Chapter 4B. Friction and Equilibrium

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# Chapter 4B. Friction and Equilibrium - PowerPoint PPT Presentation

Chapter 4B. Friction and Equilibrium. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University. © 2007.

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### Chapter 4B. Friction and Equilibrium

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

Equilibrium: Until motion begins, all forces on the mower are balanced. Friction in wheel bearings and on the ground oppose the lateral motion.

• Define and calculate the coefficients of kinetic and static friction, and give the relationship of friction to the normal force.
• Apply the concepts of static and kinetic friction to problems involving constant motion or impending motion.

P

Friction Forces

When two surfaces are in contact, friction forces oppose relative motion or impending motion.

Friction forces are parallel to the surfaces in contact and oppose motion or impending motion.

Static Friction:No relative motion.

Kinetic Friction:Relative motion.

n

12 N

n

8 N

4 N

4 N

6 N

2 N

Friction and the Normal Force

n

The force required to overcome static or kinetic friction is proportional to the normal force, n.

fs = msn

fk = mkn

4N

4N

Friction forces are independent of area.

If the total mass pulled is constant, the same force (4 N) is required to overcome friction even with twice the area of contact.

For this to be true, it is essential that ALL other variables be rigidly controlled.

4N

4 N

Friction forces are independent of temperature, provided no chemical or structural variations occur.

Heat can sometimes cause surfaces to become deformed or sticky. In such cases, temperature can be a factor.

2 N

2 N

Friction forces are independent of speed.

5 m/s

20 m/s

The force of kinetic friction is the same at 5 m/s as it is for 20 m/s. Again, we must assume that there are no chemical or mechanical changes due to speed.

When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value.

n

P

fs

W

In this module, when we use the following equation, we refer only to the maximum value of static friction and simply write:

fs = msn

The Static Friction Force

For motion that is impending and for motion at constant speed, the resultant force is zero and SF = 0. (Equilibrium)

fs

P

fk

P

Rest

Constant Speed

P – fs = 0

P – fk = 0

Constant or Impending Motion

Here the weight and normal forces are balanced and do not affect motion.

When P is greater than the maximum fs the resultant force produces acceleration.

a

fk

P

Constant Speed

Friction and Acceleration

This case will be discussed in a later chapter.

fk = mkn

Note that the kinetic friction force remains constant even as the velocity increases.

n

P

fs

+

W

EXAMPLE 1:If mk = 0.3 and ms= 0.5, what horizontal pull P is required to just start a 250-N block moving?

1. Draw sketch and free-body diagram as shown.

2. List givens and label what is to be found:

mk = 0.3; ms = 0.5; W = 250 N

Find: P = ? to just start

3. Recognize for impending motion:P – fs = 0

n

250 N

P

fs

+

EXAMPLE 1(Cont.):ms = 0.5, W = 250 N. Find P to overcome fs (max). Static friction applies.

For this case: P – fs = 0

4. To find P we need to know fs , which is:

fs = msn

n = ?

n – W = 0

5. To findn:

SFy = 0

W = 250 N

n = 250 N

(Continued)

n

250 N

P

fs

ms = 0.5

+

EXAMPLE 1(Cont.):ms = 0.5, W = 250 N. Find P to overcome fs (max). Now we know n= 250 N.

6. Next we findfsfrom:

fs = msn = 0.5 (250 N)

7. For this case: P – fs = 0

P = fs = 0.5 (250 N)

P = 125 N

This force (125 N) is needed to just start motion. Next we consider P needed for constant speed.

SFy = may = 0

n - W = 0

n = W

n

P

fk

+

mg

EXAMPLE 1(Cont.):If mk = 0.3 and ms = 0.5, what horizontal pull P is required to move with constant speed? (Overcoming kinetic friction)

mk = 0.3

Now: fk = mkn = mkW

SFx = 0; P - fk = 0

P = fk = mkW

P = 75.0 N

P = (0.3)(250 N)

The normal force is NOT always equal to the weight. The following are examples:

Here the normal force is less than weight due to upward component of P.

P

n

300

m

W

P

Here the normal force is equal to only the compo- nent of weight perpendi- cular to the plane.

n

W

The Normal Force and Weight
Review of Free-body Diagrams:

For Friction Problems:

• Read problem; draw and label sketch.
• Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.
• Dot in rectangles and label x and y compo-nents opposite and adjacent to angles.
• Label all components; choose positive direction.
For Friction in Equilibrium:
• Read, draw and label problem.
• Draw free-body diagram for each body.
• Choose x or y-axis along motion or impending motion and choose direction of motion as positive.
• Identify the normal force and write one of following:

fs = msn or fk = mkn

• For equilibrium, we write for each axis:

SFx = 0 SFy = 0

• Solve for unknown quantities.

P = ?

n

400

fk

P

Py

n

W

400

Px

fk

+

W

Example 2.A force of 60 N drags a 300-N block by a rope at an angle of 400 above the horizontal surface. If uk = 0.2, what force P will produce constant speed?

1. Draw and label a sketch of the problem.

W = 300 N

2. Draw free-body diagram.

m

P sin400

Py

The force P is to be replaced by its com- ponents Px and Py.

P cos 400

P sin400

P

n

400

fk

P cos 400

+

mg

Example 2 (Cont.).P = ?; W = 300 N; uk = 0.2.

3. Find components of P:

Px = P cos 400 = 0.766P

Py = P sin 400 = 0.643P

Px = 0.766P; Py = 0.643P

Note: Vertical forces are balanced, and for constant speed, horizontal forces are balanced.

300 N

0.643P

P

n

400

fk

0.766P

+

Example 2 (Cont.).P = ?; W = 300 N; uk = 0.2.

Px = 0.766PPy = 0.643P

4. Apply Equilibrium con- ditions to vertical axis.

SFy = 0

[Py and nare up (+)]

n + 0.643P – 300 N= 0

n = 300 N – 0.643P;

Solve for n in terms of P

n = 300 N – 0.643P

300 N

0.643P

P

n

400

fk

0.766P

+

Example 2 (Cont.).P = ?; W = 300 N; uk = 0.2.

n = 300 N – 0.643P

5. Apply SFx = 0 to con- stant horizontal motion.

SFx = 0.766P – fk = 0

fk = mk n =(0.2)(300 N - 0.643P)

fk = (0.2)(300 N - 0.643P) = 60 N – 0.129P

0.766P – fk = 0;

0.766P – (60 N – 0.129P) =0

0.643P

0.766P – (60 N – 0.129P )=0

fk

0.766P

300 N

0.766P + 0.129P = 60 N

P

n

400

+

P = 67.0 N

Example 2 (Cont.).P = ?; W = 300 N; uk = 0.2.

6.Solve for unknown P.

0.766P – 60 N + 0.129P =0

If P = 67 N, the block will be dragged at a constant speed.

0.766P + 0.129P = 60 N

0.895P = 60 N

P = 67.0 N

600

n

x

y

fk

600

Example 3:What push P up the incline is needed to move a 230-N block up the incline at constant speed if mk = 0.3?

P

Step 1: Draw free-body including forces, angles and components.

W =230 N

P

Step 2: SFy = 0

W cos 600

n – W cos 600 = 0

W sin 600

n = (230 N)cos 600

230 N

n = 115 N

n = 115 N

W = 230 N

600

n

x

y

P

fk

W cos 600

W sin 600

600

W

Step 3. Apply SFx= 0

P - fk - W sin 600= 0

fk = mkn = 0.2(115 N)

fk = 23N, P = ?

P - 23N - (230 N)sin 600 = 0

P = 222 N

P - 23N - 199 N= 0

The maximum force of static friction is the force required to just start motion.

n

P

fs

W

Equilibrium exists at that instant:

The force of kinetic friction is that force required to maintain constant motion.

n

P

W

fk

Summary: Important Points (Cont.)
• Equilibrium exists if speed is constant, but fkdoes not get larger as the speed is increased.

Choose an x or y-axis along the direction of motion or impending motion.

mk = 0.3

n

P

fk

+

In this figure, we have:

W

Summary: Important Points (Cont.)

The SFwill be zero along the x-axis and along the y-axis.

P

n

300

m

W

P

n

W

Summary: Important Points (Cont.)

It is necessary to draw the free-body diagram and sum forces to solve for the correct n value.

Summary

Static Friction:No relative motion.

Kinetic Friction:Relative motion.

fs≤ msn

fk = mkn

Procedure for solution of equilibrium problems is the same for each case: