1 / 34

PHY 113 A General Physics I 11 AM-12:15 P M MWF Olin 101 Plan for Lecture 5:

PHY 113 A General Physics I 11 AM-12:15 P M MWF Olin 101 Plan for Lecture 5: Chapter 5 – NEWTON’S GENIUS Concept of force Relationship between force and acceleration Net force Note: today we are going to neglect friction. Webassign question from Assignment #4. North. 1. East. 4.

arlen
Download Presentation

PHY 113 A General Physics I 11 AM-12:15 P M MWF Olin 101 Plan for Lecture 5:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. PHY 113 A General Physics I • 11 AM-12:15 PM MWF Olin 101 • Plan for Lecture 5: • Chapter 5 – NEWTON’S GENIUS • Concept of force • Relationship between force and acceleration • Net force • Note: today we are going to neglect friction. PHY 113 C Fall 2013 -- Lecture 5

  2. PHY 113 C Fall 2013 -- Lecture 5

  3. Webassign question from Assignment #4 North 1 East 4 2 3 PHY 113 C Fall 2013 -- Lecture 5

  4. Webassign question from Assignment #4 vi x= Given: h,H,x,q Determine: vi PHY 113 C Fall 2013 -- Lecture 5

  5. Two-dimensional motion with constant acceleration • Problem solving steps • Visualize problem – labeling variables • Determine which basic physical principle(s) apply • Write down the appropriate equations using the variables defined in step 1. • Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns). • Solve the equations. • Check whether your answer makes sense (units, order of magnitude, etc.). PHY 113 C Fall 2013 -- Lecture 5

  6. Isaac Newton, English physicist and mathematician (1642—1727) In the absence of a net force, an object remains at constant velocity or at rest. In the presence of a net force F, the motion of an object of mass m is described by the form F=ma. F12 =– F21. http://www.newton.ac.uk/newton.html PHY 113 C Fall 2013 -- Lecture 5

  7. Detail: “Inertial” frame of reference Strictly speaking, Newton’s laws work only in a reference frame (coordinate system) that is stationary or moving at constant velocity. In a non-inertial (accelerating) reference frame, there are some extra contributions. • iclicker question: • Are we (here in Winston-Salem) • In an inertial frame of reference (exactly) • In a non-inertial frame of reference Newton’s laws are only approximately true in Winston-Salem, but the corrections are very small. PHY 113 C Fall 2013 -- Lecture 5

  8. PHY 113 C Fall 2013 -- Lecture 5

  9. Example force – weight 1 lb = 4.45 N 1 N = 0.225 lb Newton PHY 113 C Fall 2013 -- Lecture 5

  10. Gravitational mass versus inertial mass – The concept of mass (“inertial” mass m) is distinct from the concept of weight mg. In fact the gravitation “constant” g varies slightly throughout the globe. The weight of mass m on the surface of the earth is different from its weight measured on the surface of the moon or on the surface of Mars. (Stay tuned to Chapter 13 – “Universal Gravitation”.) • iclicker question • Will we always need to take gravity into account? • Yes • No PHY 113 C Fall 2013 -- Lecture 5

  11. Isaac Newton, English physicist and mathematician (1642—1727) In the absence of a net force, an object remains at constant velocity or at rest. In the presence of a net force F, the motion of an object of mass m is described by the form F=ma. F12 =– F21. http://www.newton.ac.uk/newton.html PHY 113 C Fall 2013 -- Lecture 5

  12. Examples Ftension= T j Fgravity=-mg j F=Fnet=Ftension+Fgravity=(T-mg) j =0 PHY 113 C Fall 2013 -- Lecture 5

  13. Examples Ftension= T j Fgravity=-mg j Fgravity=-mg j F=Fnet=Fgravity=(-mg) j = ma F=Fnet=Ftension+Fgravity=(T-mg) j =0 a=-gj PHY 113 C Fall 2013 -- Lecture 5

  14. Two-dimensional motion with constant acceleration vi x= PHY 113 C Fall 2013 -- Lecture 5

  15. iclicker question: • An object having a mass of 2 kg is moving at constant velocity of 3 m/s in the upward direction near the surface of the Earth. What is the net force acting on the object? • 0 N. • 6 N in the upward direction. • 25.6 N in the upward direction. • 19.6 N in the downward direction. PHY 113 C Fall 2013 -- Lecture 5

  16. Example: (assume surface is frictionless) Suppose T measured a observed & measured T = m a PHY 113 C Fall 2013 -- Lecture 5

  17. Example of two dimensional motion on a frictionless horizontal surface m PHY 113 C Fall 2013 -- Lecture 5

  18. Example of two dimensional motion on a frictionless horizontal surface -- continued m PHY 113 C Fall 2013 -- Lecture 5

  19. Example – support forces Fsupport acts in direction  to surface (in direction of surface “normal”) PHY 113 C Fall 2013 -- Lecture 5

  20. Another example of support force: PHY 113 C Fall 2013 -- Lecture 5

  21. iclicker question: • Why do we care about forces in equilibrium? • Normal people don’t care. • Physicist care because they are weird. • It might be important if the materials responsible for the support forces are near their breaking point. PHY 113 C Fall 2013 -- Lecture 5

  22. Example of forces in equilibrium PHY 113 C Fall 2013 -- Lecture 5

  23. Example of forces in equilibrium -- continued PHY 113 C Fall 2013 -- Lecture 5

  24. Example of forces in equilibrium -- continued PHY 113 C Fall 2013 -- Lecture 5

  25. Newton’s second law F = m a Types of forces: FundamentalApproximateEmpirical Gravitational F=-mg j Friction Electrical Support Magnetic Elastic Elementary particles PHY 113 C Fall 2013 -- Lecture 5

  26. Examples (one dimension): (constant) PHY 113 C Fall 2013 -- Lecture 5

  27. Newton’s third law F12 = - F21 FP T T 1 2 T=m1a a=FP/(m1+m2) FP-T=m2a T=FP (m1/(m1+m2)) PHY 113 C Fall 2013 -- Lecture 5

  28. T-m1g=m1a T-m2g=-m2a => after some algebra: PHY 113 C Fall 2013 -- Lecture 5

  29. Fgravity=-mg j Fscale Fscale + Fgravity = 0; Fscale = mg Question: What if you step on the scale in the elevator? Fscale + Fgravity = m a; Fscale = mg +m a PHY 113 C Fall 2013 -- Lecture 5

  30. PHY 113 C Fall 2013 -- Lecture 5

  31. Exercise For this exercise, you should ride the elevator from the first floor of Olin Physical Laboratory up to the second or third floor. While in the elevator, stand on the scale and record the scale readings to answer the following questions. 1. What is the scale reading when the elevator is stationary? 2. What is the scale reading when the elevator just starts moving upward? 3. What is the scale reading when the elevator is coming to a stop just before your destination floor? x >x <x PHY 113 C Fall 2013 -- Lecture 5

  32. Ffloor= mg Ffloor= mg-T mg mg Mg=12 lb mg=14.5 lb T=? iclicker exercise mg Mg Not enough info T T Mg PHY 113 C Fall 2013 -- Lecture 5

  33. Example: 2-dimensional forces • A car of mass m is on an icy (frictionless) driveway, inclined at an angle t as shown. Determine its acceleration. Conveniently tilted coordinate system: PHY 113 C Fall 2013 -- Lecture 5

  34. Another example: Note: we are using a tilted coordinate frame i vf=0 vi PHY 113 C Fall 2013 -- Lecture 5

More Related