Ch 5 Friction

8/10/2019 Ch 5 Friction

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FMB 11103

Statics and Dynamics

Friction


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Objectivea) Understand the characteristics of dry friction.

b) Draw a FBD including friction.

c) Solve problems involving friction.


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Types of Friction

Dry FrictionWhen the unlubricated surfaces of two solids are incontact under a condition of sliding or tendency to

slide.

Fluid FrictionWhen adjacent layers in a fluid are moving at differentvelocities.

Internal FrictionWhen solid materials are subjected to cyclical loading.


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APPLICATIONS

In designing a brake system for abicycle, car, or any other vehicle, it isimportant to understand the frictionalforces involved.

For an applied force on the brakepads, how can we determine themagnitude and direction of theresulting friction force?


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APPLICATION

Consider pushing a box as shownhere.

How can you determine if it willslide, tilt, or stay in staticequilibrium?

What physical factors affectthe 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 retardsslipping of the body relative to a secondbody.

Experiments show that frictional forces

act tangent (parallel) to the contactingsurface in a direction opposing therelative motion or tendency for motion.


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For a body to be in equilibrium, the following must be true:F = P, N = W, and W.x = P.h.


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To study the characteristics of the friction force F, let us assumethat 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 shownin the right figure above.


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The maximum friction force is attained just before the blockbegins to move (a situation that is called impending motion).

The value of the force is found using Fs = s N, where s iscalled the coefficient of static friction. The value of s dependson the materials in contact.

Once the block begins to move, the frictional force typicallydrops and is given by Fk = k N. The value of k (coefficient

of kinetic friction) is less than s .


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Determining sExperimentallyA block with weight W is placed on aninclined plane. The plane is slowly tilteduntil the block just begins to slip.

The inclination, s, is noted. Analysis ofthe block just before it begins to movegives (using Fs = s N):

+ Fy = N W coss = 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 findthe S between two materials in contact.


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Problems Involving Dry Friction

Steps for solving equilibrium problems involving dry friction:

1. Draw the necessary free body diagrams. Make sure thatyou 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 Nunless the impending motion condition is given.

3. Apply the equations of equilibrium and appropriatefrictional equations to solve for the unknowns.


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Reading QuizA friction force always acts _____ to the contact surface.

A) normal B) at 45

C) parallel D) at the angle of static friction

Considering the forces acting on the block shown in the figure below. Ifthe pulling force (P) equals the friction force (F

friction), which of the

following is true?

a) The block will remain stationary if the friction force is static.b) The block will accelerate if the friction force is kinetic.c) The block will move at constant velocity if the friction force is static.d) The block will begin to move if the friction force is static.


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Impending Tipping versus

SlippingFor a given W and h, how can wedetermine if the block will slidefirst or tip first?

In this case, we have fourunknowns (F, N, x, and P) andonly three E-of-E.


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Impending Tipping versus Slipping

Hence, we have to make an assumption to give us anotherequation. Then we can solve for the unknowns using thethree E-of-E. Finally, we need to check if our assumptionwas correct.


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EXAMPLEGiven: A uniform ladder weighs 20 N.The vertical wall is smooth (nofriction). The floor is rough ands = 0.8.

Find: The minimum force P needed tomove ( 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 solvefor the unknowns.

e) Check assumptions, if required.


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EXAMPLE

There are four unknowns: NA, FA, NB, and P. Let us assume that theladder will tip first. Hence, NB = 0

FY = NA 20 = 0 ; so NA = 20 N

+ MA = 20 ( 3 ) P( 4 ) = 0 ; so P = 15 N

+ FX = 15 FA = 0 ; so FA = 15 N

P20 N

NB

4 m

4 m

3 m 3 mNA

FA

A FBD of the ladder


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EXAMPLE

Now check the assumption.

Fmax = s NA = 0.8 * 20 N = 16 NIs FA = 15 N Fmax = 16 N? Yes, hence our assumption of tipping iscorrect.

Remember that if you assume that the ladder slips first (F = s NA),and find NB and P, then your P would be 17N and the situation at which

P= 15N happens before P=17, so the ladder tips first.

P20 N

NB

4 m

4 m

3 m 3 mNA

FA

A FBD of the ladder


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Concept Quiz

1. A 100 N box with wide base is pulled by a force P and s = 0.4.Which force orientation requires the least force to beginsliding?

A) P(A) B) P(B)

C) P(C) D) Can not be determined

P(A)

P(B)

P(C)100 N

2. A ladder is positioned as shown. Indicate thedirection of the friction force on the ladder at B.

A) B)

C) D) A

B


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Example 2

Given: Drum weight = 100 N,s = 0.5 , a = 3 m and b = 4 m.

Find: The smallest magnitude of Pthat 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 equation as appropriate) to solve forthe unknowns.

e) Check assumptions, as required.


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Example 2

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

First, lets assume the drum slips. Then the frictionequation is F = s N = 0.5 N.

X

3

4

51.5 m 1.5 m

100 N

0

4 m

F

A FBD of the drum:

P


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Example

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

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

These two equations give:P = 100 N and N = 160 N

+ 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 N

X

3

4

5 1.5 m 1.5 m

100 N

0

4 m

F

A FBD of the drum:

P


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Attention Quiz

1. A 10 N block is in equilibrium. What is themagnitude of the friction force betweenthis block and the surface?

A) 0 N B) 1 N

C) 2 N D) 3 N

S = 0.3

2 N

2. The ladder AB is postioned as shown. What is thedirection of the friction force on the ladder at B.

A) B)

C) D)

B

Writing the equation of equilibrium along x, you will notice that friction forceis equal to the force applied or 2lb. The maximum friction force is (0.3)N =0.3(W) = 3lb which would exist when the block is on the verge of slipping.

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Ch 5 Friction