lecture8x

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7/31/2019 lecture8x

2/24Engr221 Chapter 8

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 the

magnitude and direction of the

resulting friction force?

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?

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Characteristics of Dry Friction

Friction is defined as a force of resistanceacting 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.

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 with

P, as shown in the right figure above.

Characteristics of Dry Friction - continued

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The maximum friction force is attained just before the

block begins to move (a situation called impending

motion). The value of the force is found using Fs = s N,where s is called the coefficient of static friction. The valueofs 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 .

Characteristics of Dry Friction - continued

Determining s

Experimentally

A block with weight Wis 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 = (Wsin s ) / (W cos s ) = tan sThis simple experiment allows us to find

the S between two materials in contact.

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Procedure for Analysis

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 that

F = SN unless the impending motion condition is given.

3. Apply the equations of equilibrium and appropriate

frictional equations to solve for the unknowns.

Impending Tipping vs. Slipping

For a given weightand height,

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 E-of-E.

Hence, we have to make an

assumption to give us another

equation. Then we can solve

for the unknowns using the

three E-of-E. Finally, we need

to check if our assumption was

correct.

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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

Impending Tipping vs. Slipping - continued

Example A

Given: A uniform ladder weighs 20 lb.

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 E-of-E (and friction equations, if appropriate) to

solve for the unknowns

e) Check assumptions, if required

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There are four unknowns: NA, FA, NB, and P. Assume that

the ladder will tip first so NB = 0

+ FY = NA 20 = 0 ; NA = 20 lb+ MA = 20(3) P(4) = 0 ; P = 15 lb

+ FX = 15 FA = 0 ; FA = 15 lb

P20 lb

NB

4 ft

4 ft

3 ft 3 ftNA

FA

A FBD of the ladder

Example A - continued

Now check the assumption.

Fmax = s NA = 0.8 * 20 lb = 16 lb

Is FA = 15 lb Fmax = 16 lb? Yes, hence our assumptionof tipping is correct.

P20 lb

NB

4 ft

4 ft

3 ft 3 ftNA

FA

A FBD of the ladder


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Example B

Given: Drum weight = 100 lb, s

=

0.5, a = 3 ft, b = 4 ft

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 necessaryd) Apply E-of-E (and friction equations as appropriate) to

solve for the unknowns

e) Check assumptions, as required

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

Example B - continued

X

3

4

51.5 ft 1.5 ft

100 lb

0

4 ft

F

A FBD of the drum

P

N

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

+ FY = N (3/5) P 100 = 0

These two equations give:

P = 100 lb and N = 160 lb

+ MO = (3/5) 100 (1.5) (4/5) 100 (4) + 160 (x) = 0

Check: x = 1.44 1.5 so OK!

Drum slips as assumed at P = 100 lb

X

34

5

1.5 ft 1.5 ft

100 lb

0

4 ft

F

A FBD of the drum

P

N


Questions

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) sNB) = sN

C) sN

D) = kN

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Question

A 100 lb box with wide base is pulled by a force P and s = 0.4Which force orientation requires the least force to begin sliding?

A) A

B) B

C) CD) Can not be determined

P(A)

P(B)

P(C)100 lb

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

the magnitude of the friction force

between this block and the surface?

A) 0 lb B) 1 lb

C) 2 lb D) 3 lb

Questions

2. The ladder AB is positioned 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 lb

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Example Problem

The refrigerator has a weight of 180 lb and rests on a tile floor for

which s = 0.25 Also, the man has a weight of 150 lb and thecoefficient of static friction between the floor and his shoes is s = 0.6If he pushes horizontally on the refrigerator, determine if he can move

it. If so, does the refrigerator tip or slip?

Tips: P = 67.5 lb

Slips: P = 45 lbs

Man slips at 90 lbs

F = 45 lbs

Yes, he can move it

Textbook Problem 8.53

The 50-lb board is placed across the channel and a 100-lb boy

attempts to walk across. If the coefficient of static friction atA andB

is s = 0.4, determine if he can make the crossing; and if not, how farwill he get fromA before the board slips?

NA = 60.3 lb

NB = 86.2 lb

d = 6.47 ft

No, the board will slip

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Example Problem

The 5-kg cylinder is suspended from two equal-length cords. Theend of each cord is attached to a ring of negligible mass, which

passes along a horizontal shaft. If the coefficient of static friction

between each ring and the shaft is s = 0.5, determine the greatestdistance dby which the rings can be separated and still support

the cylinder.

Textbook Problem 8.44 (HW)

The crate has a weight of 300 lb and a center of gravity at G. If the

coefficient of static friction between the crate and floor is s = 0.4,determine the smallest weight of the man so he can push the crate to the

left. The coefficient of static friction between his shoes and the floor is

s = 0.4. Assume the man exerts only a horizontal force on the crate.

Wman = 171 lb

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Textbook Problem 8.50 (HW)

Determine the angle at which P should act on the block so that the

magnitude ofP is as small as possible to begin pulling the block up

the incline. What is the corresponding value ofP? The block weighs

Wand the slope is known.

= tan-1

P = Wsin(+ )

Textbook Problem 8.58

The carpenter slowly pushes the uniform board horizontally over

the top of the saw horse. The board has a uniform weight of 3

lb/ft, and the saw horse has a weight of 15 lb and a center of

gravity at G. Determine if the saw horse will stay in position, slip,

or tip if the board is pushed forward when d = 14 ft. The

coefficients of static friction are shown in the figure.

NA = 60.3 lb

NB = 86.2 lb

d = 6.47 ftNo, the board will slip

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Understand the characteristics of dry friction

Draw a FBD including friction

Solve problems involving friction

Summary

Announcements

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Wedges and Belts

Todays Objectives

Determine the forces on a wedge

Determine the tensions in a belt

Class Activities

Applications

Analysis of a wedge

Analysis of a belt

Questions

Wedges are used to adjust the

elevation or provide stability for

heavy 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 on its own?

Under what physical conditions

will it come out?

Applications

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Belt drives are commonly

used for transmitting power

from one shaft to another.

How can we decide that the

belts will function properly,

i.e., without slipping or

breaking?


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, the

applied force P and the torqueM, related?


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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:

1) The friction forces are always in the

direction opposite to the motion, or

impending motion, of the wedge

2) The friction forces are along the contacting

surfaces

3) The normal forces are perpendicular to the

contacting surfaces

W

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

is drawn. 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

b) all other forces acting on the object should

be shownTo determine the unknowns, we must apply E-of-E, Fx = 0 and

Fy = 0, to the wedge and the object as well as the impendingmotion frictional equation, F = SN

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.

Analysis of a Wedge - continued

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If the object is to be lowered, then the

wedge needs 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 ofP 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

Analysis of a Wedge - continued

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.

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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

when using 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 fixed

curved surface with the total angle of

contact equal to radians.

Belt Analysis - continued

Given: The load weighs 100 lb and the

S between surfacesACandBDis 0.3 Smooth rollers are placed

between wedgesA andB.

Assume the rollers and the

wedges have negligible weights.

Find: The force P needed to lift the

load.

Plan:

1. Draw a FBD of wedgeA. Why doA first?

2. Draw a FBD of wedgeB.

3. Apply the E-of-E to wedgeB. Why doB first?

4. Apply the E-of-E to wedgeA.

Example A

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The FBDs of wedges A and B are shown inthe figures. Applying the E-of-E to wedge

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 lb and N3 = 18.6 lb

Applying the E-of-E to the wedge A, we get

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

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

10N2

AP

N1

F1= 0.3N1

N2 10

B

F3= 0.3N3

N3

100 lb


Given: BlocksA andB weigh 50 lb and 30 lb, respectively.

Find: The smallest weight of cylinderD which will causethe loss of static equilibrium.

Example B

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Plan:

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

b) blockB slides over blockA

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

3. For each case, apply the E-of-E 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 blockD.


Case a: blocks slide together.

+ FY = N 80 = 0

N = 80 lb

+ FX = 0.4 (80) P = 0

P = 32 lb

PB

A

N

F=0.4 N

30 lb

50 lb


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+ Fy = N cos 20 + 0.6 N sin 20 30 = 0 N = 6.20 lb

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

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 lb

Block D weight of 12.7 lb will cause block B to slide over block A.

20

30 lb

0.6 NP

NCase b:


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

Questions

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2. The boy (hanging) in the picture weighs

100 lb and the woman weighs 150 lb. The

coefficient of static friction between her

shoes 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 frictionforce on object B at the contact point

between A and B.

A) B)

C) D)

Questions

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

Questions

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Determine the forces on a wedge

Determine the tensions in a belt

Summary

lecture8x

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