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Physics 111: Lecture 5 Today’s Agenda

Physics 111: Lecture 5 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). Review: Newton's Laws.

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Physics 111: Lecture 5 Today’s Agenda

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  1. Physics 111: Lecture 5Today’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)

  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, FNET = ma Where FNET = 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.

  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/s2. • Fg = mg = (55 kg)x(9.81 m/s2 ) • Fg = 540 N = WEIGHT FS,E = Fg = mg FE,S = -mg

  4. Lecture 5, Act 1Mass 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. His foot hurts... Ouch! • (a) more (b) less (c) the same

  5. Lecture 5, Act 1Solution • 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!

  6. Lecture 5, Act 1Solution Wow! That’s light. • However the weights of the bowling ball and the astronaut are less: • Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. W= mgMoon gMoon < gEarth

  7. The Free Body Diagram • Newton’s 2nd Law says that for an object F = ma. • Key phrase here isfor an object. • So before we can apply F = ma to any given object we isolate the forces acting on this object:

  8. FP,W FW,P FP,F FP,E FF,P FE,P The Free Body Diagram... • Consider the following case • What are the forces acting on the plank ? P = plank F = floor W = wall E = earth

  9. FP,W FW,P FP,F FP,E FF,P FE,P 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.

  10. FP,W FP,F FP,E The Free Body Diagram... • The forces acting on the plank should reveal themselves...

  11. FP,W FP,F FP,E Aside... • In this example the plank is not moving... • It is certainly not accelerating! • So FNET = mabecomes FNET = 0 • This is the basic idea behind statics, which we will discuss in a few weeks. FP,W + FP,F + FP,E = 0

  12. Example • 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= ? x m

  13. Example... • Draw a picture showing all of the forces y FB,F F x FB,E FF,B FE,B

  14. Example... • Draw a picture showing all of the forces. • Isolate the forces acting on the block. y FB,F F x FB,E = mg FF,B FE,B

  15. Example... • Draw a picture showing all of the forces. • Isolate the forces acting on the block. • Draw a free body diagram. y FB,F x F mg

  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. • FX = maX • FB,F - mg = maY y x FB,F F mg

  17. Example... • FX = maX • So aX = FX / m = (10 N)/(2 kg) = 5 m/s2. • FB,F - mg = maY • But aY = 0 • So FB,F = mg. • The vertical component of the forceof the floor on the object (FB,F ) isoften called the Normal Force (N). • Since aY = 0, N = mg in this case. N y FX x mg

  18. Example Recap N = mg y FX aX= FX / m x mg

  19. a Lecture 5, Act 2Normal 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 m

  20. Lecture 5, Act 2Solution All forces are acting in the y direction, so use: Ftotal =ma N - mg= ma N=ma +mg therefore N > mg N m a mg

  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

  22. Tools: Ropes & Strings... • Consider a horizontal segment of rope having mass m: • Draw a free-body diagram (ignore gravity). • Using Newton’s 2nd law (in x direction): FNET = T2 - T1 = ma • So if m = 0 (i.e. the rope is light) then T1 =T2 m T1 T2 a x

  23. Tools: Ropes & Strings... 2 skateboards • 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... • We will deal mostly with ideal massless ropes. T T T = Tg T = 0

  24. Tools: Ropes & Strings... • 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

  25. Lecture 5, Act 3Force 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/s2. What is the mass of the fish? snap ! (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg a = 12.2 m/s2 m = ?

  26. Use Newton’s 2nd lawin the upward direction: a = 12.2 m/s2 m = ? FTOT= ma Lecture 5, Act 3Solution: T • Draw a Free Body Diagram!! T - mg = ma mg T = ma + mg =m(g+a)

  27. ideal peg or pulley 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: F1 | F1 | = | F2 | F2

  28. T m mg 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: FW,S = mg T = mg

  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. • FX = -k xWhere xis the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = 0 x

  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. • FX = -k x Where xis the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = -kx > 0 x • x  0

  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. • FX = -k x Where xis the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = - kx < 0 x • x > 0

  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 0 2 4 6 8

  33. ? Lecture 5, Act 4Force 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 (2) (1) (a)0 lbs.(b)4 lbs.(c)8 lbs.

  34. Lecture 5, Act 4Solution: • Draw a Free Body Diagram of one of the blocks!! T • Use Newton’s 2nd Lawin the y direction: m T = mg a = 0 since the blocks are stationary mg FTOT= 0 T - mg = 0 T = mg = 4 lbs.

  35. Lecture 5, Act 4Solution: • 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

  36. Problem: Accelerometer • A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle  with respect to the vertical (y) axis. Solve for  in terms of a and g. a  i

  37. T (string tension)  m mg (gravitational force) Accelerometer... • Draw a free body diagram for the mass: • What are all of the forces acting? i

  38. j i Accelerometer... • Using components (recommended): i: FX = TX = T sin = ma j: FY = TY - mg = T cos - mg = 0 TX  TY T  m ma mg

  39. j i Accelerometer... • Using components : i: T sin = ma j: T cos - mg = 0 • Eliminate T : TX TY T  m ma T sin = ma T cos = mg mg

  40. Accelerometer... • Alternative solution using vectors (elegant but not as systematic): • Find the total vector force FNET: T (string tension) T  mg  m FTOT mg (gravitational force)

  41. Accelerometer... • Alternative solution using vectors (elegant but not as systematic): • Find the total vector force FNET: • Recall that FNET= ma: • So T (string tension)  T mg  m ma mg (gravitational force)

  42. Accelerometer... Cart w/ accelerometer • Let’s put in some numbers: • Say the car goes from 0 to 60 mph in 10 seconds: • 60 mph = 60 x 0.45 m/s = 27 m/s. • Acceleration a = Δv/Δt = 2.7 m/s2. • So a/g = 2.7 / 9.8 = 0.28 . • = arctan(a/g) = 15.6 deg a 

  43. Problem: Inclined plane • A block of mass m slides down a frictionless ramp that makes angle  with respect to the horizontal. What is its acceleration a ? m a 

  44. j i Inclined plane... • Define convenient axes parallel and perpendicular to plane: • Acceleration ais in x direction only. m a 

  45. j i Inclined plane... Incline • Consider x and y components separately: • i: mg sin =ma. a = g sin  • j: N - mg cos  = 0. N = mg cos  ma mg sin  N  mg cos  mg

  46. j i Inclined plane... • Alternative solution using vectors: m N  mg a = g sin i N = mg cosj

  47. ma = mg sin   Angles of an Inclined plane The triangles are similar, so the angles are the same! N  mg

  48. x = 0 k M Lecture 6, Act 2Forces and Motion • A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30owhat is the new equilibrium position of the block x1?(a) x1 = 20cm (b) x1 = 25cm (c) x1 = 30cm x1 = ? k M q = 30o

  49. Choose the x-axis to be along downward direction of ramp. • FBD: The total force on the block is zero since it’s at rest. • Consider x-direction: Force of gravity on block is Fx,g = Mg sinq Force of spring on block is Fx,s = -kx1 N Fx,s = -kx1 y x • Fx,g = Mg sinq Mg q Lecture 6, Act 2Solution x1 k M q

  50. Mg sinq - kx1 = 0 y x Lecture 6, Act 2Solution • Since the total force in the x-direction must be 0: x1 Fx,s = -kx1 k M • Fx,g = Mg sinq q

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