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

2007

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

Objectives: After completing this module, you should be able to: • 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.

Friction Forces When two surfaces are in contact, friction forces oppose relative motion or impending motion. P Friction forces are parallel to the surfaces in contact and oppose motion or impending motion. Static Friction: No relative motion.

Kinetic Friction: Relative motion. motion

Friction and the Normal Force 12 N n 8N n 4N n2N 4N

6N

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

fk = kn

Friction forces are independent of area. 4N

4N

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.

Friction forces are independent of temperature, provided no chemical or structural variations occur. 4N

4N

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

Friction forces are independent of speed. 5 m/s

2 N

20 m/s

2 N

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.

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

n

P

fs

f s  s

n

W

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

Constant or Impending Motion For motion that is impending and for motion at constant speed, the resultant force is zero and F = 0. (Equilibrium) fs

P Rest

P – fs = 0

fk

P Constant Speed

P – fk = 0

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

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

a fk

P Constant Speed

This case will be discussed in a later chapter. fk = kn

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

EXAMPLE 1: If kk = 0.3 and ss = 0.5, what horizontal pull P is required to just start a 250-N block moving? 1. Draw sketch and freebody diagram as shown.

n fs

P +

W

2. List givens and label what is to be found: k = 0.3;  s = 0.5; W = 250 N

Find: P = ? to just start

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

EXAMPLE 1(Cont.): ss = 0.5, W = 250 N. Find P to overcome f s (max). Static friction applies.

n

P

fs

+

For this case: P – fs = 0 4. To find P we need to know fs , which is: f s = sn

250 N

5. To find n:

Fy = 0

W = 250 N

n=? n–W=0

n = 250 N

(Continued)

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

6. Next we find fs from: fs = sn = 0.5 (250 N)

n fs

P +

7. For this case: case P – fs = 0 P = fs = 0.5 (250 N)

250 N

s = 0.5

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

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

Fy = may = 0 k = 0.3

fk

n

n-W=0 P

n=W

Now: fk = k n = kW

+ mg P = (0.3)(250 N)

Fx = 0;

P - fk = 0

P = fk = k W P = 75.0 N

The Normal Force and Weight

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

n

P

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

300

m

W

P

n W 

Here the normal force is equal to only the component of weight perpendicular to the plane.

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 components opposite and adjacent to angles. • Label all components; choose positive direction.

For Friction in Equilibrium: • Read, draw and label problem. • Draw free-body body diagram diagram for for each each body. 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 = sn or f k = kn

• For equilibrium, we write for each axis: Fx = 00

Fy = 0

• Solve for unknown quantities.

Example 2. A force of 60 N drags0a 300--N block by a rope at an angle of 40 above the horizontal surface. If uk = 0.2, what force P will produce constant speed? W = 300 N

fk

n

P=? 400

m

W

The force P is to be replaced by its components Px and Py.

1. Draw and label a sketch of the problem. 2. Draw free-body diagram. P sin 400 Py

n

P 400

fk W

Py

Pxx

P cos 400

+

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

3. Find components of P:

P sin 400

n

400

Px = P cos 400 = 0.766P Py = P sin

400

= 0.643P

Px = 0.766 P; Py = 0.643P

P

fk

P cos 4000

mg

+

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

F

x

0

F

y

0

Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2. Px = 0.766P 0.643P Py = 0.643P P

4. Apply Equilibrium conditions to vertical axis. Fy = 0

n

400

fk 300 N

0.766P

+

n + 0.643P – 300 N= 0 [Py and n are up (++)] n = 300 N – 0.643P; Solve for n in terms of P n = 300 N – 0.643P

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

n = 300 N – 0.643P

0.643P

n

5. Apply Fx = 0 to con-

stant horizontal motion.

Fx = 0.766 P – fk = 0

P 400

fk 300 N

0.766P

+

f k = k n = (0.2)(300 N - 0.643P) fk = (0.2)(300 N - 0.643 P) = 60 N – 0.129P 0.766P – fk = 0; 0.766P – (60 N – 0.129 P) = 0

Example 2 (Cont.). P = ?; W = 300 N; u k = 0.2. 0.643P P 0.766P – (60 N – 0.129 P )=0

n

400

fk

0.766P

300 N

+

6. Solve for unknown P. 0.766P – 60 N + 0.129P =0

0.766P + 0.129P = 60 N 0.766P + 0.129P = 60 N 0.895P = 60 N

P = 67.0 N

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

P = 67.0 N

Example 3: What push P up the incline is needed to move a 230-N block up the incline at constant speed if k = 0.3? P Step 1: Draw free-body including forces, angles and components.

n

y W sin

600

fk

P

230 N

600

x

W cos 600 600

W =230 N

Step 2:

Fy = 0

n – W cos 600 = 0 n = (230 N) cos 600 n = 115 N

Example 3 (Cont.): Find P to give move up the incline (W = 230 N).

n

y W sin 600

x

P

fk

W cos 600 600

W

600

n = 115 N

W = 230 N

Step 3. Apply Fx= 0 P - fk - W sin 600 = 0 fk = k n = 0.2(115 N) fk = 23 N, P = ?

P - 23 N - (230 (230 N)sin 600 = 0 P - 23 N - 199 N= 0

P = 222 N

Summary: Important Points to Consider When Solving Friction Problems. • The maximum force of static friction is the force required to just start motion.

n

P

fs

f s  s

n

W

Equilibrium exists at that instant:

Fx  0;

Fy  0

Summary: Important Points (Cont.) • The force of kinetic friction is that force required to maintain constant motion.

n fk

P

fk  k

n

W

• Equilibrium exists if speed is constant, but f k does not get larger as the speed is increased.

Fx  0;

Fy  0

Summary: Important Points (Cont.) • Choose an x or y-axis along the direction of motion or impending motion.

k = 0.3

fk

n

P +

W

The F will be zero along the x -axis and along the y -axis. In this figure, we have:

Fx  0;

Fy  0

Summary: Important Points (Cont.) • Remember the normal force n is not always equal to the weight of an object.

n

P It is necessary to draw the free-body diagram and sum forces to solve

300

m

W

P

n W 

for the correct

 Fx  0;

n value.

Fy  0

Summary Static Friction: No relative motion.

Kinetic Friction: Relative motion.

fs ≤ sn

fk = kn

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

Fx  0 Fy  0

CONCLUSION: Chapter 4B Friction and Equilibrium...