announcements dry friction

Engr221 Chapter 8 1

Announcements

Dry Friction

Today’s Objectives• Understand the characteristics of dry friction

• Draw a FBD including friction

• Solve problems involving friction

Class Activities• Applications

• Characteristics of dry friction

• Problems involving dry friction

• Examples

Engr221 Chapter 8 2

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?

Engr221 Chapter 8 3

Characteristics of Dry Friction

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.

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

Engr221 Chapter 8 4

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 = µsN, where µs is called the coefficient of static friction. The value of µsdepends on the materials in contact.

Once the block begins to move, the frictional force typically drops and is given by Fk = µk N. The value ofµk

(coefficient of kinetic friction) is less than µs .

Characteristics of Dry Friction - continued

Determining µµµµs 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, θs, is noted. Analysis of the block just before it begins to move gives (using Fs = µs N):+ ∑ Fy = N – W cosθs = 0

+ ∑ FX = µS N – W sin θs = 0

Using these two equations, we get µs = (W sin θs ) / (W cos θs ) = tan θsThis simple experiment allows us to find the µS between two materials in contact.

Engr221 Chapter 8 5

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 = µSN 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 weight and 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.

Engr221 Chapter 8 6

Assume: Slipping occurs

Known: F = µs 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 ≤ µs 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 µs = 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

Engr221 Chapter 8 7

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

+↑ Σ FY = NA – 20 = 0 ; NA = 20 lb

+ Σ MA = 20(3) – P(4) = 0 ; P = 15 lb

+ → Σ 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 = µs NA = 0.8 * 20 lb = 16 lb

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

P20 lb

NB

4 ft

4 ft

3 ft 3 ftNA

FA

A FBD of the ladder


Engr221 Chapter 8 8

Example B

Given: Drum weight = 100 lb, µs = 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 necessary

d) 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, let’s assume the drum slips. Then the friction equation is F = µs N = 0.5 N

Example B - continued

X

34

51.5 ft 1.5 ft

100 lb

0

4 ft

F

A FBD of the drum

P

N

Engr221 Chapter 8 9

+ → ∑ FX = (4/5) P – 0.5 N = 0

+ ↑ ∑ FY = N – (3/5) P – 100 = 0

These two equations give:

P = 100 lband 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

51.5 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) ≤ µs N

B) = µs N

C) ≥ µsN

D) = µk N

Engr221 Chapter 8 10

Question

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

A) A

B) B

C) C

D) 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

µ S = 0.3

2 lb


Example Problem

The refrigerator has a weight of 180 lb and rests on a tile floor for which µs = 0.25 Also, the man has a weight of 150 lb and the coefficient of static friction between the floor and his shoes is µs = 0.6 If 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 at A and Bis µs = 0.4, determine if he can make the crossing; and if not, how far will he get from A before the board slips?

NA = 60.3 lb

NB = 86.2 lb

d = 6.47 ft

No, the board will slip


Example Problem

The 5-kg cylinder is suspended from two equal-length cords. The end 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 µs = 0.5, determine the greatest distance d by 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 µs = 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 µs = 0.4. Assume the man exerts only a horizontal force on the crate.

Wman= 171 lb


Textbook Problem 8.50 (HW)

Determine the angle φ at which P should act on the block so that the magnitude of P is as small as possible to begin pulling the block up the incline. What is the corresponding value of P? The block weighs W and the slope α is known.

φ= tan-1 µP = Wsin(φ + α)

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 ft

No, the board will slip


• Understand the characteristics of dry friction

• Draw a FBD including friction

• Solve problems involving friction

Summary

Announcements


Wedges and Belts

Today’s 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


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 torque M , related?



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 wedgesince 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 wedge2) The friction forces are along the contacting surfaces3) 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 surfacesbetween 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 shown

To determine the unknowns, we must apply E-of-E, ∑ Fx = 0 and ∑ Fy = 0, to the wedge and the object as well as the impending motion frictional equation, F = µSN

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


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 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 wedge angle θ 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 designinga belt drive or a brake system.


Detailed analysis (please refer to your textbook) shows that T2 = T1 e µµµµ ββββ where µ is the coefficient of static friction between the belt and the surface. Be sure to use radianswhen 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 ββββ radians.

Belt Analysis - continued

Given: The load weighs 100 lb and the µS 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 E-of-E to wedge B. Why do B first?

4. Apply the E-of-E to wedge A.

Example A


The FBDs of wedges A and B are shown in the figures. Applying the E-of-E to wedgeB, 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

10ºN2

AP

N1F1= 0.3N1

N2 10º

BF3= 0.3N3

N3

100 lb


Given: Blocks A and B weigh 50 lb and 30 lb, respectively.

Find: The smallest weight of cylinder D which will cause the loss of static equilibrium.

Example B


Plan:

1. Consider two cases: a) both blocks slide together, and b) block B slides over block A

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 block D.


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



→ + ∑ 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 µ β = 5.812 e0.5 ( 0.5 π ) = 12.7 lb

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

20º

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 eµ

W

Questions


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 friction force 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 µ β

In this equation, β 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


• Determine the forces on a wedge

• Determine the tensions in a belt

Summary

announcements dry friction

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