1 / 13

Chapter 4 – Part 2 Force & motion with friction

Chapter 4 – Part 2 Force & motion with friction. Sue Ramlo, PhD Part 2 for Panopto / Springboard Including frictional forces… what is friction? How do we calculate it?.

ralph
Download Presentation

Chapter 4 – Part 2 Force & motion with friction

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. Chapter 4 – Part 2Force & motion with friction Sue Ramlo, PhD Part 2 for Panopto / Springboard Including frictional forces… what is friction? How do we calculate it?

  2. The force of friction always opposes the direction of motion (or of the direction the motion would be in the absence of friction). Depending on the circumstances, friction may be desirable or undesirable.

  3. Two force of friction types: • ABOUT to move… Maximum..STATIC • Fs max= s FN • If on incline, only mg, FN, Fs acting then… tan θ = s • MOVING… kinetic friction force, Fk • Fk = k FN • If on incline, only mg, FN, Fk acting & constant speed then… tan θ = k

  4. 4.6 Friction The coefficients of friction depend on both materials involved.

  5. Consider this system with friction… 20

  6. Fk FN 2 1 T T m2g m1g sin 20° m1g cos 20° Re-draw and find Fnet from picture • In motion so Fk  m1g sin θ– T + T - m2g - Fk = Fnet • Here == > Fk = kmg cos θ • Newton #2 - Fnet = ma  Fnet = (m1+ m2) a • Bring together: m1g sin θ - m2g - kmg cos θ = (m1+ m2) a • Solve for a… + direction

  7. Acceleration of system with friction… • Solve for a • Plug in values with units (k=0.030 … VERY small) • Answer: a = 0.15 m/s2 (about half as much as w/o friction!)

  8. 108. In loading a fish delivery truck, a person pushes a block of ice up a 20° incline at constant speed. The push is 150 N in magnitude and parallel to the incline. The block has a mass of 35.0 kg. (a) Is the incline frictionless? (b) If not, what is the force of kinetic friction on the block of ice?  (a) no (b) 33 N

  9. 104. Suppose the slope conditions for the skier shown in Fig. 4.40 are such that the skier travels at a constant velocity. From the photo, could you find the coefficient of kinetic friction between the snowy surface and the skis? If so, describe how this would be done.  yes;

  10. 4.4 Newton’s Third Law of Motion For every force (action), there is an equal and opposite force (reaction). Note that the action and reaction forces act on different objects. This image shows how a block exerts a downward force on a table; the table exerts an equal and opposite force on the block, called the normal force N.

  11. 4.4 Newton’s Third Law of Motion This figure illustrates the action-reaction forces for a person carrying a briefcase. Is there a reaction force in (b)? If so, what is it?

  12. 52. In an Olympic figure-skating event, a 65-kg male skater pushes a 45-kg female skater, causing her to accelerate at a rate of At what rate will the male skater accelerate? What is the direction of his acceleration?   opposite to her’s

More Related