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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
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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
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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
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GENERAL APPROACH FOR ANALYZING RIGID BODIES GENERAL APPROACH FOR
ANALYZING RIGID BODIES SUBJECTED TO DRY FRICTIONSUBJECTED TO DRY
FRICTION
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:
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GENERAL APPROACH FOR ANALYZING RIGID BODIES GENERAL APPROACH FOR
ANALYZING RIGID BODIES SUBJECTED TO DRY FRICTIONSUBJECTED TO DRY
FRICTION
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=
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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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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.
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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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
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.
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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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
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)
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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
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
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
= + + =
+ + = = = =
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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
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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:
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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.
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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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 8Example # 8
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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 9Example # 9
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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 9Example # 9
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 = =
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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.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 10Example # 10
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.
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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.
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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)