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PROBLEM 3.1
A 13.2N force
P
is applied to the lever which controls the auger of a
snowblower. Determine the moment of
P
about
A
when
α
is equal to 30°.
SOLUTION
First note
( )
sin
13.2 N sin30
6.60 N
x
PP
==
°
=
( )
cos
13.2 N cos30
11.4315 N
°
=
y
Noting that the direction of the moment of each force component about
A
is counterclockwise,
//
AB
A
yB
A
x
M
xP yP
=+
( )( ) ( )( )
0.086 m 11.4315 N
0.122 m
6.60 N
1.78831 N m
=
⋅
or
1.788 N m
A
=⋅
M
W
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View Full DocumentPROBLEM 3.2
The force
P
is applied to the lever which controls the auger of a
snowblower. Determine the magnitude and the direction of the smallest
force
P
which has a 2.20 N m
⋅
counterclockwise moment about
A
.
SOLUTION
For
P
to be a minimum, it must be perpendicular to the line joining points
A
and
B
.
()
(
)
22
86 mm
122 mm
149.265 mm
=+=
AB
r
11
122 mm
tan
tan
54.819
86 mm
αθ
−−
=
==
=
°
y
x
Then
min
=
AA
B
M
rP
or
min
=
A
AB
M
P
r
2.20 N m
1000 mm
149.265 mm
1 m
⋅
=
14.7389 N
=
min
14.74 N
∴=
P
54.8
°
or
min
14.74 N
=
P
35.2
°
W
PROBLEM 3.3
A 13.1N force
P
is applied to the lever which controls the auger of a
snowblower. Determine the value of
α
knowing that the moment of
P
about
A
is counterclockwise and has a magnitude of 1.95 N m.
⋅
SOLUTION
By definition
/
sin
θ
=
AB
A
Mr
P
where
( )
90
φα
=+ °
−
and
1
122 mm
tan
54.819
86 mm
φ
−
=
=°
Also
()
(
)
22
/
86 mm
122 mm
149.265 mm
=+=
BA
r
Then
( )( ) ( )
1.95 N m
0.149265 m 13.1 N sin 54.819
90
⋅=
°
+°
−
or
( )
sin 144.819
0.99725
°−
=
or
144.819
85.752
=
°
and
144.819
94.248
=
°
50.6 , 59.1
∴= °
°
W
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View Full DocumentPROBLEM 3.4
A foot valve for a pneumatic system is hinged at
B
. Knowing that
α
=
28°, determine the moment of the 4lb force about point
B
by
resolving the force into horizontal and vertical components.
SOLUTION
Note that
20
28
20
8
θ
=
−°
=°
=
°
and
( )
4 lb cos8
3.9611 lb
=
x
F
( )
4 lb sin8
0.55669 lb
=
y
F
Also
( )
6.5 in. cos20
6.1080 in.
=
x
( )
6.5 in. sin 20
2.2231 in.
=
y
Noting that the direction of the moment of each force component about
B
is counterclockwise,
B
yx
M
xF
yF
=+
( )( ) ( )( )
6.1080 in. 0.55669 lb
2.2231 in. 3.9611 lb
12.2062 lb in.
=
⋅
or
12.21 lb in.
B
=⋅
M
W
PROBLEM 3.5
A foot valve for a pneumatic system is hinged at
B
. Knowing that
α
=
28°, determine the moment of the 4lb force about point
B
by
resolving the force into components along
ABC
and in a direction
perpendicular to
ABC
.
SOLUTION
First resolve the 4lb force into components
P
and
Q
, where
( )
4.0 lb sin 28
1.87787 lb
Q
=°
=
Then
/
=
B
AB
M
rQ
( )( )
6.5 in. 1.87787 lb
=
12.2063 lb in.
=
⋅
or
12.21 lb in.
B
=⋅
M
W
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View Full DocumentPROBLEM 3.6
It is known that a vertical force of 800 N 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
which creates the same moment about
B
if
α
= 10°, (
c
) the
smallest force
P
which creates the same moment about
B
.
SOLUTION
(
a
) Have
/
=
B
CB N
M
rF
( )( )
0.1 m 800 N
=
80.0 N m
=
⋅
or
80.0 N m
B
=⋅
M
W
(
b
) By definition
/
sin
θ
=
BA
B
Mr
P
where
( )
90
90
70
=
°−
° −
90
20
10
=
°
60
=
°
( )
80.0 N m
0.45 m
sin 60
P
∴
⋅=
°
205.28 N
=
P
or
205 N
P
=
W
(
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.
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 Spring '08
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