Solucionario Mecanica Vectorial para ingenieros Estatica - Edicion 9

192 Pages·2013·4.1 MB·English

by  Beer, Johnston

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
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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Inc. All rights reserved. M> ,«,#/ «///,& MW ,»«> 6* rfwpfe^
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165