6 friction

7/28/2019 6 Friction

1/19

FRICTIONCHAPTER

Suriati Akmal

6 BMFG 1823Static and Strengthof Material

STATIC & STRENGTH OF MATERIALS

FRICTION (Sections 8.1 - 8.2)Todays Objective:

Students will be able to:

a) Understand the characteristics of

dry friction.

b) Draw a FBD including friction.

c) Solve problems involving friction.

In-Class Activities:

Check homework, if any

Reading quiz

Applications

Characteristics of dry friction

Problems involving dry friction

Concept quiz

Group problem solving

Attention quiz


2/19


READING QUIZ

1. A friction force always acts _____ to the contact surface.

A) normal B) at 45

C) parallel D) at the angle of static friction

2. If a block is stationary, then the friction force acting on it is

________ .

A) sN B) = sN

C) sN

D) = kN


APPLICATIONS

In designing a brake system for a

bicycle, car, or any other vehicle, it is

important to understand the frictional

forces involved.

For an applied force on the brake

pads, how can we determine themagnitude and direction of the

resulting friction force?


3/19


APPLICATIONS (continued)

Consider pushing a box as

shown here. How can you

determine if it will slide, tilt, or

stay in static equilibrium?

What physical factors affect

the answer to this question?


CHARACTERISTICS OF DRY FRICTION (Section 8.1)

Friction is defined as a force of resistance

acting on a body which prevents or

retards slipping of the body relative to a

second body.

Experiments show that frictional forces

act tangent (parallel) to the contacting

surface in a direction opposing the

relative motion or tendency for motion.

For the body shown in the figure to be in

equilibrium, the following must be true:

F = P, N = W, and Wx = Ph.


4/19


CHARACTERISTICS OF FRICTION (continued)

To study the characteristics of the friction force F, let us assume

that tipping does not occur (i.e., h is small or a is large).

Then we gradually increase the magnitude of the force P.

Typically, experiments show that the friction force F varies withP, as shown in the left figure above.


FRICTION CHARACERISTICS (continued)

The maximum friction force is attained just before the block

begins to move (a situation that is called impending

motion). The value of the force is found using Fs

= s

N,where s is called the coefficient of static friction. Thevalue ofs depends on the materials in contact.

Once the block begins to move, the frictional force

typically drops and is given by Fk= kN. The value ofk (coefficient of kinetic friction) is less than s .


5/19


DETERMING s

EXPERIMENTALLY

A block with weight w is placed on an

inclined plane. The plane is slowly

tilted until the block just begins to slip.

The inclination, s, is noted. Analysis ofthe block just before it begins to move

gives (using Fs = s N):+ Fy = N W cos s = 0

+ FX = S N W sin s = 0

Using these two equations, we get s =(W sin

s

) / (W cos s

) = tan sThis simple experiment allows us to find

the S between two materials in contact.


PROCEDURE FOR ANALYSIS (Section 8.2)

Steps for solving equilibrium problems involving dry friction:

1. Draw the necessary free body diagrams. Make sure that

you show the friction force in the correct direction (it

always opposes the motion or impending motion).

2. Determine the number of unknowns. Do not assume

F = S N unless the impending motion condition is given.

3. Apply the equations of equilibrium and appropriate

frictional equations to solve for the unknowns.


6/19


IMPENDING TIPPING versus SLIPPING

For a given W and h, how can

we determine if the block will

slide first or tip first? In this

case, we have four unknowns

(F, N, x, and P) and only three

EofE.

Hence, we have to make an

assumption to give us another

equation. Then we can solve for

the unknowns using the three

EofE. Finally, we need to check

if our assumption was correct.


IMPENDING TIPPING versus SLIPPING (continued)

Assume: Slipping occurs

Known: F = s N

Solve: x, P, and N

Check: 0 x b/2

Or

Assume: Tipping occurs

Known: x = b/2

Solve: P, N, and F

Check: F s N


7/19


EXAMPLE

Given: A uniform ladder weighs 100 N. The

vertical wall is smooth (no friction).The floor is rough and s = 0.8.

Find: The minimum force P needed to

move ( tip or slide) the ladder.

Plan:

a) Draw a FBD.

b) Determine the unknowns.

c) Make any necessary friction assumptions.

d) Apply EofE (and friction equations, if appropriate ) to solve for theunknowns.

e) Check assumptions, if required.


EXAMPLE (continued)

There are four unknowns: NA, FA, NB, and P. Let usassume that the ladder will tip first. Hence, NB = 0

+ FY = NA 100 = 0 ; so NA = 100 N

+ MA = 100 ( 0.9 ) P( 1.2 ) = 0 ; so P = 75 N

+ FX = 75 FA = 0 ; so FA = 75 N

1.2 m

P100 N

NB

1.2 m

0.9 m 0.9 mNA

FA

A FBD of the ladder


8/19


EXAMPLE (continued)

Now check the assumption.

Fmax = s NA = 0.8 * 100 N = 80 N

Is FA = 75 N Fmax = 80 N? Yes, hence our assumption oftipping is correct.

P100 N

NB

1.2 m

0.9 m 0.9 mNA

FA

A FBD of the ladder


CONCEPT QUIZ

1. A 100 N box with wide base is pulled by a

force P and s = 0.4. Which force orientationrequires the least force to begin sliding?

A) A B) B

C) C D) Can not be determined

2. A ladder is positioned as shown. Please indicate

the direction of the friction force on the ladder atB.

A) B)

C) D)

P(A)

P(B)

P(C)100 N

A

B


9/19


GROUP PROBLEM SOLVING

Given: Drum weight = 500 N,

s = 0.5, a = 0.75 m and b = 1m.Find: The smallest magnitude

of P that will cause impending

motion (tipping or slipping) of

the drum.

Plan:

a) Draw a FBD of the drum.

b) Determine the unknowns.

c) Make friction assumptions, as necessary.

d) Apply EofE (and friction eqn. as appropriate) to solve for theunknowns.

e) Check assumptions, as required.


GROUP PROBLEM SOLVING (continued)

There are four unknowns: P, N, F and x.

First, lets assume the drum slips. Then the friction

equation is F = s N = 0.5 N.

A FBD of the drum:

X

3

4

50.375 m 0.375 m

500 N

0

1 m

F

P

N


10/19


+ FX = (4 / 5) P 0.5 N = 0

+ FY = N (3 / 5) P 500 = 0

These two equations give:


P = 500 N and N = 800 N

+

MO = (3 /5) 500 (0.375) (4 / 5) 500 (1) + 800 (x) = 0Check: x = 0.359 0.375 so OK!

Drum slips as assumed at P = 500 N

A FBD of the drum:

X

3

4

50.375 m 0.375 m

500 N

0

1 m

F

P

N


1. A 10 N block is in equilibrium. What is

the magnitude of the friction force

between this block and the surface?

A) 0 N B) 1 N

C) 2 N D) 3 N

ATTENTION QUIZ

2. The ladder AB is postioned as shown. What is the

direction of the friction force on the ladder at B.

A) B)

C) D) A

B

S = 0.3

2 N


11/19


FRICTION IN PRACTICALENGINEERING


READING QUIZ

1. A wedge allows a ______ force P to lift

a _________ weight W.

A) (large, large) B) (small, small)

C) (small, large) D) (large, small)

2. Considering friction forces and the

indicated motion of the belt, how are belt

tensions T1 and T2 related?

A) T1 > T2 B) T1 = T2

C) T1 < T2 D) T1 = T2 e

W


12/19


APPLICATIONS

Wedges are used to adjust the

elevation or provide stability forheavy objects such as this large

steel vessel.

How can we determine the

force required to pull the

wedge out?

When there are no applied

forces on the wedge, will it

stay in place (i.e., be self-

locking) or will it come out onits own? Under what physical

conditions will it come out?


APPLICATIONS (continued)

In the design of a band brake, it

is essential to analyze the

frictional forces acting on the

band (which acts like a belt).

How can we determine the

tensions in the cable pulling on

the band?

How are these tensions, theapplied force P and the torque M,

related?


13/19


ANALYSIS OF A WEDGE

A wedge is a simple machine in which a

small force P is used to lift a large weight W.

To determine the force required to push the

wedge in or out, it is necessary to draw FBDs

of the wedge and the object on top of it.

It is easier to start with a FBD of the wedge

since you know the direction of its motion.

Note that:

a) the friction forces are always in the

direction opposite to the motion, or impending

motion, of the wedge;

b) the friction forces are along the contacting

surfaces; and,

c) the normal forces are perpendicular to the

contacting surfaces.

W


WEDGE ANALYSIS (continued)

Next, a FBD of the object on top of the wedge

is drawn. Please note that:

a) at the contacting surfaces between the

wedge and the object the forces are equal in

magnitude and opposite in direction to those

on the wedge; and, b) all other forces acting on

the object should be shown.

To determine the unknowns, we must apply EofE, Fx = 0 and

Fy = 0, to the wedge and the object as well as the impendingmotion frictional equation, F = S N.

Now of the two FBDs, which one should we start analyzing first?

We should start analyzing the FBD in which the number of

unknowns are less than or equal to the number of equations.


14/19


WEDGE ANALYSIS (continued)

If the object is to be lowered, then the wedgeneeds to be pulled out. If the value of the

force P needed to remove the wedge is

positive, then the wedge is self-locking, i.e.,

it will not come out on its own.

However, if the value of P is negative, or

zero, then the wedge will come out on its

own unless a force is applied to keep the

wedge in place. This can happen if the

coefficient of friction is small or the wedgeangle is large.

W


BELT ANALYSIS

Belts are used for transmitting power or

applying brakes. Friction forces play an

important role in determining the

various tensions in the belt. The belt

tension values are then used for

analyzing or designing a belt drive or a

brake system.


15/19


BELT ANALYSIS (continued)

Detailed analysis (please refer to your textbook) shows that

T2 = T1 e

where is the coefficient of static frictionbetween the belt and the surface. Be sure to use radians whenusing this formula!!

If the belt slips or is just about to slip,

then T2 must be larger than T1 and the

friction forces. Hence, T2 must be

greater than T1.

Consider a flat belt passing over a fixedcurved surface with the total angle of

contact equal to radians.


EXAMPLEGiven: The load weighs 100 N and the

S between surfaces AC and BDis 0.3. Smooth rollers are placed

between wedges A and B.

Assume the rollers and the

wedges have negligible weights.

Find: The force P needed to lift the load.

Plan:

1. Draw a FBD of wedge A. Why do A first?

2. Draw a FBD of wedge B.

3. Apply the EofE to wedge B. Why do B first?

4. Apply the EofE to wedge A.


16/19


EXAMPLE (continued)

The FBDs of wedges A and B are shown

in the figures. Applying the EofE towedge B, we get

+ FX = N2 sin 10 N3 = 0

+ FY = N2 cos 10 100 0.3 N3 = 0

Solving the above two equations, we get

N2 = 107.2 N and N3 = 18.6 N

Applying the EofE to the wedge A, we get

+ FY = N1 107.2 cos 10 = 0; N1 = 105.6 N

+ FX = P 107.2 sin 10 0.3 N1 = 0; P = 50.3 N

10N2

AP

N1F1= 0.3N1

N2 10

B

F3= 0.3N3

N3

100 N


CONCEPT QUIZ

2. The boy (hanging) in the picture weighs

500 N and the woman weighs 750 N. The

coefficient of static friction between hershoes and the ground is 0.6. The boy will

______ ?

A) be lifted up. B) slide down.

C) not be lifted up. D) not slide down.

1. Determine the direction of the friction

force on object B at the contact point

between A and B.

A) B)

C) D)


17/19


GROUP PROBLEM SOLVING

Given: Blocks A and B weigh 50 N and

30 N, respectively.

Find: The smallest weight of cylinder D

which will cause the loss of static

equilibrium.


Plan:

1. Consider two cases: a) both blocks slide together, and,

b) block B slides over the block A.

2. For each case, draw a FBD of the block(s).

3. For each case, apply the EofE to find the force needed to

cause sliding.

4. Choose the smaller P value from the two cases.

5. Use belt friction theory to find the weight of block D.



18/19


GROUP PROBLEM SOLUTION

Case a:

+ FY = N 80 = 0

N = 80 N

+ FX = 0.4 (80) P = 0

P = 32 N

PB

A

N

F=0.4 N

30 N

50 N


GROUP PROBLEM SOLUTION

+ Fy = N cos 20 + 0.6 N sin 20 30 = 0N = 26.20 N

+ Fx = P + 0.6 ( 26.2 ) cos 20 26.2 sin 20 = 0P = 5.812 N

Case b has the lowest P and will occur first. Next, using the

frictional force analysis of belt, we get

WD = P e = 5.812 e 0.5 ( 0.5 ) = 12.7 N

A Block D weighing 12.7 N will cause the block B to slide over

the block A.

20

30 N

0.6 NP

N

Case b:


19/19


ATTENTION QUIZ

1. When determining the force P needed to lift

the block of weight W, it is easier to draw a

FBD of ______ first.

A) the wedge B) the block

C) the horizontal ground D) the vertical wall

2. In the analysis of frictional forces on a flat belt, T2 = T1 e .

In this equation, equals ______ .

A) angle of contact in degrees B) angle of contact in radians

C) coefficient of static friction D) coefficient of kinetic friction

W


6 friction

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