F R2 Sin 67.5 = V V2 98 N = F R1 Sin F R2 Sin N = F R1 Sin F R1 Sin 67.5
118 Because of the symmetry: 98 N = 2( F R1 Sin 67.5 ) 98 N = 1.85 F R1 F R1 = 53 N F R2 = 53 N
119 The End
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