10 19
10 19
10 19
10
The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts
a 125-N force directed along its center line on the ball and socket at B,
determine the moment of the force about A.
SOLUTION
First note
Then
and
dCB =
( 344 mm )2 + (152.4 mm )2
cos =
344 mm
376.25 mm
152.4 mm
376.25 mm
Now
sin =
= 376.25 mm
125 N
( 344 mm ) i + (152.4 mm ) j
376.25 mm
M A = rB/ A FCB
where
Then
125 N
M A = ( 410 mm ) i ( 87.6 mm ) j
( 344i 152.4 j)
376.25
= ( 30770 N mm ) k
= ( 30.770 N m ) k
or M A = 30.8 N m
PROBLEM 3.11
A winch puller AB is used to straighten a fence post. Knowing that the
tension in cable BC is 260 lb, length a is 8 in., length b is 35 in., and
length d is 76 in., 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.
SOLUTION
Slope of line EC =
(a)
Then
and
Then
TABx =
35 in.
5
=
76 in. + 8 in. 12
12
(TAB )
13
12
( 260 lb ) = 240 lb
13
TABy =
5
( 260 lb ) = 100 lb
13
or M D = 7600 lb in.
(b) Have
M D = TABx ( y ) + TABy ( x )
PROBLEM 3.12
It is known that a force with a moment of 7840 lb in. about D is required
to straighten the fence post CD. If a = 8 in., b = 35 in., and d = 112
in., determine the tension that must be developed in the cable of winch
puller AB to create the required moment about point D.
SOLUTION
Slope of line EC =
35 in.
7
=
112 in. + 8 in. 24
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
Have
M D = TABx ( y ) + TABy ( x )
7840 lb in. =
24
7
TAB ( 0 ) +
TAB (112 in.)
25
25
TAB = 250 lb
or TAB = 250 lb
PROBLEM 3.13
It is known that a force with a moment of 1152 N m about D is required
to straighten the fence post CD. If the capacity of the winch puller AB is
2880 N, determine the minimum value of distance d to create the
specified moment about point D knowing that a = 0.24 m and
b = 1.05 m.
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y ( d )
M D = 1152 N m
where
(TAB max ) y
sin =
Now
(d
1152 N m = 2880 N
or
or
or
+ 0.24 ) + (1.05 ) m
2
1.05
( d + 0.24 )2 + (1.05)2
( d + 0.24 )2 + (1.05)2
(d
(d )
= 2.625d
Using the quadratic equation, the minimum values of d are 0.48639 m and 0.40490 m.
Since only the positive value applies here, d = 0.48639 m
or d = 486 mm
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 580 N is exerted on
the alternator B. Determine the moment of that force about bolt C if its
line of action passes through O.
SOLUTION
M C = rB/C FB
Have
x = 144 mm 78 mm = 66 mm
y = 86 mm + 108 mm = 194 mm
and
FBx =
FBy =
78
( 78)
+ ( 86 )
86
( 78) + (86 )
2
( 580 N ) = 389.65 N
( 580 N ) = 429.62 N
PROBLEM 3.15
Form the vector products B C and B C, where B = B, and use the
results obtained to prove the identity
sin cos =
SOLUTION
1
sin
2
( + ) +
1
sin
2
( ) .
B = B ( cos i + sin j)
First note
B = B ( cos i sin j)
C = C ( cos i + sin j)
By definition
Now
B C = BC sin ( )
(1)
B C = BC sin ( + )
(2)
(3)
(4)
(5)
(6)
1
1
sin ( + ) + sin ( )
2
2
PROBLEM 3.16
A line passes through the points (420 mm, 150 mm) and (140 mm,
180 mm). Determine the perpendicular distance d from the line to the
origin O of the system of coordinates.
SOLUTION
d = AB rO/ A
Have
AB =
where
rB/ A
rB/ A
and
= ( 560 mm ) i + ( 330 mm ) j
rB/ A =
AB =
1
( 56i 33j) ( 420 mm ) i + (150 mm ) j = 84.0 mm
65
d = 84.0 mm
PROBLEM 3.17
A plane contains the vectors A and B. Determine the unit vector normal
to the plane when A and B are equal to, respectively, (a) 4i 2j + 3k and
2i + 6j 5k, (b) 7i + j 4k and 6i 3k + 2k.
SOLUTION
=
(a) Have
AB
AB
A = 4i 2 j + 3k
where
B = 2i + 6 j 5k
Then
i j k
A B = 4 2 3 = (10 18 ) i + ( 6 + 20 ) j + ( 24 4 ) k = 2 ( 4i + 7 j + 10k )
2 6 5
AB = 2
and
=
( 4 )2 + ( 7 )2 + (10 )2
2 ( 4i + 7 j + 10k )
or =
2 165
=
(b) Have
= 2 165
1
( 4i + 7 j + 10k )
165
AB
AB
A = 7i + j 4k
where
B = 6i 3j + 2k
Then
and
i j k
A B = 7 1 4 = ( 2 12 ) i + ( 24 14 ) j + ( 21 + 6 ) k = 5 ( 2i + 2 j 3k )
6 3 2
AB = 5
=
( 2 )2 + ( 2 )2 + ( 3)2
5 ( 2i + 2 j 3k )
5 17
= 5 17
or =
1
( 2i + 2 j 3k )
17
PROBLEM 3.18
The vectors P and Q are two adjacent sides of a parallelogram.
Determine the area of the parallelogram when (a) P = (8 in.)i + (2 in.)j
(1 in.)k and Q = (3 in.)i + (4 in.)j + (2 in.)k, (b) P = (3 in.)i + (6 in.)j +
(4 in.)k and Q = (2 in.)i + (5 in.)j (3 in.)k.
SOLUTION
A = PQ
(a) Have
where
Then
) (
) (
= 8 in 2 i 13 in 2 j + 38 in 2 k
=
= 40.951 in 2
or A = 41.0 in 2
A = PQ
(b) Have
where
Then
i j k
P Q = 3 6 4 in 2 = ( 18 20 ) i + ( 8 9 ) j + ( 15 12 ) k in 2
2 5 3
) (
) (
= 38 in 2 i 1 in 2 j 27 in 2 k
=
( 38)2 + ( 1)2 + ( 27 )2 in 2
= 46.626 in 2
or A = 46.6 in 2
PROBLEM 3.19
Determine the moment about the origin O of the force F = (5 N)i (2
N)j + (3 N)k which acts at a point A. Assume that the position vector of
A is (a) r = (4 m)i (2 m)j (1 m)k, (b) r = (8 m)i + (3 m)j + (4 m)k,
(c) r = (7.5 m)i + (3 m)j (4.5 m)k.
SOLUTION
MO = r F
(a) Have
F = (5 N ) i ( 2 N ) j + (3 N ) k
where
r = ( 4 m ) i ( 2 m ) j (1 m ) k
MO
i j k
= 4 2 1 N m = ( 6 2 ) i + ( 5 12 ) j + ( 8 10 ) k N m
5 2 3
= ( 8i 7 j 18k ) N m
or M O = ( 8 N m ) i ( 7 N m ) j (18 N m ) k
MO = r F
(b) Have
F = (5 N ) i ( 2 N ) j + (3 N ) k
where
r = (8 m ) i + ( 3 m ) j ( 4 m ) k
MO
i j k
= 8 3 4 N m = ( 9 + 8 ) i + ( 20 + 24 ) j + (16 + 15 ) k N m
5 2 3
= (17i + 4 j + 31k ) N m
or M O = (17 N m ) i + ( 4 N m ) j + ( 31 N m ) k
(c) Have
where
MO = r F
F = (5 N ) i ( 2 N ) j + (3 N ) k
r = ( 7.5 m ) i + ( 3 m ) j ( 4.5 m ) k
i
j
k
= 7.5 3 4.5 N m = ( 9 9 ) i + ( 22.5 22.5 ) j + ( 15 + 15 ) k N m
5 2 3
or M O = 0