Phys 207. Quiz 2 average score = 1.7/4 (N=110)

Transcription

1 Phys 207 Announcements Hwk3 submission deadline = 1 st thing Monday morning Quiz 2 average score = 1.7/4 (N=110) oday s Agenda More discussion of dynamics Recap he Free Body Diagram he tools we have for making & solving problems:» Ropes & Pulleys (tension) 1 opics 1. Position, speed, acceleration 2. Motion constant acceleration 3. Scalar vs. Vector 4. Decomposing vectors into components along x, y, z 5. Adding subtracting vectors 6. Decompose 2D motion into x and y components 7. angential vs. radial velocity, acceleration 8. Applying 1D kinematics to x or y motion 9. Centripetal acceleration 10.Systems of equations for solving 2D motion problems 2 Page 1

2 A Office Hours R: Philip Castro / 7:00 9:00 pm / 227 SHL F: Dan Pajerowski / 1:30 3:30 pm / 131B SHL F: Xing Chen / 5:00 7:00 pm / 225 SHL (Commons Room),R: Nowak / 9:30 10:30am / 228 SHL R: Nowak / 1:30 2:30pm / 228 SHL and by appt. 3 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 NE = ma Where F NE = 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. 4 Page 2

3 Gravity: What is the force of gravity exerted by the earth on a typical physics student? ypical student mass m = 55kg g = 9.81 m/s 2. F g = = (55 kg)x(9.81 m/s 2 ) F g = 540 N = WEIGH F E,S = F g = F S,E = - 5 Lecture 8, 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 6 Page 3

4 Lecture 8, Act 1 Solution he 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! 7 Lecture 8, Act 1 Solution However the weights of the bowling ball and the astronaut are less: Wow! hat s light. W= Moon g Moon < g Earth hus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. 8 Page 4

5 he 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: 9 he Free Body Diagram... Consider the following case What are the forces acting on the plank? P = plank F = floor W = wall E = earth F W,P F P,W F F,P F E,P F P,F F P,E 10 Page 5

6 he 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 W,P F P,W F F,P F E,P F P,F F P,E 11 he Free Body Diagram... he forces acting on the plank should reveal themselves... F W,P F F,P F E,P 12 Page 6

7 Aside... In this example the plank is not moving... It is certainly not accelerating! So F NE = ma becomes F NE = 0 F W,P F W,P + F F,P + F E,P = 0 F F,P F E,P his is the basic idea behind statics, which we will discuss in a few weeks. 13 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 14 Page 7

8 Example... Draw a picture showing all of the forces y F F F,B x F B,F F E,B F B,E 15 Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. y F F F,B x F B,F F E,B = F B,E 16 Page 8

9 Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. y F F F,B x 17 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 y F F,B - = ma Y F F F,B x 18 Page 9

10 Example... F X = ma X So a X = F X / m = (10 N)/(2 kg) = 5 m/s 2. - = ma Y But a Y = 0 So F F,B =. F F,B F X N y x he vertical component of the force of the floor on the object (F F,B ) is often called the Normal Force (N). Since a Y = 0, N = in this case. 19 Example Recap F X N = a X = F X / m y x 20 Page 10

11 Lecture 8, 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 > (b) N = (c) N < a m 21 Lecture 8, Act 2 Solution All forces are acting in the y direction, so use: F total =ma N - = ma N=ma + m N a therefore N > 22 Page 11

12 ools: Ropes & Strings Can be used to pull from a distance. ension () at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position. he force you would feel if you cut the rope and grabbed the ends. An action-reaction pair. cut 23 ools: Ropes & Strings... Consider a horizontal segment of rope having mass m: Draw a free-body diagram (ignore gravity). m 1 a 2 x Using Newton s 2nd law (in x direction): F NE = 2-1 = ma So if m = 0 (i.e. the rope is light) then 1 = 2 24 Page 12

13 ools: Ropes & Strings... An ideal (massless) rope has constant tension along the rope. If a rope has mass, the tension can vary along the rope For example, a heavy rope hanging from the ceiling... = g = 0 We will deal mostly with ideal massless ropes. 25 ools: Ropes & Strings... he direction of the force provided by a rope is along the direction of the rope: Since a y = 0 (box not moving), m = 26 Page 13

14 Lecture 8, 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. he 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 27 Draw a Free Body Diagram!! Lecture 8, Act 3 Solution: Use Newton s 2nd law in the upward direction: F O = ma - = ma = ma + = m(g+a) a = 12.2 m/s 2 m =? m = g a 180 N m = = 8. 2 kg 2 m s + ( ) 28 Page 14

15 ools: 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 29 ools: 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 S,W = m = 30 Page 15

16 Lecture 8, Act 4 Force and acceleration A block weighing 4 lbs is hung from a rope attached to a scale. he 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. 31 Lecture 8, Act 4 Solution: Draw a Free Body Diagram of one of the blocks!! Use Newton s 2nd Law in the y direction: a= 0since the blocks are stationary F O = 0 - = 0 = = 4 lbs. m = 32 Page 16

17 Lecture 8, Act 4 Solution: he scale reads the tension in the rope, which is = 4 lbs in both cases! m m m 33 Page 17