Table Of ContentCHAPTER 3
80 mm
PROBLEM 3.1
A foot valve for a pneumatic system is hinged at B. Knowing
that a - 28°, determine the moment of the
1 6-N force about
Point B by resolving the force
into horizontal and vertical
components.
SOLUTION
Note that
and
= a- 20° = 28° -20° = 8°
Fx = (1 6 N)cos 8° = 1 5.8443 N
Fv =(16N)sin8° = 2.2268N
^£*.
\©* — tL>
Kl
* U^r^-^T^P,,
/|t^^^u^
d
^r^^^^-,
Also
x = (0. 1 7 m) cos 20° = 0. 1 59748 m
y = (0.17 m)sin 20° = 0.058143 m.
>n^y
c
Noting that the direction of the moment of each force component about B is counterclockwise,
MB =xFy +yFx
= (0.1 59748 m)(2.2268N)
+(0.058143 m)(l 5.8443 N)
= 1.277 N-m
or MB =1.277N-m^)4
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153
PROBLEM 3.2
A foot valve for a pneumatic system
is hinged at B. Knowing
that a = 28°, determine the moment of the
1 6-N force about
Point B by resolving the force into components along ABC and
in a direction perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into components P and Q, where
g = (16 N) sin 28°
= 7.5115 N
Then
MB = rmQ
°^7^
= (0.17m)(7.5115N)
= 1.277N-m
or M B = 1.277 N-m^^
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154
200 in..,
25"
-100 mm-
200 mm
i
'"> mm
PROBLEM 3.3
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a)
O.2.
PC
0>T-**>
Fv =(300N)cos25°
= 27.1.89 N
Fy =(300 N) sin 25°
= 126.785 N
F = (27 1 .89 N)i + (1 26.785 N)
j
r = ZM = -(0.1m)i-(0.2m)j
M D =rxF
M D = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j]
= -(12.6785 N
• m)k + (54.378 N
• m)k
= (41.700 N-m)k
M
=41.7N-nO^
(b)
The smallest force Q at B must be perpendicular to
DB at 45°^£L
MD =Q(DB)
41 .700 N m = (2(0.28284 m)
Q = 147.4 N ^L 45° <
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155
200 mn.
25°
•-lOOnim-*
-200 mm
*.
125 il
• C
H
PROBLEM3.4
A 300-N force
is applied at A as shown. Determine (a) the
moment of the 300-N force about D, (b) the magnitude and
sense of the horizontal force applied at C that creates the
same moment about D, (c) the smallest force applied at C that
creates the same moment about D.
SOLUTION
(a)
See Problem 3.3 for the figure and analysis leading to the determination of Md
M
=41.7N-m^H
Cl^n
0>\7.Siy\
c
= at.
(b)
Since C is horizontal C = Ci
r = DC = (0.2 m)i - (0. 1 25 m)j
M D =rxCi = C(0.l25m)k
4l.7N-m = (0.l25m)(C)
C = 333.60 N
(c)
The smallest force C must be perpendicular to DC; thus, it forms a with the vertical
C = 334N <
tan6^
0.125 m
0.2 m
a = 32.0°
M D = C(£>C); DC = V(
-2 m)
2 + (°- 1 25 m )'
= 0.23585 m
41.70 Nm = C(0.23585m)
C = 176.8 N^L 5HX)°<
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156
PROBLEM 3.5
An 8-1b force P is applied to a shift lever. Determine the moment of V about B
when a is equal to. 259
.
SOLUTION
First note
Px = (8 lb) cos 25°
= 7.2505 lb
/^ =(8 lb) sin 25°
= 3.3809 lb
Noting that the direction of the moment of each force component about B is
clockwise, have
= -(8in.)(3.3809 1b)
- (22 in.)(7.2505 lb)
= -186.6 lb -in.
i*22.
•**.
or Mj =186.6 lb -in. J) ^
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157
PROBLEM 3.6
For the shift lever shown, determine the magnitude and the direction of the smallest
force P that has a 21 0-lb
• in. clockwise moment about B.
22 in.
SOLUTION
For P to be minimum
it must be perpendicular to the line joining Points A. and B. Thus,
a =
e
,
8
in
^T^s*. K
22 in.
/
= 19.98°
i
and
MB =dP^n
p
ZZ
i«.
Where
d = rAIB
fl
$ u
= ^m.y+(22m.y
JB w
= 23.409 in.
Then
_210Ib-in.
' min " 23.409 in.
-8.97 lb
Pmin =8.97 lb^ 19.98° <
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158
PROBLEM 3.7
An
1 1 -lb force P is applied to a shift lever. The moment of P about B is clockwise
and has a magnitude of 250 lb
• in. Determine the value of a.
22 in.
SOLUTION
By definition
where
and
also
Then
or
or
and
MB =rmPsm.6
= a + (9Q°-tf>)
_i
8 in.
<p - tan"
19.9831'
22 in.
r f/fl =V(8in.)
2 +(22in.)
2
= 23.409 in.
250lb-in = (23.409in.)(lllb)
xsin(tf + 90° -19.9831°)
sin (« + 70.01 69°) = 0.97088
a + 70.0 169° = 76. 1391°
a+ 70.0169° = 103.861°
a = 6.12°
33.8° <
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159
PROBLEM 3.8
It is known that a vertical force of 200 lb is required to remove the nail
at C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
the force P that creates the same moment about B if a ~
.1 0°,
(c) the
smallest force P that creates the same moment about B.
V—A
r 4 in. n
SOLUTION
(a)
We have
MB =raBFN
(4 in.)(200 lb)
800 lb -in.
or MB =H00\b-m.)<
A- m.
(/;)
By definition
MB ~rA/BP sin
6» = 10° + (180°~70°)
= 120°
Then
800 lb
• in. = (18 in.) x Psin 1 20°
or P = 51.3 lb <
(c)
For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
or
or
A*W™.
cl = fA IB
800 lb
• in. = (18 in.)Pm
^i„=44.4 1b
Pmitl =44.41b^l20 ^
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160
PROBLEM 3.9
A winch puller AB
is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is
1 ,90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
0.2 m
0.875 m
SOLUTION
(a)
Slope of line
Then
EC =
0.875 m
1.90 m + 0.2 m
12
*abx - ,~ (Tab)
12
13
960 N
(1040 N)
0,1«y
and
*W=-0040N)
= 400 N
Then
(b)
We have
MD - TABx (0.875 m)~TABy (0.2 m)
= (960 N)(0.875 m) - (400 N)(0.2 m)
= 760 N
•
m
MD ~rm (y) + TABx (x)
= (960 N)(0) + (400 N)(I .90 m)
= 760 N
•
m
or
M./J =760N-m
>)^
or M 7> = 760N-m
v
)<«
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161
PROBLEM 3.10
It is known that a force with a moment of 960 N
• m about D is required to straighten the fence post CD. If
d- 2,80 m, determine the tension
that must be developed
in the cable of winch puller AB
to create the
required moment about Point D.
0.875 m
0.2 j.i»
SOLUTION
\*.ȣ^
*AB
'My
o. a? 5^i
z-acw
OiZ^
Slope of line
Then
and
We have
EC =
0.875 m
7
7'
2.80 m + 0.2 m
24
24
My
r
/ffl)>
25
1
25
T/)B
rAH
24
7
960N-m=—7^(0)+— 7^(2.80*)
7^= 1224 N
or r^=1224N ^
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162
PROBLEM 3.11
It is known that a force with a moment of 960 N
• m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D.
fe
0.2 in
0.875 in
SOLUTION
o.mtn
czom
The minimum value ofd can be found based on the equation relating the moment of the force TAB about D:
MD ={TABmK ),(d)
where
Now
MD =960N-m
(^flmax )y = TAIHmx sin & = (2400 N)sfo
.
.
0.875m
sin #
960 N
• m = 2400 N
^(t/ + 0.20)
2 +(0.875)
2m
0.875
(d)
+ 0.20)
2 +(0.875)
2
or
^ + 0.20)
2 + (0.875)
2 = 2. ! 875d
or
(J + 0.20)
2 + (0.875)
2 = 4.7852rf
2
or
3 .7852</
2 - 0.40c/ - .8056 =
Using the quadratic equation, the minimum values ofd are 0.51719 m and -.41151 m.
Since only the positive value applies here, d — 0.5 1 7 1 9 m
or d ~ 5
1 7 mm ^
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163
5.3 in.
PROBLEM 3.12
12.0 in.
2.33 in.
mmM
1 >iii. i
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
SOLUTION
First note
Then
and
Now
where
Then
dcB
== 7(12.0 in.)
2
= 12.224 lin.
+ (2.33 in.)
2
cos 9 =
12.0 in.
12.2241 in.
sin 9-
2.33 in.
12.2241 in.
*cb = FCB cos 9\ -FCB $m9l
1251b
12.2241 in.
[(12.0 in.)i- (2.33 in.) j]
M.A = VB/A X ^CB
XBIA = (15.3in.)i-- (12.0 in. + 2.33 in.) j
M
= (15.3 in.) i--(14.33 in.) j
ru-nimn*
1251b
\«5»,a>
im.
\Z.& >N.
2.33
ttvi.
— (12.01 — 2.331)
12.2241 in.
(1393.87 lb
in.)k
(116.156 lb -ft)k
or M„ = 116.2 lb- ft *)<
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164
20.5 in.
h—
-
4.38 in
i
*Ti
7.62
1
t§|
1.7.2 hi.
PROBLEM 3.13
The tailgate of a car is supported by the hydraulic lift EC. If the lift
exerts a 125-lb force directed along
its centerline on the ball and
socket at B, determine the moment of the force about A.
SOLUTION
First note
Then
dCB
:= 7(17.2 in.)2
= 18.8123 in.
+ (7.62 in.)
2
cos
-
17.2 in.
18.8123 in.
sin 6 -
7.62 in.
18.8123 in.
Z.O.'S
«w.
and
Now
where
Then
fcd = (Ft:v* cos 0)\ - (FCB sin 6>)j
=S
(,7 '2i"-)i + (7 '62i^
r^ = (20.5 in.)i- (4.38 in.)j
M, = [(20.5 in.)i - (4.38 in.)j] x
t
.
1251b
(1 7.21 - 7.62j)
18.8 123 in.
n.z
.«»».
4.?.e. >Ni.
XKvZ
>KJ.
(1538.53 lb -in.)k
(128.2 lb -ft)k
or M^ =128.2 lb- ft ^H
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165
