Forces and two masses.
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- Helen Moody
- 5 years ago
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1 Forces and two masses.
2 Atwood's Machine frictionless, massless pulley +a +a M 1 M 2
3 Atwood's Machine frictionless, massless pulley +a T +a m 1 a=t m 1 g 1 M 1 M 2 T m 2 a=m 2 g T 2 w 1 =m 1 g w 2 =m 2 g Solve 1 for T and plug into 2.
4 Atwood's Machine frictionless, massless pulley +a T +a m 1 a=t m 1 g T =m 1 a m 1 g 1 M 1 M 2 T m 2 a=m 2 g T 2 w 1 =m 1 g w 2 =m 2 g Solve 1 for T and plug into 2.
5 Atwood's Machine frictionless, massless pulley +a T +a m 1 a=t m 1 g T =m 1 a m 1 g 1 M 1 T m 2 a=m 2 g T 2 M 2 w 1 =m 1 g m 2 a=m 2 g m 1 a m 1 g w 2 =m 2 g Solve 1 for T and plug into 2.
6 Atwood's Machine frictionless, massless pulley +a T +a m 1 a=t m 1 g T =m 1 a m 1 g 1 M 1 T m 2 a=m 2 g T 2 M 2 w 2 =m 2 g w 1 =m 1 g m 2 a=m 2 g m 1 a m 1 g m 2 a=m 2 g m 1 a m 1 g Solve 1 for T and plug into 2.
7 Atwood's Machine frictionless, massless pulley +a T +a m 1 a=t m 1 g T =m 1 a m 1 g 1 M 1 T m 2 a=m 2 g T 2 M 2 w 2 =m 2 g w 1 =m 1 g m 2 a=m 2 g m 1 a m 1 g m 2 a=m 2 g m 1 a m 1 g m 2 m 1 a= m 2 m 1 g Solve 1 for T and plug into 2.
8 Atwood's Machine frictionless, massless pulley +a T +a m 1 a=t m 1 g T =m 1 a m 1 g 1 M 1 T m 2 a=m 2 g T 2 M 2 w 2 =m 2 g w 1 =m 1 g m 2 a=m 2 g m 1 a m 1 g m 2 a=m 2 g m 1 a m 1 g m 2 m 1 a= m 2 m 1 g Solve 1 for T and plug into 2. a= m 2 m 1 m 2 m 1 g
9 Atwood's Machine frictionless, massless pulley This is how an elevator works. +a T T Counter Weight +a Pulley is replaced by a motor. One mass is the car. The other mass is a counter-balance that is equal to the mass of the car + the mass of the people in a fully packed car. car w 1 =m 1 g If car is full and going up, the masses are equal and the motor does not have to do much work. If car is empty and going down, the motor has to do a lot of work, lifting the Counter Weight. w 2 =m 2 g Solve 1 for T and plug into 2.
10 Frictionless, massless pulley m M
11 Frictionless, massless pulley +a m M +a
12 Frictionless, massless pulley +a m T M T +a
13 Frictionless, massless pulley F N m +a T M T +a W m = mg W M = Mg F N m: M: T T F N mg mg
14 Frictionless, massless pulley F N m +a T W m = mg T +a M W M = Mg m: F N mg cos =0 F N =mg cos T mg sin =ma T =ma mg sin M: Mg T =Ma F N m: M: T T F N mg mg
15 Frictionless, massless pulley F N F N m +a T W m = mg m: M: T +a M W M = Mg m: F N mg cos =0 F N =mg cos T mg sin =ma T =ma mg sin M: Mg T =Ma Mg ma mg sin =Ma T T F N mg mg
16 Frictionless, massless pulley F N F N m +a T W m = mg m: M: T +a M W M = Mg m: F N mg cos =0 F N =mg cos T mg sin =ma T =ma mg sin M: Mg T =Ma Mg ma mg sin =Ma Mg ma mg sin =Ma T T F N mg mg
17 Frictionless, massless pulley F N F N m +a T W m = mg m: M: T +a M W M = Mg m: F N mg cos =0 F N =mg cos T mg sin =ma T =ma mg sin M: Mg T =Ma Mg ma mg sin =Ma Mg ma mg sin =Ma T T F N M m sin a= M m g mg mg
18 Using Newton's third law: Find the acceleration. F app = 10 N M 1 = 3 kg M 2 = 2 kg
19 Using Newton's third law: Find the acceleration. You can probably guess that the acceleration will be: F app = 10 N M 1 = 3 kg M 2 = 2 kg a= F app = 10 N M 1 M 2 3kg 2kg =2 m s 2
20 Using Newton's third law: Find the acceleration. You can probably guess that the acceleration will be: F app = 10 N M 1 = 3 kg M 2 = 2 kg a= F app = 10 N M 1 M 2 3kg 2 kg =2 m s 2 F N1 F N2 F app = 10 N F 21 F 12 M 1 = 3 kg M 2 = 2 kg W 1 =M 1 g W 2 =M 2 g
21 Using Newton's third law: Find the acceleration. You can probably guess that the acceleration will be: F app = 10 N M 1 = 3 kg M 2 = 2 kg a= F app = 10 N M 1 M 2 3kg 2 kg =2 m s 2 In y direction: F N1 F N2 F N 1 =M 1 g F app = 10 N F 21 M 1 = 3 kg F 12 M 2 = 2 kg F N 2 =M 2 g W 1 =M 1 g W 2 =M 2 g
22 Using Newton's third law: Find the acceleration. F app F 21 =M 1 a F 12 =M 2 a But according to Newton's Third Law: F 21 = - F 12 In y direction: F N1 F N2 F N 1 =M 1 g F app = 10 N F 21 M 1 = 3 kg F 12 M 2 = 2 kg F N 2 =M 2 g W 1 =M 1 g W 2 =M 2 g
23 Using Newton's third law: Find the acceleration. F app F 21 =M 1 a 1 F 12 =M 2 a F 21 =M 2 a 2 But according to Newton's Third Law: F 21 = - F 12 In y direction: F N1 F N2 F N 1 =M 1 g F app = 10 N F 21 M 1 = 3 kg F 12 M 2 = 2 kg F N 2 =M 2 g W 1 =M 1 g W 2 =M 2 g
24 Using Newton's third law: Find the acceleration. F app F 21 =M 1 a F 12 =M 2 a F 21 =M 2 a But according to Newton's Third Law: F 21 = - F F app F 21 =M 1 a F app M 2 a=m 1 a a= F app M 1 M 2 =2 m s 2 F N1 F N2 In y direction: F N 1 =M 1 g F app = 10 N F 21 M 1 = 3 kg F 12 M 2 = 2 kg F N 2 =M 2 g W 1 =M 1 g W 2 =M 2 g
25 Spring:
26 Spring: F s m 1 W= m 1 g
27 Spring: y y y F s F s m 1 F s m 2 W= m 1 g m 3 W= m 2 g W= m 3 g
28 Force due to a Spring: F s = k s The spring constant k depends on the make-up of the spring. The minus sign means that the force due to the spring is always opposite the the displacement.
29 F s = k x x Slope = -k
30 F s =0 x 0 X=0 F s 0 F app F s 0 x 0 F app
31 Spring Find k: y=0.02m F s M 1 =10 kg W= m 1 g
32 Spring Find k: y=0.02 m F s F s = w k y=mg k= mg y k=4900 N /m=4900 kg / s 2 M 1 =10 kg W= m 1 g
33 Work = done on an object by a constant force is defined as the product of the displacement times the component of the force parallel to the displacement. W =F d=f cos d 1 Joule = 1 J = 1 Nm F 1 = 10 N W 1 =F 1 d=10 N 2m =20 J F 1 = 10 N 2 m 60 o W 1 =F 1 cos d=10 N cos60 o 2m W 1 =5 N 2m =10 J 2 m
34 Work done by a constant force: F W =F cos d=f cos x f x i F cos x i x f x
35 Work done by a constant force: F W =F cos d=f cos x f x i F cos WORK x i x f x Work = The area under an Force versus displacement plot.
36 Work done by a variable force: F x i x f x Work = The area under an Force versus displacement plot.
37 Work done by a variable force: F W 1 W 2 W 3 W 4 W 5 W 6 x i x f x Work = The area under an Force versus displacement plot. n W = net i=1 W i W net =W 1 W 2 W 3 W 4 W 5 W 6
38 F area of triangle= 1 2 b h 10 N 0m 30 m x Work is equal to the area under a force versus displacement plot.
39 F area of triangle= 1 2 b h 10 N 0m 30 m x Work is equal to the area under a force versus displacement plot.
40 F area of triangle= 1 2 b h 10 N 0m 30 m x Work is equal to the area under a force versus displacement plot. W = m 10 N =150 N
41 Work and Energy: A constant force applied to a mass through a displacement. F = constant v o x o
42 Work and Energy: A constant force applied to a mass through a displacement. F = constant v o F = constant v=? x o x x W =F x W =ma x
43 Work and Energy: A constant force applied to a mass through a displacement. F = constant v o F = constant v=? x o x x W =F x W =ma x For a constant Force, acceleration is a constant. v 2 =v o 2 2a x x= v2 2 v o 2a
44 Work and Energy: A constant force applied to a mass through a displacement. F = constant v o F = constant v=? x o x x W =F x W =ma =m W v2 v 2 o 2 For a constant Force, acceleration is a constant. v 2 =v o 2 2 a x a x= v2 2 v o 2
45 Work and Energy: A constant force applied to a mass through a displacement. F = constant v o F = constant v=? x o x x W =F x W =ma =m W v2 v 2 o 2 For a constant Force, acceleration is a constant. v 2 =v o 2 2a x a x= v2 2 v o 2 W = 1 2 m v2 1 2 m v 2 o
46 Work and Energy: A constant force applied to a mass through a displacement. F = constant v o F = constant v=? x o x x W =F x W =ma =m W v2 v 2 o 2 For a constant Force, acceleration is a constant. v 2 =v o 2 2a x a x= v2 2 v o 2 W = 1 2 m v2 1 2 m v 2 =K K o o
47 Work Kinetic Energy Theorem W = K =K K o K = 1 2 m v2 [ K ]=J
48 F 1 =20 N 10 kg 60 o v o =0m/ s ma x =F 1 cos x =10m v=? a= F 1cos m 20 N cos 60o = 10 kg =1 m s 2 v 2 =v 2 2a x=0 2 1 m m2 10 m =20 o s 2 s 2 v=4.5 m/s
49 F 1 =20 N 10 kg 60 o v o =0 m/s W = K F cos x=k K o x=10m v=? F cos x= 1 2 m v2 v= 2 F cos x m = 2 20N cos 60o 10m 10kg =4.5m/s
50 Gravitational Potential Energy: Move mass up at a constant velocity, i.e zero acceleration. What applied force do you need to apply? y F app h W=mg y o
51 Gravitational Potential Energy: y Move mass up at a constant velocity, i.e zero acceleration. What applied force do you need to apply? Correct! The applied force needs to be equal to the weight! F app h W=mg y o
52 Gravitational Potential Energy: y Move mass up at a constant velocity, i.e zero acceleration. What applied force do you need to apply? Correct! The applied force needs to be equal to the weight! F app h W=mg y o
53 Gravitational Potential Energy: Let's look at the work done by the gravitational force, i.e., the weight. y W g =mg cos d h F app v=constant W g =mg cos180 o h W g = mgh W g = mg y y o W g = mgy mgy o W g = PE PE o W=mg W g = PE y o
54 Gravitational Potential Energy W g = PE PE=mgh [ PE ]=J
55 Mechanical Energy ( No dissipative forces, such as friction) : W = KE= PE KE= PE KE KE o = PE PE o KE PE=KE o PE o E total =KE PE=constant E total =E total o
56 Drop a mass from rest: 10 kg v o =0m/ s E o =E KE o PE o =KE PE 20 m 10 kg v=?
57 Drop a mass from rest: E o =E 10 kg v o =0 m/s KE o PE o =KE PE 1 2 m v 2 mgh = 1 o o 2 m v2 mgh 20 m 1 2 m v 2 mgh = 1 o o 2 m v2 mgh 10 kg v=?
58 Drop a mass from rest: E o =E 10 kg v o =0 m/s KE o PE o =KE PE 1 2 m v 2 mgh = 1 o o 2 m v2 mgh 20 m 1 2 m v 2 mgh = 1 o o 2 m v2 mgh v= 2 g h o 10 kg v=? v= m/s 2 20m=19.8 m/s
59 v o =0 m/s F N No friction! h m w=mg x v=? ma x =m g sin a x =g sin v 2 =v o 2 2 g sin x v 2 =v o 2 2 g sin h sin sin = opp hyp sin = h x x= h sin
60 v o =0 m/s F N No friction! h m w=mg x v=? m a x =m g sin a x =g sin v 2 =v o 2 2 g sin x v 2 =v o 2 2 g sin h sin sin = opp hyp sin = h x x= h sin v 2 =2gh v= 2gh
61 v o =0 m/s F N No friction! h m w=mg x v=? E o =E 1 2 m v 2 mgy = 1 o o 2 m v2 mgy
62 v o =0 m/s F N No friction! h m w=mg x v=? E o =E 1 2 m v 2 mgy = 1 o o 2 m v2 mgy 1 2 m v 2 mgh= 1 o 2 m v2 mgy v= 2gh
63 w
64 F N w y w x w
65 F N w y w x w Can not use constant acceleration equations, net force and acceleration are not constant. Can use conservation of energy.
66 F N w y w x w x Can not use constant acceleration equations, net force and acceleration are not constant. Can use conservation of energy.
67 F N w y w x w w x w x Can not use constant acceleration equations, net force and acceleration are not constant. Can use conservation of energy.
68 F N w y w x w w x w x w x Can not use constant acceleration equations, net force and acceleration are not constant. Can use conservation of energy.
69 w y w F N w x w x E i =E f KE i PE i =KE f PE f 1 2 m v 2 i mgh i = 1 2 m v 2 f mgh f w x w x Can not use constant acceleration equations, net force and acceleration are not constant. Can use conservation of energy.
70 Roller Coaster ( No Friction): KE= 1 2 mv2 =0.0 J 20 kg PE=mgh=1960 J E total =1960 J 10 m 6 m
71 Roller Coaster ( No Friction): KE= 1 2 mv2 =0.0 J PE=mgh=1960 J E total =1960 J 10 m 20 kg 6 m KE= 1 2 mv2 =1960 J PE =mgh=0.0 J E total =1960 J v= 2 KE m =14 m/ s
72 Roller Coaster ( No Friction): KE= 1 2 mv2 =0.0 J PE=mgh=1960 J E total =1960 J E total =1960 J PE=mgh=20 kg 9.8 m/ s 2 6m=1176 J KE= 1 2 mv2 =1960 J 1176 J =784 J 10 m 20 kg v= J 20 kg =8.85m/ s 6 m KE= 1 2 mv2 =1960 J PE=mgh=0.0 J E total =1960 J v= 2 KE m =14 m/s
73 F s =k x x
74 F s =k x Work = area under force vs. displacement curve. Work=W =area under curve W = 1 2 base height W = 1 2 kx x=1 2 kx 2 x PE spring = 1 2 k x 2
75 Mechanical Energy ( No dissipative forces, such as friction) : KE PE PE sp =KE o PE o PE spo E total =KE PE PE sp =constant E total =E totalo
76 No Friction! Release from rest! 10 kg 1 m K = 600 N/m
77 KE= 1 2 m v2 =0 J No Friction! Release from rest! 10 kg PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x2 =0 J 1 m v=? K = 600 N/m
78 KE= 1 2 m v 2 =0 J No Friction! Release from rest! PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x 2 =0 J 1 m v=? 10 kg KE= 1 2 m v 2 =98 J K = 600 N/m PE=mgh=0 J PE sp = 1 2 k x 2 =0 J v= 2 KE m = 2 98 J 10kg =4.4 m s
79 KE= 1 2 m v2 =0 J No Friction! Release from rest! PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x2 =0 J 1 m v=? Dx 10 kg KE= 1 2 m v 2 =98 J K = 600 N/m PE=mgh=0 J PE sp = 1 2 k x 2 =0 J v= 2 KE m = 2 98 J 10kg =4.4 m s
80 KE= 1 2 m v2 =0 J PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x2 =0 J KE= 1 2 m v 2 =0 J PE=mgh=0 J PE sp = 1 2 k x 2 =98 J x = 2 PE k = 2 98 J 600 N /m =0.57m 1 m Dx 10 kg KE= 1 2 m v 2 =98 J K = 600 N/m PE=mgh=0 J PE sp = 1 2 k x 2 =0 J No Friction! Release from rest! v= 2 KE m = 2 98 J 10kg =4.4 m s
81 KE= 1 2 m v2 =0 J PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x 2 =0 J KE= 1 2 m v2 =0 J PE=mgh=0 J PE sp = 1 2 k x 2 =98 J x= 2 PE k = 2 98 J 600 N /m =0.57m 1 m v=4.4 m/s 10 kg KE= 1 2 m v2 =98 J K = 600 N/m PE=mgh=0 J PE sp = 1 2 k x 2 =0 J No Friction! Release from rest! v= 2 KE m = 2 98 J 10kg =4.4 m s
82 KE= 1 2 m v2 =0 J KE= 1 2 m v2 =0 J 10 kg PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x2 =0 J PE=mgh=0 J PE sp = 1 2 k x 2 =98 J 1 m x= 2 PE k = 2 98 J 600 N /m =0.57m KE= 1 2 m v2 =98 J K = 600 N/m PE=mgh=0 J PE sp = 1 2 k x 2 =0 J No Friction! Release from rest! v= 2 KE m = 2 98 J 10kg =4.4 m s
83 KE= 1 2 m v2 =0 J PE=mgh=10kg 9.8 m 2 1 m =98 J s PE sp = 1 2 k x2 =0 J KE= 1 2 m v2 =0 J PE=mgh=0 J PE sp = 1 2 k x 2 =98 J x= 2 PE k = 2 98 J 600 N /m =0.57m 1 m v=4.4 m/ s 10 kg KE= 1 2 m v2 =98 J K = 600 N/m PE=mgh=0 J PE sp = 1 2 k x 2 =0 J v= 2 KE m = 2 98 J 10kg =4.4 m s
84 KE= 1 2 m v2 =0 J PE=mgh=10kg 9.8 m 1 m =98 J 2 s PE sp = 1 2 k x2 =0 J KE= 1 2 m v2 =0 J PE=mgh=0 J PE sp = 1 2 k x 2 =98 J x= 2 PE k = 2 98 J 600 N /m =0.57m 1 m Dx 10 kg If there is no friction, no energy is dissipated, and this continues forever!!! KE= 1 2 m v2 =98 J PE=mgh=0 J PE sp = 1 2 k x 2 =0 J v= 2 KE m = 2 98 J 10kg =4.4 m s K = 600 N/m
85 Conservative Forces Work done by a force is independent of the path taken. Non-Conservative Forces Work done by a force depends on the path taken.
86 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. x B A m h C Lift a box from point A to point C.
87 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. B x A m Path 1 W G Path1 =W A B W BC Path 1 h Path 2 C Lift a box from point A to point C.
88 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. B x m A Path 1 W G Path 1 =W A B W BC W G Path 1 =mg cos90 o x W BC Work=W =F cos d Path 1 mg h Path 2 C Lift a box from point A to point C.
89 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. h Path 1 B m mg x A Path 2 Work=W =F cos d Path 1 W G Path 1 =W A B W BC W G Path 1 =mg cos90 o x mg cos0 o h W G Path 1 =0 mgh=mgh C Lift a box from point A to point C.
90 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. h Path 1 B C x m mg A Path 2 h 2 x 2 Work=W =F cos d Path 1 W G Path 1 =W A B W BC W G Path 1 =mg cos90 o x mg cos0 o h W G Path 1 =0 mgh=mgh Path 2 Lift a box from point A to point C.
91 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. h Path 1 B C x m mg A Path 2 h 2 x 2 Work=W =F cos d Path 1 W G Path 1 =W A B W BC W G Path 1 =mg cos 90 o x mg cos 0 o h W G Path 1 =0 mgh=mgh Path 2 W G Path 2 =W A C Lift a box from point A to point C.
92 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. h Path 1 B C x m mg A Path 2 h 2 x 2 Work=W =F cos d Path 1 W G Path 1 =W A B W BC W G Path 1 =mg cos 90 o x mg cos 0 o h W G Path 1 =0 mgh=mgh Path 2 W G Path2 =W A C Lift a box from point A to point C.
93 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. h Path 1 B C x m mg A Path 2 h 2 x 2 Work=W =F cos d Path 1 W G Path1 =W A B W BC W G Path1 =mg cos180 o h mg cos 90 o x W G Path1 = mgh 0= mgh Path 2 W G Path 2 =W A C W G Path 2 =mg cos h 2 x 2 Lift a box from point A to point C.
94 Conservative Forces Work done by a force is independent of the path taken. Example: Gravitational Force. h Path 1 B x m mg A Path 2 h 2 x 2 Path 1 W G Path 1 =W A B W BC W G Path 1 =mg cos 90 o x mg cos 0 o h W G Path 1 =0 mgh=mgh Path 2 W G Path 2 =W A C W G Path 2 =mg cos h 2 x 2 Work=W =F cos d C W G Path 2 =mg h h 2 x 2 W G Path 2 =mgh h 2 x 2 Lift a box from point A to point C. W G Path 1 =W GPath 2
95 Non-Conservative Forces Work done by a force depends on the path taken. Example: friction W f = f cos 180 o s= mg s F n m s f = F N = mg mg
96 Non-Conservative Forces Work done by a force depends on the path taken. Example: friction Work =W =F cos d B m x f A Path 1 W f Path 1 =W A B W BC W f Path 1 = f cos180 o x W BC Path 1 s W f Path 1 = mg cos180 o x W BC y Path 2 h 2 x 2 C Move a box from point A to point C. Looking down on table.
97 Non-Conservative Forces Work done by a force depends on the path taken. Example: friction Work=W =F cos d x Path 1 Path 1 B f A W f Path 1 =W A B W BC W f Path 1 = f cos180 o x W BC W f Path 1 = mg cos180 o x mg cos180 o y W f Path 1 = mg x y y m s Path 2 h 2 x 2 C Move a box from point A to point C. Looking down on table.
98 Non-Conservative Forces Work done by a force depends on the path taken. Example: friction Work=W =F cos d x Path 1 Path 1 B f A W f Path 1 =W A B W BC W f Path 1 = f cos 180 o x W BC W f Path 1 = mg cos180 o x mg cos 180 o y W f Path 1 = mg x y y s m Path 2 h 2 x 2 W f Path1 =W A C W f Path1 = f cos180 o x 2 y 2 C W f Path1 = mg cos180 o x 2 y 2 W f Path1 = mg x 2 y 2 W f Path1 W f Path2 Move a box from point A to point C. Looking down on table.
99 Mechanical Energy ( With dissipative forces, such as friction) : KE o PE o PE spo =KE PE PE sp W nc
100 Mechanical Energy ( With dissipative forces, such as friction) : KE o PE o PE spo =KE PE PE sp W nc Block slides to a stop because of friction. v = 20 m/s v = 0 m/s 10 kg 10 kg x=? =0.1
101 Mechanical Energy ( With dissipative forces, such as friction) : KE o PE o PE spo =KE PE PE sp W nc Block slides to a stop because of friction. F N v = 20 m/s v = 0 m/s f = F N 10 kg 10 kg mg OLD WAY USING FORCES x=? =0.1 F y =ma y F N mg=0 F N =mg F x =ma x f =ma x F N =ma x mg=ma x a x = g=.98 m/s 2
102 Mechanical Energy ( With dissipative forces, such as friction) : KE o PE o PE spo =KE PE PE sp W nc Block slides to a stop because of friction. F N v = 20 m/s v = 0 m/s f = F N 10 kg 10 kg mg x=? =0.1 OLD WAY USING FORCES F y =m a y F N mg=0 F N =mg F x =ma x f =ma x F N =m a x mg=ma x a x = g=.98 m/s 2 v 2 =v o 2 2a x x= v2 2 v o m/ s 2 =0 20 =204 m 2a.98 m/s 2
103 Mechanical Energy ( With dissipative forces, such as friction) : KE o PE o PE spo =KE PE PE sp W nc Block slides to a stop because of friction. F N v = 20 m/s v = 0 m/s f = F N 10 kg 10 kg mg OLD WAY USING ENERGY x=? KE o PE o PE spo =KE PE PE sp W nc KE o = W nc 1 2 m v 2 o= f cos x 1 2 m v o 2 = m g cos180 o x 1 2 v 2 o= g x 2 x= v o 2 g =204 m =0.1
104 Power rate of doing work. P= W t [ P]=Watt W I W =1 J / s
105 If velocity equals a constant: P= W t = F x t =F x t =F v
106 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
107 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. F Look at is force. No matter how large the force, the bolt will not turn. The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
108 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. F Same here. No matter how large the force, the bolt will not turn. The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
109 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. F Same here. No matter how large the force, the bolt will not turn. The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
110 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. F In order to get the bolt to turn, the force should be oriented in the proper direction. Here the force turns the bolt in the positive direction. The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
111 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. In order to get the bolt to turn, the force should be oriented in the proper direction. Here the force turns the bolt in the negative direction. F The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
112 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. r F = F X r The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
113 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. r F = F X r = F r The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
114 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. r F = F X r = F r The magnitude of the torque will depend on two things: 1. The angle at which the force is applied. 2. The distance from the axis of rotation to the force. (the MOMENT ARM or LEVER ARM)
115 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. r F = F X r = F r
116 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. r F F = F X r = F r
117 Torque result of a force causing rotation. Bolt and wrench Use torque to tighten or loosen the bolt. r F F = F X r = F r = F r sin
118 Torque Another Look. Bolt and wrench Use torque to tighten or loosen the bolt. Line of Action r C F C = F X r r = F r = F r sin
119 N =... = NET 1 2 N i=1 Now for static equilibrium: N NET = =0 i=1 N F x net = F x =0 i=1 N F ynet = i=1 F y =0
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