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# Cap 3, Novena Edc

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```CHAPTER 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
F
v
=(16N)sin8° = 2.2268N
* 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
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 MB = 1.277 N-m^^
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers andeducators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
you are using it without permission.
154
200 in..,
25&quot;
-100 mm- 200 mm
i
'&quot;> 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-**>
F
v
=(300N)cos25°
= 27.1.89 N
F
y =(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
MD =rxF
MD = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j]
= -(12.6785 N \u2022 m)k + (54.378 N \u2022 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° <
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
you are using it without permission.
155
200 mn.
25°
\u2022-lOOnim-* -200 mm *.
125 il
\u2022
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 ofMd
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. 125 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
+ (°- 125 m)'
= 0.23585 m
41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)°<
PROPRIETARY MATERIAL. \u20ac5 20)0 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 P
x
= (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. \u2022**.
or Mj =186.6 lb -in. J) ^
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
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 \u2022 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
&quot; 23.409 in.
-8.97 lb Pmin =8.97 lb^ 19.98° <
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,
you are using it without permission.
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 of250 lb \u2022 in. Determine the value ofa.
22 in.
SOLUTION
By definition
where
and
also
Then
or
or
and
MB =rmPsm.6
= a + (9Q°-tf>)
_i 8 in.
<p - tan&quot; 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° <
reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,
you are using it without permission.
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\u2014Ar 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/BPsin
6» = 10° + (180°~70°)
= 120°
Then 800 lb \u2022 in. = (18 in.) x Psin 1 20°
or P = 51.3 lb <
(c) For P to be minimum, it must be perpendicular to the```