chapter 4 – part 2 force & motion with friction

13
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?

Upload: ralph

Post on 11-Jan-2016

40 views

Category:

Documents


3 download

DESCRIPTION

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?. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chapter 4  – Part 2 Force  &  motion with friction

Chapter 4 – Part 2Force & motion with friction

Sue Ramlo, PhDPart 2 for Panopto / Springboard

Including frictional forces… what is friction? How do we calculate it?

Page 2: Chapter 4  – Part 2 Force  &  motion with friction

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.

Page 3: Chapter 4  – Part 2 Force  &  motion with friction

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

Page 4: Chapter 4  – Part 2 Force  &  motion with friction
Page 5: Chapter 4  – Part 2 Force  &  motion with friction

4.6 FrictionThe coefficients of friction depend on both materials involved.

Page 6: Chapter 4  – Part 2 Force  &  motion with friction

Consider this system with friction…

20

Page 7: Chapter 4  – Part 2 Force  &  motion with friction

Re-draw and find Fnet from picture• In motion so Fk m1g sin θ – T + T - m2g - Fk = Fnet

• Here == > Fk = k mg cos θ

• Newton #2 - Fnet = ma Fnet = (m1 + m2) a

• Bring together: m1g sin θ - m2g - k mg cos θ = (m1 + m2) a

• Solve for a…

m1g cos 20°

FN

m1g sin 20°

12

m2gTT

+ direction

Fk

Page 8: Chapter 4  – Part 2 Force  &  motion with friction

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!)

6.5kg)20 cos N) (49 .0300( -(14.7N)20sin (49N)

= a

21

1k21

m + m)20 cos gm ( -gm20sin gm

= a

6.5kg)20 cos )m/s kg)(9.80 (5.0 ( -)m/s kg)(9.80 (1.520sin )m/s kg)(9.80 (5.0

= a2

k22

6.5kgN) (1.4 -(14.7N) (17N)

= a

Page 9: Chapter 4  – Part 2 Force  &  motion with friction

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

Page 10: Chapter 4  – Part 2 Force  &  motion with friction

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;

Page 11: Chapter 4  – Part 2 Force  &  motion with friction

4.4 Newton’s Third Law of MotionFor 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.

Page 12: Chapter 4  – Part 2 Force  &  motion with friction

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

Page 13: Chapter 4  – Part 2 Force  &  motion with friction

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