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Transcript
Phys 207 Announcements z Hwk3 submission deadline = 1st thing Monday morning Quiz 2 average score = 1.7/4 (N=110) z Today’s Agenda More discussion of dynamics ÍRecap Í The Free Body Diagram ÍThe tools we have for making & solving problems: » Ropes & Pulleys (tension) 1 Topics 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. Tangential 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 TA Office Hours z z z z z 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) T,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, FNET = ma Where FNET = Σ F Law 3: Forces occur in action-reaction pairs, FA ,B = - FB ,A. Where FA ,B is the force acting on object A due to its interaction with object B and vice-versa. 4 Page 2 Gravity: z What is the force of gravity exerted by the earth on a typical physics student? ÍTypical student mass m = 55kg Íg = 9.81 m/s2. ÍFg = mg = (55 kg)x(9.81 m/s2 ) ÍFg = 540 N = WEIGHT FE,S = Fg = mg FS,E = -mg 5 Lecture 8, Act 1 Mass vs. Weight z 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) more (b) less (c) the same 6 Page 3 Lecture 8, Act 1 Solution z 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! 7 Lecture 8, Act 1 Solution z However the weights of the bowling ball and the astronaut are less: W = mgMoon z Wow! That’s light. gMoon < gEarth Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. 8 Page 4 The Free Body Diagram z Newton’s 2nd Law says that for an object F = ma. z Key phrase here is for an object. z So before we can apply F = ma to any given object we isolate the forces acting on this object: 9 The Free Body Diagram... z Consider the following case ÍWhat are the forces acting on the plank ? P = plank F = floor W = wall E = earth FW,P FP,W FF,P FP,F FE,P FP,E 10 Page 5 The Free Body Diagram... z Consider the following case ÍWhat are the forces acting on the plank ? Isolate the plank from the rest of the world. FW,P FP,W FF,P FE,P FP,F FP,E 11 The Free Body Diagram... z The forces acting on the plank should reveal themselves... FW,P FF,P FE,P 12 Page 6 Aside... z In this example the plank is not moving... ÍIt is certainly not accelerating! ÍSo FNET = ma becomes FNET = 0 FW,P FW,P + FF,P + FE,P = 0 FF,P FE,P ÍThis is the basic idea behind statics, which we will discuss in a few weeks. 13 Example z Example dynamics problem: A box of mass m = 2 kg slides on a horizontal frictionless floor. A force Fx = 10 N pushes on it in the x direction. What is the acceleration of the box? y F = Fx i a=? m x 14 Page 7 Example... z Draw a picture showing all of the forces y FF,B F x FB,F FE,B FB,E 15 Example... z z Draw a picture showing all of the forces. Isolate the forces acting on the block. y FF,B F x FB,F FE,B = mg FB,E 16 Page 8 Example... z z z Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. y FF,B F x mg 17 Example... z z z z 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. y Í FX = maX Í FF,B - mg = maY FF,B x F mg 18 Page 9 Example... z z FX = maX Í So aX = FX / m = (10 N)/(2 kg) = 5 m/s2. FF,B - mg = maY Í But aY = 0 Í So FF,B = mg. N y FX x mg z z The vertical component of the force of the floor on the object (FF,B ) is often called the Normal Force (N). Since aY = 0 , N = mg in this case. 19 Example Recap N = mg y FX aX = FX / m mg x 20 Page 10 Lecture 8, Act 2 Normal Force z 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 a (c) N < mg m 21 Lecture 8, Act 2 Solution All forces are acting in the y direction, so use: N Ftotal = ma a m N - mg = ma mg N = ma + mg therefore N > mg 22 Page 11 Tools: Ropes & Strings z z 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. ÍThe force you would feel if you cut the rope and grabbed the ends. ÍAn action-reaction pair. T cut T T 23 Tools: Ropes & Strings... z Consider a horizontal segment of rope having mass m: ÍDraw a free-body diagram (ignore gravity). m T1 a T2 z Using Newton’s 2nd law (in x direction): FNET = T2 - T1 = ma z So if m = 0 (i.e. the rope is light) then T1 = T2 x 24 Page 12 Tools: Ropes & Strings... z An ideal (massless) rope has constant tension along the rope. T z z T If a rope has mass, the tension can vary along the rope Í For example, a heavy rope hanging from the ceiling... T = Tg T=0 We will deal mostly with ideal massless ropes. 25 Tools: Ropes & Strings... z The direction of the force provided by a rope is along the direction of the rope: T Since ay = 0 (box not moving), m T = mg mg 26 Page 13 Lecture 8, Act 3 Force and acceleration z 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/s2. What is the mass of the fish? snap ! (a) 14.8 kg (b) 18.4 kg (c) a = 12.2 m/s2 8.2 kg m=? 27 Lecture 8, Act 3 Solution: z Draw a Free Body Diagram!! z Use Newton’s 2nd law in the upward direction: a = 12.2 m/s2 T m=? FTOT = ma T - mg = ma mg T = ma + mg = m(g+a) m= T g +a m = (9 . 8 180 N + 12 . 2 ) m s 2 = 8 . 2 kg 28 Page 14 Tools: Pegs & Pulleys z Used to change the direction of forces ÍAn ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: F1 | F1 | = | F2 | ideal peg or pulley F2 29 Tools: Pegs & Pulleys z Used to change the direction of forces ÍAn ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: FS,W = mg T T = mg m mg Page 15 30 Lecture 8, Act 4 Force and acceleration z 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 (2) (1) (a) 0 lbs. (b) 4 lbs. (c) 8 lbs. 31 Lecture 8, Act 4 Solution: z z Draw a Free Body Diagram of one of the blocks!! Use Newton’s 2nd Law in the y direction: T m T = mg a = 0 since the blocks are stationary mg FTOT = 0 T - mg = 0 T = mg = 4 lbs. 32 Page 16 Lecture 8, Act 4 Solution: z The scale reads the tension in the rope, which is T = 4 lbs in both cases! T T T T m T T T m m 33 Page 17