Lecture 8 - Faculty Server Contactfaculty.uml.edu/Andriy_Danylov/Teaching/documents/LECTURE8.pdf · 95.141, Fall 2013, Lecture 8 . A new contact force…friction . Friction is always

Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 8

Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

Lecture Capture:

http://echo360.uml.edu/danylov2013/physics1fall.html

Lecture 8

Chapter 5

Friction force

09.30.2013 Physics I

http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

http://echo360.uml.edu/danylov2013/physics1fall.html


Exam I.

Wed Oct. 2; 9 - 9.50 am; Olney 150

Have your student ID Remember your recitation section Calculators are allowed A sheet with formulae will be provided

Review Session Mon 6-8:30 pm, Ball 210


Chapter 5. Sections 5.1 Friction More Newton’s Laws

Outline


A new contact force…friction

Friction is always present when two solid surfaces slide along each other.

The microscopic details are not yet fully understood.

Presenter

Presentation Notes

Figure 5-1. Caption: An object moving to the right on a table or floor. The two surfaces in contact are rough, at least on a microscopic scale.


Friction

m

mg

m

FN FA

mg

FN

mg

m

FN

FA

Ffr Ffr no motion

static no motion

static

Static. Limit. Motion is about to start

0=− frA FF

Afr FF =

0 ANSfr FFF == µ

frF

AF

NS Fµ

45

m

FN

FA

Ffr Sliding.

Friction (kinetic) is constant

NKfr FF µ=

Afr FF =

Kinetic friction

NKfr FF µ=Static friction

frictionstaticoftcoefficienS −µ

frictionkineticoftcoefficienK −µ

NS Fµ<

NSfr FF µ≤ If a horizontal force is applied on an object ,

the object does not move. It means a second force, Static Friction, must be

opposing the applied force.


Static friction

Static friction applies when two surfaces are at rest with respect to each other (such as a book sitting on a table).

The static frictional force is as big as it needs to be to prevent slipping, up to a maximum value.

Usually it is easier to keep an object sliding than it is to get it started.

NSfr FF µ≤


Kinetic friction

Sliding friction is called kinetic friction.

Approximation of the frictional force:

FN is the normal force

μk is the coefficient of kinetic friction, which is different for each pair of surfaces.

NKfr FF µ=


Friction

Friction is the contact force parallel to the surfaces Force opposite to a direction of motion. Magnitude of friction depends on:

– Properties (roughness) of the two surfaces – Normal Force between the two surfaces

Presenter

Presentation Notes

Friction increases with applied parallel force till object starts moving, then it adjusts to a constant value A microscopic effect: roughness of surfaces sliding against one another provides


Coefficients of Friction

Note that, in general, μs > μk.


Static and Kinetic Friction


Example A 10-kg box moves on a horizontal surface. The system has coefficients of friction μK=0.5. [Use g=10 m/s2] A horizontal force of 60N is applied. What is the acceleration?

x

y

Fg=mg

m Ffr

FN

=μKFN mgFN =

NKfr FF µ= mgKµ=

maFFF frAx =−+=∑

FA

mamgF KA =− µm

mgFa KA µ−= 2

2 110

10105.060s

msm

kgkgN

=⋅⋅−

=

The static friction force has a maximum of µsN = 50 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction.

ConcepTest 1. Will It move? 1) moves to the left

2) moves to the right

3) moves up

4) moves down

5) the box does not move

A box of weight 100 N is at rest on a floor where µs = 0.5. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

T=30N m

Follow-up: What happens if the tension is 55 N?

NSfr FF µ= forceNmgS max501005.0 =⋅== µ


mg

Fn

θ θ

How to measure μS?

Given: m, μs, with friction

Find: θ

A block lies on an inclined plane with μk. The angle of the incline is slowly increased. At what angle does the block start to move?

Ffr

NSfr FF µ= θµ cosmgS=

0

About to start moving

By measuring angle, when an object gets lose, we can find static friction coefficient

Presenter

Presentation Notes

Typically, equations simplify if one axis chosen to be along observed accelerated motion


Demonstrations

Two Interleaved Books

Simply lay the pages of two phone books on top of each other one by one before attempting

to pull them apart.

mg

Fn

θ θ

Ffr

(I)

(II)

(III)


What is going to happen with a car parked on a street with a slope of more than 45 degrees if µs=1?

sµθ =tanSince , then at 45 deg the car will be about to get loos and start sliding down.

1=sµ

Example: A car parked on a hill

Die Hard


Conceptual Example: A box against a wall

You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving vertically?

You have to press the box hard so that the force of static friction, which depends on the normal force, would be equal to box’s weight.

Presenter

Presentation Notes

Figure 5-4. Answer: Friction, of course! The normal force points out from the wall, and the frictional force points up.


Example 5-7: A ramp, a pulley, and two boxes.

Presenter

Presentation Notes

Figure 5-9. Caption: Example 5–7. Note choice of x and y axes. Solution: Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns. The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down). 0.78 m/s2


Example 5-7: A ramp, a pulley, and two boxes

Presenter

Presentation Notes


ConcepTest 2. Going Sledding

1

2

In case 1, the force F is pushing down (in addition to mg), so the normal force is larger. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. Recall that the frictional force is proportional to the normal force.

1) pushing her from behind

2) pulling her from the front

3) both are equivalent

4) it is impossible to move the sled

5) tell her to get out and walk

Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?


Example 5-7: A ramp, a pulley, and two boxes

If m1 moves up the incline plane. Friction force will change its direction

Presenter

Presentation Notes



Similar: Example 5-5: Two boxes and a pulley. Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We wish to find the acceleration, a, of the system. As box B moves down, box A moves to the right.

Fg=mA g

Ffr

FN

FT

Fg=mB g

FT

Presenter

Presentation Notes

Figure 5-7. Solution: For box A, the normal force equals the weight; we therefore know the magnitude of the frictional force, but not of the tension in the cord. Box B has no horizontal forces; the vertical forces on it are its weight (which we know) and the tension in the cord (which we don’t). Solving the two free-body equations for a gives 1.4 m/s2 (and the tension as 17 N).


Similar: Example 5-6: The skier.

This skier is descending a slope, at constant speed. What can you say about the coefficient of kinetic friction?

Presenter

Presentation Notes

Figure 5-8. Caption: Example 5–6. A skier descending a slope; FG = mg is the force of gravity (weight) on the skier. Solution: Since the speed is constant, there is no net force in any direction. This allows us to find the normal force and then the frictional force; the coefficient of kinetic friction is 0.58.


Thank you See you on Wednesday

Exam I

Please!!! Remember your recitation section number!!!

Documents

Lecture 8 - Faculty Server Contactfaculty.uml.edu/Andriy_Danylov/Teaching/documents/LECTURE8.pdf · 95.141, Fall 2013, Lecture 8 . A new contact force…friction . Friction is always