Physics 100: Lecture 4b Chapter 4

Transcription

1 Physics 100: Lecture 4b Chapter 4 Today s Agenda More discussion of dynamics Recap The Free Body Diagram The tools we have for making & solving problems:» Ropes & Pulleys (tension)» Hooke s Law (springs) Physics 100: Lecture 4b, Pg 1

2 Review: Newton's Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, F NET = ma Where F NET = Σ F Law 3: Forces occur in action-reaction pairs, F A,B = - F B,A. Where F A,B is the force acting on object A due to its interaction with object B and vice-versa. Physics 100: Lecture 4b, Pg 2

3 Gravity: What is the force of gravity exerted by the earth on a typical physics student? Typical student mass m = 55kg g = 9.81 m/s 2. F g = mg = (55 kg)x(9.81 m/s 2 ) F g = 540 N = WEIGHT F S,E = F g = mg F E,S = -mg Physics 100: Lecture 4b, Pg 3

4 Lecture 4b, Act 1 Mass vs. Weight An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force. Ouch! His foot hurts... (a) (b) (c) more less the same Physics 100: Lecture 4b, Pg 4

5 Lecture 4b, Act 1 Solution The masses of both the bowling ball and the astronaut remain the same, so his foot will feel the same resistance and hurt the same as before. Ouch! Physics 100: Lecture 4b, Pg 5

6 Lecture 4b, Act 1 Solution However the weights of the bowling ball and the astronaut are less: Wow! That s light. W = mg Moon g Moon < g Earth Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. Physics 100: Lecture 4b, Pg 6

7 The Free Body Diagram Newton s 2nd Law says that for an object F = ma. Key phrase here is for an object. So before we can apply F = ma to any given object we isolate the forces acting on this object: Physics 100: Lecture 4b, Pg 7

8 The Free Body Diagram... Consider the following case What are the forces acting on the plank? P = plank F = floor W = wall E = earth F P,W F W,P F P,F F P,E F F,P F E,P Physics 100: Lecture 4b, Pg 8

9 The Free Body Diagram... Consider the following case What are the forces acting on the plank? Isolate the plank from the rest of the world. F P,W F W,P F P,F F P,E F F,P F E,P Physics 100: Lecture 4b, Pg 9

10 The Free Body Diagram... The forces acting on the plank should reveal themselves... F P,W F P,F F P,E Physics 100: Lecture 4b, Pg 10

11 Aside... In this example the plank is not moving... It is certainly not accelerating! So F NET = ma becomes F NET = 0 F P,W F P,W + F P,F + F P,E = 0 F P,F F P,E This is the basic idea behind statics, which we will discuss in a few weeks. Physics 100: Lecture 4b, Pg 11

12 Example Example dynamics problem: A box of mass m = 2 kg slides on a horizontal frictionless floor. A force F x = 10 N pushes on it in the x direction. What is the acceleration of the box? y F = F x i a =? m x Physics 100: Lecture 4b, Pg 12

13 Example... Draw a picture showing all of the forces y F F B,F x F F,B F B,E F E,B Physics 100: Lecture 4b, Pg 13

14 Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. y F F B,F x F F,B F B,E = mg F E,B Physics 100: Lecture 4b, Pg 14

15 Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. y F F B,F x mg Physics 100: Lecture 4b, Pg 15

16 Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. Solve Newton s equations for each component. F X = ma X F B,F - mg = ma Y F F B,F y x mg Physics 100: Lecture 4b, Pg 16

17 Example... F X = ma X So a X = F X / m = (10 N)/(2 kg) = 5 m/s 2. F B,F - mg = ma Y But a Y = 0 So F B,F = mg. F X N y x The vertical component of the force of the floor on the object (F B,F ) is often called the Normal Force (N). Since a Y = 0, N = mg in this case. mg Physics 100: Lecture 4b, Pg 17

18 Example Recap F X N = mg a X = F X / m y mg x Physics 100: Lecture 4b, Pg 18

19 Lecture 4b, Act 2 Normal Force A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? (a) N > mg (b) N = mg (c) N < mg a m Physics 100: Lecture 4b, Pg 19

20 Lecture 4b, Act 2 Solution All forces are acting in the y direction, so use: F total = ma N - mg = ma N = ma + mg m N mg a therefore N > mg Physics 100: Lecture 4b, Pg 20

21 Tools: Ropes & Strings Can be used to pull from a distance. Tension (T) at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position. The force you would feel if you cut the rope and grabbed the ends. An action-reaction pair. T cut T T Physics 100: Lecture 4b, Pg 21

22 Tools: Ropes & Strings... Consider a horizontal segment of rope having mass m: Draw a free-body diagram (ignore gravity). m T 1 T a 2 x Using Newton s 2nd law (in x direction): F NET = T 2 - T 1 = ma So if m = 0 (i.e. the rope is light) then T 1 = T 2 Physics 100: Lecture 4b, Pg 22

23 Tools: Ropes & Strings... 2 skateboards An ideal (massless) rope has constant tension along the rope. T T If a rope has mass, the tension can vary along the rope For example, a heavy rope hanging from the ceiling... T = T g T = 0 We will deal mostly with ideal massless ropes. Physics 100: Lecture 4b, Pg 23

24 Tools: Ropes & Strings... The direction of the force provided by a rope is along the direction of the rope: T Since a y = 0 (box not moving), m T = mg mg Physics 100: Lecture 4b, Pg 24

25 Lecture 4b, Act 3 Force and acceleration A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s 2. What is the mass of the fish? a = 12.2 m/s 2 snap! m =? (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg Physics 100: Lecture 4b, Pg 25

26 Draw a Free Body Diagram!! Lecture 4b, Act 3 Solution: T Use Newton s 2nd law in the upward direction: F TOT = ma T - mg = ma T = ma + mg = m(g+a) a = 12.2 m/s 2 m =? mg m = T g a 180N m = = 8. 2kg 2 m s + ( ) Physics 100: Lecture 4b, Pg 26

27 Tools: Pegs & Pulleys Used to change the direction of forces An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: F 1 ideal peg or pulley F 1 = F 2 F 2 Physics 100: Lecture 4b, Pg 27

28 Tools: Pegs & Pulleys Used to change the direction of forces An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: F W,S = mg T m T = mg mg Physics 100: Lecture 4b, Pg 28

29 Springs Hooke s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. F X = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = 0 x Physics 100: Lecture 4b, Pg 29

30 Springs... Hooke s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. F X = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = -kx > 0 x < 0 x Physics 100: Lecture 4b, Pg 30

31 Springs... Horizontal springs Hooke s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. F X = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = - kx < 0 x > 0 x Physics 100: Lecture 4b, Pg 31

32 Scales: Spring/string Springs can be calibrated to tell us the applied force. We can calibrate scales to read Newtons, or... Fishing scales usually read weight in kg or lbs. 1 lb = 4.45 N Physics 100: Lecture 4b, Pg 32

33 Lecture 4b, Act 4 Force and acceleration? Scale on a skate A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs? m m m (1) (2) (a) 0 lbs. (b) 4 lbs. (c) 8 lbs. Physics 100: Lecture 4b, Pg 33

34 Lecture 4b, Act 4 Solution: Draw a Free Body Diagram of one of the blocks!! Use Newton s 2nd Law in the y direction: a = 0 since the blocks are stationary F TOT = 0 T - mg = 0 T = mg = 4 lbs. m mg T T = mg Physics 100: Lecture 4b, Pg 34

35 Lecture 4b, Act 4 Solution: The scale reads the tension in the rope, which is T = 4 lbs in both cases! T T T T T T T m m m Physics 100: Lecture 4b, Pg 35

36 Recap of today s lecture.. More discussion of dynamics. Recap (Text: 4-1 to 4-6) The Free Body Diagram (Text: 4-7) The tools we have for making & solving problems:» Ropes & Pulleys (tension) (Text: 4-7)» Hooke s Law (springs). (Text: 4-7) Look at Textbook problems Chapter 4: # Physics 100: Lecture 4b, Pg 36