2- Handouts Lecture-38 to 40

COURSE:COURSE: CE 201 (STATICS)CE 201 (STATICS)

LECTURE NO.:LECTURE NO.: 38 to 4038 to 40

FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD

DEPARTMENT:DEPARTMENT: CIVIL ENGINEERINGCIVIL ENGINEERING

UNIVERSITY:UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA& MINERALS, DHAHRAN, SAUDI ARABIA

TEXT BOOK:TEXT BOOK: ENGINEERING MECHANICSENGINEERING MECHANICS--STATICS STATICS by R.C. HIBBELER, PRENTICE HALLby R.C. HIBBELER, PRENTICE HALL

LECTURE NO. 38 to 40LECTURE NO. 38 to 40PROBLEMS INVOLVING DRY FRICTIONPROBLEMS INVOLVING DRY FRICTION

Objectives:Objectives: To show how to analyze the equilibrium of rigid To show how to analyze the equilibrium of rigid

bodies subjected to the dry frictionbodies subjected to the dry friction

GENERAL APPROACH FOR ANALYZING RIGID BODIES GENERAL APPROACH FOR ANALYZING RIGID BODIES SUBJECTED TO DRY FRICTIONSUBJECTED TO DRY FRICTION

Following steps may be used to analyze rigid bodies subjected to dry friction:

1. Draw the free-body diagrams for the given rigidbody in the manner similar to that used in theprevious chapters showing the frictional force F at rough contact points in addition to usual reactiveforces

2. Apply equilibrium conditions to the free-body diagrams to form equations for determining thevalues of unknown by solving the equations ofequilibrium simultaneously in the manner similar tothat used in the previous chapters


3. However, following additional points should be noted for analyzing rigid bodies subjected to dry friction:

a.If the state of the body at a rough contact point ismentioned to be under impending, then at such contact point take s sF F N= =

b. If the state of the body at a rough contact point is notmentioned, then at such contact point consider F and Nas unknowns independently and determine their values by applying equilibrium conditions

The value of F obtained this way is to be considered asthe force required for keeping the body in equilibrium

The state of the body and actual value of the frictionalforce may be decided using the calculated values of Fand N in the following manner:


If sF F< , the body will be in the state of stable equilibrium and actual frictional force will betaken equal to the calculated value of F

If sF F= , the body will be in the state ofimpending and actual frictional force will betaken equal to the calculated value of F or

( )s sF N=

If sF F> , the body will be in the state motion and actual frictional force will be taken equal to

( )k kF N=

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1

Determine the value of the force Prequired to cause sliding of the 100 Nblock as shown below.

PW

34

PW = 100 N

34

s = 0.5

PW

34

PW = 100 N

34

s = 0.5


PW

34

PW = 100 N

34

s = 0.5

PW

34

PW = 100 N

34

s = 0.5

PW

34

N

PW

34

Fs = N= 0.5N N

PW

34

N

PW

34

Fs = sN= 0.5N N

5PW

34

N

PW

34

Fs = N= 0.5N N

PW

34

N

PW

34

Fs = sN= 0.5N N

PW

34

N

PW

34

Fs = N= 0.5N N

PW

34

N

PW

34

Fs = sN= 0.5N N

5

The force required to cause slidingof the block will be determinedassuming the block to be in thestate of impending as shown in the free-body diagram of the blockbelow:

Fx = 0 45 P 0.5 N = 0 N = 1.6 P (1) Fy = 0

N W 35 P = 0 N = 100 + 0.6P (2) Equating Eq. (1) and Eq. (2)

1.6 P = 100 + 0.6P P = 100 N Ans.


Determine the value of the forceP required to cause sliding of the100 N block as shown below.

P3

4

P

W = 100 N

34

s= 0.5

P3

4

P

W = 100 N

34

s= 0.5


P3

4

P

W = 100 N

34

s= 0.5

P3

4

P

W = 100 N

34

s= 0.5

The force required to cause slidingof the block will be determinedassuming the block to be in thestate of impending as shown in thefree-body diagram of the blockbelow:

PW

34

N

PW

34

NNFs = s N = 0.5N

PW

34

N

PW

34

NNFs = s N = 0.5N

5P

W

34

N

PW

34

NNFs = s N = 0.5N

PW

34

N

PW

34

NNFs = s N = 0.5N

5

Fx = 0 45 P 0.5 N = 0 N = 1.6 P (1) Fy = 0

N W + 35 P = 0 N = 100 0.6P (2) Equating Eq. (1) and Eq. (2)

1.6 P = 100 0.6P P = 45.45 N Ans.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3Block A having weight of 80 N is placed onbloc B having a weight of 120 N, as shown inthe figure below. Both blocks are in the stateof impending motion under the action of theapplied force P. If the applied force Pincreases state whether the block A will slide over block B or both blocks will slidetogether.

A

B

P s = 0.5s = 0.25

A

B

P s = 0.5s = 0.25


A

B

P s = 0.5s = 0.25

A

B

P s = 0.5s = 0.25

Case I: Let us assume that block A slides over B: P required for starting sliding of block A over block B can be determined by applyingequilibrium conditions to theFBD of the block A, as follows:

P

A

NF=Fs

P

P

A

NF=Fs

80 N

=sN

P

A

NF=Fs

P

P

A

NF=Fs

80 N

=sN

Fx = 0 P s N = 0 P = 0.5 80 = 40 N

Fy = 0 N 80 = 0 N = 80 N

Case II: Let us assume that both blocks slide together: P required for starting sliding ofboth blocks together can bedetermined by applying equilibriumconditions to the FBD of bothblocks together, as follows:

P200N

Fs N

P200N

Fs N

Fy = 0 N 200 = 0 N = 200 N Fx = 0 P s N = 0 P = 0.25 200 = 50 N

Since P required in Case I is lessthan the P required in Case II, an increase in the value of P will causesliding of block A over block B.


If P = 140 lb, determine thenormal and frictional forcesacting on the 300-lb pipe, shown in the figure below. Take s = 0.3 and k = 0.2.


The normal force N and frictionalforce can be determined byapplying equilibrium conditions tothe FBD of the pipe, as follows:

20o

20o

20o

P=140lb

W=300lb xy

F

N20o

20o

20o

P=140lb

W=300lb xy

F

N

Fx = 0 140 cos 20 F 300 sin 20 =0 F = 28.95 lb F is the force required to keep the pipe in equilibrium

Fy = 0 N 140 sin 20 300 cos 20 = 0 N = 329.79 lb Ans. Fs = s N = 0.3 329.79 = 98.93 lb Since F < Fs, The frictional force F = 28.95 lb Ans.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5One end of a pole is supported on arough floor at A and another end on a smooth wall at B, as shown in thefigure below. If weight of the pole is 30-lb and d = 10 ft, will the pole remain in this position when it isreleased? Take s = 0.3.


A

FA

NAd=10ft.

13Cosft5

261013 ==

26 ft.

W=30 lb

NB B

ft241026 22

=

A

FA

NAd=10ft.

13Cosft5

261013 ==

26 ft.

W=30 lb

NB B

ft241026 22

=

M about A = 0 24NB 30 5 = 0 NB = 6.25 lb. Fx = 0 NB FA = 0 FA = 6.25 lb. Fy = 0 NA 30 = 0 NA = 30 lb.

Fs = s NA = 0.30 30 = 9 lb. The pole will remain in the sate of equilibrium if FA Fs Since FA < Fs, the pole will remain in the above position Ans.

FA = 6.25 lb (Force required to keep in equilibrium)

PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6One end of a 400 N ladder is supported ona rough floor at A and another end on a rough wall at B, as shown in the figure below. Determine the distance x up to which a 700 N man can climb on theladder without causing slipping of theladder. Take A = 0.4, B = 0.25.

A

B

4m

3m

x

A

B

4m

3m

x


A

B

4m

3m

x

A

B

4m

3m

x

x

700NFB

FA

NB

400N

x

700NFB

FA

NB

400N

1.5 m

35x

3 m

4 m

= 0.25 NB

= 0.4NANA

x

700NFB

FA

NB

400N

x

700NFB

FA

NB

400N

1.5 m

35x

3 m

4 m

= 0.25 NB

= 0.4NANA

x

700NFB

FA

NB

400N

x

700NFB

FA

NB

400N

1.5 m

35x

3 m

4 m

= 0.25 NB

= 0.4NANA

Since the state ofthe ladder will beconsidered to be impending, FA = 0.4 NA FB = 0.25 NB

0 0.4 0 (1)

0

0.25 400 700 00.25 1100 (2)

x

A B

y

A B

A B

FN N

FN NN N

= = = + = + =

By solving Equations (1) and(2), values of NA and NB can be obtained as: NA = 1000 N and NB = 400 N

0

34 3 0.25 700 400 1.5 0

53

4 400 3 0.25 400 700 400 1.5 05

420 13001300

3.1 m Ans.420

about A

B B

M

xN N

x

x

x

= + + =

+ + = = = =

PROBLEM SOLVING: PROBLEM SOLVING: Example # 7Example # 7A box weighing 100 N is placed on arough inclined plane as shown below.Will the box trip or slide down if theangle of inclination of plane, , increases? Take s = 0.45.

0.2

0.4

0.2 m

0.4 m

0.2

0.4

0.2 m

0.4 m

PROBLEM SOLVING: PROBLEM SOLVING: Example # 7Example # 70.2

0.4

0.2 m

0.4 m

0.2

0.4

0.2 m

0.4 m

Let us assume that the boxwould be tipping:

T

W

y'

x'

W Cos

W Sin0.2

OF < FS

e=0.1

N

T

W

y'

x'

W Cos

W Sin0.2m

OF < FS

e=0.1m

N

T

W

y'

x'

W Cos

W Sin0.2

OF < FS

e=0.1

N

T

W

y'

x'

W Cos

W Sin0.2m

OF < FS

e=0.1m

N

Fy' = 0 N W CosT = 0 N = W CosT M about o = 0 N e + (W SinT) 0.2 = 0 W CosT 0.1 +W SinT 0.2 = 0 TanT = T

T

SinCos

=

0.1 1 0.50.2 2

= = 1(0.5) 26.57T Tan = = D

Value of the angle of inclinationof plane corresponding totipping condition, T, can be determined by applyingequilibrium conditions to the FBD of the box, as follows:

PROBLEM SOLVING: PROBLEM SOLVING: Example # 7Example # 70.2

0.4

0.2 m

0.4 m

0.2

0.4

0.2 m

0.4 m

Let us assume that the boxwould be sliding:

Value of the angle of inclinationof plane corresponding tosliding condition, S, can be determined by applyingequilibrium conditions to theFBD of the box, as follows:

S

W

y'

x'

W Cos

W Sin

FS=

e=0.1

N

S

W

y'

x'

W CosS

W SinS

FS=sN

e=0.1m

N

S

W

y'

x'

W Cos

W Sin

FS=

e=0.1

N

S

W

y'

x'

W CosS

W SinS

FS=sN

e=0.1m

N

Fx' = 0 FS W SinS = 0 FS = W SinS sN = W SinS s W CosS = W SinS TanS = S = 0.45 1(0.45) 24.23S Tan = = D

Fy' = 0 N W CosS = 0 N = W CosS

Since S < T (sliding occurs at an inclination less than that fortipping), the box will slide first.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 8Example # 8If s between drum and brake = 0.3 and M= 35 N-m, determine the smallest force Pthat is to be applied to prevent the drumfrom rotating. Also calculate reactions atpin O, Ox and Oy. Take mass of drum as 25 kg and neglect the weight and thickness ofthe brake.


Determination of Ox and Oy andNB:

OO

O

35N-m

B

NB

FB=SNB=0.4NB

W=25 9.81=245N0.125m

OOx

Oy

35N-m

B

NB

FB=SNB=0.4NB

W=259.81=245N0.125m

OO

O

35N-m

B

NB

FB=SNB=0.4NB

W=25 9.81=245N0.125m

OOx

Oy

35N-m

B

NB

FB=SNB=0.4NB

W=259.81=245N0.125m

M about O = 0 0.125 FB 35 = 0 FB = 280 N

NB = 2800.4 0.4BF = = 700 N

Fx = 0 FB Ox = 0 Ox = 280 N Ans. Fy = 0 Oy NB W = 0 Oy = NB + W Oy = 700 + 245 = 945 N Ans.

Determination of P:

0.5

0.70.3

P

BFB=280N

NB=700N

AAx

Ay

0.5 m

0.7m0.3 m

P

BFB=280N

NB=700N

AAx

Ay

0.5

0.70.3

P

BFB=280N

NB=700N

AAx

Ay

0.5 m

0.7m0.3 m

P

BFB=280N

NB=700N

AAx

Ay

M about A = 0 P 1 280 0.5 + 700 0.7 = 0 P = 350 N Ans.


If each book has a mass of 0.95 kg,determine the greatest number ofbooks that can be supported in thestack. The s between the mans handsand a book is 0.6 and between any twobooks is 0.4.


Number of books that can be supported in the stack can bedetermined by applyingequilibrium conditions to theFBD of the stack, as follows:

Fhand=0.6120=72N Fbook=0.959.810.4 (n-1)= 3.728 (n-1)

N=120N

Wb = n0.959.81=9.32n

Fhand=0.6120=72N Fbook=0.959.810.4 (n-1)= 3.728 (n-1)

N=120N

Wb = n0.959.81=9.32n

n = number of books that can be held

Fy = 0 72 + 3.728 (n1) 9.32n = 0

68.272 12.20, say 12 An5.59222

n = =

PROBLEM SOLVING: PROBLEM SOLVING: Example # 10Example # 10Determine how far the man can walk upthe plank without causing the plank toslip. The s at A and B is 0.3. The man hasa weight of 200 lb and a center of gravityat G. Neglect the thickness and weight of the plank.


Distance d over which the man canwalk up the plank without causingthe plank to slip can be determined by applying equilibrium conditionsto the FBD of the plank, as follows:

y

x

20o

30o

30o

B

NB90o

FB=0.3 NB15 Sin20= 5.13ft

d

(15200 lb

3 ft

G

A

NA

FA=0.3 NA

15 Cos20=14.09 ft

y

x

y

x

20o

30o

30o

B

NB90o

FB=0.3 NB15 Sin20= 5.13ft

d

(15 d)

200 lb

3 ft

G

A

NA

FA=0.3 NA

15 Cos20=14.09 ft

y

x

20o

30o

30o

B

NB90o

FB=0.3 NB15 Sin20= 5.13ft

d

(15200 lb

3 ft

G

A

NA

FA=0.3 NA

15 Cos20=14.09 ft

y

x

y

x

20o

30o

30o

B

NB90o

FB=0.3 N

y

x

20o

30o

30o

B

NB90o

FB=0.3 NB15 Sin20= 5.13ft

d

(15200 lb

3 ft

G

A

NA

FA=0.3 NA

15 Cos20=14.09 ft

y

x

y

x

20o

30o

30o

B

NB90o

FB=0.3 NB15 Sin20= 5.13ft

d

(15 d)

200 lb

3 ft

G

A

NA

FA=0.3 NA

15 Cos20=14.09 ft

M about B = 0 14.09 NA + (0.3 NA) 5.13 + 200Cos 20

(15d) + (200 Sin 20)3 = 0 12.551 NA 187.938 d = 3024.29 NA + 14.974 d = 240.96 NA = 240.96 14.974 d (1) Fx = 0 0.3 NA NB Sin 30 + 0.3 NB Cos 30 = 0 0.3 NB = 0.240 NB NB = 1.25 NA = 301.2 18.7175 d (2)

Fy = 0 NA 200+NB Cos 30+0.3 NB Sin 30 = 0 NA + 1.016 NB = 200 (3) Substituting NA and NB in Eq. 3 240.96 14.974 d + 1.016 (301.2 18.7175 d) = 200

d = 346.9833.99 =10.20 ft Ans.

Multiple Choice Problems1. A block weighing 100 N placed on a rough

surface and subjected to a force 20 N, as shown below. If s = 0.3 and k = 0.2, the friction forceF developing between block and the roughsurface will be

(a) 4 N (b) 6 N (c) 20 N (d) 30 N Ans: (c)

Feedback: By applying equilibrium condition Fx = 0, we get F = 20 N. Since Fs (= sN = 0.3 100 = 30 N)> F, value of F = 20 N is correct.

Multiple Choice Problems2. A block weighing 50 N placed on a rough

surface and subjected to a force 20 N, as shownbelow. If s = 0.3 and k = 0.2, the friction forceF developing between block and the rough surface will be

(a) 10 N (b) 15 N (c) 20 N (d) 30 N Ans: (a)

Feedback: By applying equilibrium condition Fx = 0, we get F = 20 N. Since Fs (= sN = 0.3 50 = 15 N)

2- Handouts Lecture-38 to 40

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Transcript of 2- Handouts Lecture-38 to 40