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# Resolução do Capítulo 3 - Corpos Rígidos

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```PROBLEM 3.1
A 13.2-N force P is applied to the lever which controls the auger of a
snowblower. Determine the moment of P about A when \u3b1 is equal to 30°.

SOLUTION

First note
( )sin 13.2 N sin 30 6.60 NxP P \u3b1= = ° =
( )cos 13.2 N cos30 11.4315 N\u3b1= = ° =yP P
Noting that the direction of the moment of each force component about A
is counterclockwise,
/ /A B A y B A xM x P y P= +
( )( ) ( )( )0.086 m 11.4315 N 0.122 m 6.60 N= +
1.78831 N m= \u22c5
or 1.788 N mA = \u22c5M

W

PROBLEM 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\u22c5 counterclockwise moment about A.

SOLUTION

For P to be a minimum, it must be perpendicular to the line joining points
A and B.
( ) ( )2 286 mm 122 mm 149.265 mm= + =ABr
1 1 122 mmtan tan 54.819
86 mm
\u3b1 \u3b8 \u2212 \u2212\uf8eb \uf8f6 \uf8eb \uf8f6= = = = °\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
y
x

Then min=A ABM r P
or min = A
AB
MP
r

2.20 N m 1000 mm
149.265 mm 1 m
\u22c5 \uf8eb \uf8f6= \uf8ec \uf8f7\uf8ed \uf8f8
14.7389 N=
min 14.74 N\u2234 =P 54.8°
or min 14.74 N=P 35.2°W

PROBLEM 3.3
A 13.1-N force P is applied to the lever which controls the auger of a
snowblower. Determine the value of \u3b1 knowing that the moment of P
about A is counterclockwise and has a magnitude of 1.95 N m.\u22c5

SOLUTION

By definition / sin\u3b8=A B AM r P
where ( )90\u3b8 \u3c6 \u3b1= + ° \u2212
and 1 122 mmtan 54.819
86 mm
\u3c6 \u2212 \uf8eb \uf8f6= = °\uf8ec \uf8f7\uf8ed \uf8f8
Also ( ) ( )2 2/ 86 mm 122 mm 149.265 mm= + =B Ar
Then ( )( ) ( )1.95 N m 0.149265 m 13.1 N sin 54.819 90 \u3b1\u22c5 = ° + ° \u2212
or ( )sin 144.819 0.99725\u3b1° \u2212 =
or 144.819 85.752\u3b1° \u2212 = °
and 144.819 94.248\u3b1° \u2212 = °
50.6 , 59.1\u3b1\u2234 = ° °W

PROBLEM 3.4
A foot valve for a pneumatic system is hinged at B. Knowing that
\u3b1 = 28°, determine the moment of the 4-lb force about point B by
resolving the force into horizontal and vertical components.

SOLUTION

Note that 20 28 20 8\u3b8 \u3b1= \u2212 ° = ° \u2212 ° = °
and ( )4 lb cos8 3.9611 lb= ° =xF
( )4 lb sin8 0.55669 lb= ° =yF
Also ( )6.5 in. cos 20 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 y xM xF yF= +
( )( ) ( )( )6.1080 in. 0.55669 lb 2.2231 in. 3.9611 lb= +
12.2062 lb in.= \u22c5
or 12.21 lb in.B = \u22c5M

W

PROBLEM 3.5
A foot valve for a pneumatic system is hinged at B. Knowing that
\u3b1 = 28°, determine the moment of the 4-lb 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
( )4.0 lb sin 28 1.87787 lbQ = ° =
Then /=B A BM r Q
( )( )6.5 in. 1.87787 lb=
12.2063 lb in.= \u22c5
or 12.21 lb in.B = \u22c5M

W

PROBLEM 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 \u3b1 = 10°, (c) the
smallest force P which creates the same moment about B.

SOLUTION

(a) Have /=B C B NM r F
( )( )0.1 m 800 N=
80.0 N m= \u22c5
or 80.0 N mB = \u22c5M

W
(b) By definition
/ sin\u3b8=B A BM r P
where ( )90 90 70\u3b8 \u3b1= ° \u2212 ° \u2212 ° \u2212
90 20 10= ° \u2212 ° \u2212 °
60= °
( ) 80.0 N m 0.45 m sin 60P\u2234 \u22c5 = °
205.28 N=P
or 205 NP =

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.
Thus min / min= =B A BM dP r P
or ( ) min80.0 N m 0.45 m\u22c5 = P
min 177.778 N\u2234 =P
or min 177.8 N=P 20°W

PROBLEM 3.7
A sign is suspended from two chains AE and BF. Knowing that the
tension in BF is 45 lb, determine (a) the moment about A of the force
exert by the chain at B, (b) the smallest force applied at C which creates

SOLUTION

(a) Have /A B A BF=M r T×
Noting that the direction of the moment of each force component
= +A BFy BFxM xT yT
( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + \u2212 °
282.56 lb ft= \u22c5
or 283 lb ftA = \u22c5M

W
(b) Have ( )/ minA C A C=M r F×
For CF to be minimum, it must be perpendicular to the
line joining points A and C.
( )
minA CM d F\u2234 =
where ( ) ( )2 2/ 6.5 ft 4.4 ft 7.8492 ftC Ad r= = + =
( )( )
min 282.56 lb ft 7.8492 ft CF\u2234 \u22c5 =
( )
min 35.999 lbCF =
1 4.4 fttan 34.095
6.5 ft
\u3c6 \u2212 \uf8eb \uf8f6= = °\uf8ec \uf8f7\uf8ed \uf8f8
90 90 34.095 55.905\u3b8 \u3c6= ° \u2212 = ° \u2212 ° = °
or ( )
min 36.0 lbC =F 55.9°W

PROBLEM 3.8
A sign is suspended from two chains AE and BF. Knowing that the
tension in BF is 45 lb, determine (a) the moment about A of the force
exerted by the chain at B, (b) the magnitude and sense of the vertical
force applied at C which creates the same moment about A, (c) the
smallest force applied at B which creates the same moment about A.

SOLUTION

(a) Have /A B A BF=M r T×
Noting that the direction of the moment of each force component
= +A BFy BFxM xT yT
( )( ) ( )( )6.5 ft 45 lb sin 60 4.4 ft 3.1 ft 45 lb cos60= ° + \u2212 °
282.56 lb ft= \u22c5
or 283 lb ftA = \u22c5M

W
(b) Have /A C A C=M r F×
or =A CM xF
282.56 lb ft 43.471 lb
6.5 ft
A
C
MF
x
\u22c5\u2234 = = =
or 43.5 lbC =F

W
(c) Have ( )/ minA B A B=M r F×
For BF to be minimum, it must be perpendicular to the line joining
points A and B.
( )
min \u2234 =A BM d F
where ( ) ( )2 26.5 ft 4.4 ft 3.1 ft 6.6287 ftd = + \u2212 =
( )
min
282.56 lb ft
42.627 lb
6.6287 ft
\u22c5\u2234 = = =AB MF d
and 1 6.5 fttan 78.690
4.4 ft 3.1 ft
\u3b8 \u2212 \uf8eb \uf8f6= = °\uf8ec \uf8f7\u2212\uf8ed \uf8f8
or ( )
min 42.6 lb=BF 78.7°W

PROBLEM 3.9
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 ( ) ( )2 2240 mm 46.6 mmCBd = +
244.48 mm=
Then 240 mmcos
244.48 mm
\u3b8 =

46.6 mm
sin
244.48 mm
\u3b8 =
and cos sinCB CB CBF F\u3b8 \u3b8= \u2212F i j
( ) ( )125 N 240 mm 46.6 mm
244.48 mm
\uf8ee \uf8f9= \u2212\uf8f0 \uf8fbi j
Now /A B A CB=M r F×
where ( ) ( )/ 306 mm 240 mm 46.6 mmB A = \u2212 +r i j
( ) ( )306 mm 286.6 mm= \u2212i j
Then ( ) ( ) ( )125 N306 mm 286.6 mm 240 46.6
244.48A
\uf8ee \uf8f9= \u2212 \u2212\uf8f0 \uf8fbM i j i j×
( ) ( )27878 N mm 27.878 N m= \u22c5 = \u22c5k k
or 27.9 N mA = \u22c5M

W

PROBLEM 3.10
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 ( ) ( )2 2344 mm 152.4 mm 376.25 mmCBd = + =
Then 344 mmcos
376.25 mm
\u3b8 = 152.4 mmsin
376.25 mm
\u3b8 =
and ( ) ( )cos sinCB CB CBF F\u3b8 \u3b8= \u2212F i j
( ) ( )125 N 344 mm 152.4 mm
376.25 mm
\uf8ee \uf8f9= +\uf8f0 \uf8fbi j
Now /A B A CB=M r F×
where ( ) ( )/ 410 mm 87.6 mmB A = \u2212r i j
Then ( ) ( ) ( )125 N410 mm 87.6 mm 344 152.4
376.25A
\uf8ee \uf8f9= \u2212 \u2212\uf8f0 \uf8fbM i j i j×
( )30770 N mm= \u22c5 k
( )30.770 N m= \u22c5 k
or 30.8 N mA = \u22c5M

W```