Forças e Momentos em um Bloco de Metal

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PROBLEM 3.101 
Five separate force-couple systems act at the corners of a metal block, 
which has been machined into the shape shown. Determine which of 
these systems is equivalent to a force ( )10 N=F j and a couple of 
moment ( ) ( )6 N m 4 N m= ⋅ + ⋅M i k located at point A. 
 
SOLUTION 
 
 
 
 
The equivalent force-couple system at A for each of the five force-couple 
systems will be determined. Each will then be compared to the given 
force-couple system to determine if they are equivalent. 
Force-couple system at B 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A B B AΣ Σ + =M M r F M× 
( ) ( ) ( ) ( )4 N m 2 N m 0.2 m 10 N⋅ + ⋅ + =i k i j M× 
( ) ( )4 N m 4 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Not Equivalent W 
Force-couple system at C 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A C C AΣ + =M M r F M× 
( ) ( ) ( ) ( )8.5 N m 0.2 m 0.25 m 10 N ⋅ + + = i i k j M× 
( ) ( )6 N m 2.0 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Not Equivalent W 
 
 
 
 
 
 
 
 
 
PROBLEM 3.101 CONTINUED 
Force-couple system at E 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A E E AMΣ + =M r F M× 
( ) ( ) ( ) ( )6 N m 0.4 m 0.08 m 10 N ⋅ + − = i i j j M× 
( ) ( )6 N m 4 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Equivalent W 
Force-couple system at G 
Have ( ) ( ): 10 N 10 NΣ + =F i j F 
or ( ) ( )10 N 10 N= +F i j 
F has two force components 
 force-couple system at G∴ 
 Is Not Equivalent W 
Force-couple system at I 
Have ( ): 10 NΣ =F j F 
or ( )10 N=F j 
and ( )/:A I I AΣ Σ + =M M r F M× 
 ( ) ( )10 N m 2 N m⋅ − ⋅i k 
 ( ) ( ) ( ) ( )0.4 m 0.2 m 0.4 m 10 N + − + = i j k j M× 
or ( ) ( )6 N m 2 N m= ⋅ + ⋅M i k 
 Comparing to given force-couple system at A, 
 Is Not EquivalentW 
 
 
 
 
 
PROBLEM 3.102 
The masses of two children sitting at ends A and B of a seesaw are 38 kg 
and 29 kg, respectively. Where should a third child sit so that the 
resultant of the weights of the three children will pass through C if she 
has a mass of (a) 27 kg, (b) 24 kg. 
 
SOLUTION 
 
 
 
 
First ( )38 kgA AW m g g= = 
 ( )29 kgB BW m g g= = 
(a) ( )27 kgC CW m g g= = 
 For resultant weight to act at C, 0CMΣ = 
 Then ( ) ( ) ( ) ( ) ( ) ( )38 kg 2 m 27 kg 29 kg 2 m 0g g d g     − − =      
76 58 0.66667 m
27
d −∴ = = 
 or 0.667 md = W 
(b) ( )24 kgC CW m g g= = 
 For resultant weight to act at C, 0CMΣ = 
 Then ( ) ( ) ( ) ( ) ( ) ( )38 kg 2 m 24 kg 29 kg 2 m 0g g d g     − − =      
76 58 0.75 m
24
d −∴ = = 
 or 0.750 md = W 
 
 
 
 
 
PROBLEM 3.103 
Three stage lights are mounted on a pipe as shown. The mass of each 
light is 1.8 kgA Bm m= = and 1.6 kgCm = . (a) If 0.75d = m, 
determine the distance from D to the line of action of the resultant of the 
weights of the three lights. (b) Determine the value of d so that the 
resultant of the weights passes through the midpoint of the pipe. 
 
SOLUTION 
 
 
 
 
First ( )1.8 kgA B AW W m g g= = = 
 ( )1.6 kgC CW m g g= = 
(a) 0.75 md = 
 Have A B CR W W W= + + 
( )1.8 1.8 1.6 kgR g = + +  
 or ( )5.2 Ng=R 
 Have 
( ) ( ) ( ) ( ): 1.8 0.3 m 1.8 1.3 m 1.6 2.05 m 5.2DM g g g g DΣ − − − = − 
 1.18462 mD∴ = 
 or 1.185 mD = W 
(b) 1.25 m
2
LD = = 
 Have 
( )( ) ( )( ) ( )( ): 1.8 0.3 m 1.8 1.3 m 1.6 1.3 mDM g g g dΣ − − − + 
 ( )( )5.2 1.25 mg= − 
 0.9625 md∴ = 
 or 0.963 md = W 
 
 
 
 
 
PROBLEM 3.104 
Three hikers are shown crossing a footbridge. Knowing that the weights 
of the hikers at points C, D, and E are 800 N, 700 N, and 540 N, 
respectively, determine (a) the horizontal distance from A to the line of 
action of the resultant of the three weights when 1.1 m,a = (b) the value 
of a so that the loads on the bridge supports at A and B are equal. 
 
SOLUTION 
 
(a) 
 
(b) 
 
 
 (a) 1.1 ma = 
 Have : C D EF W W W RΣ − − − = 
 800 N 700 N 540 NR∴ = − − − 
2040 NR = 
 or 2040 N=R 
 Have 
( )( ) ( )( ) ( )( ): 800 N 1.5 m 700 N 2.6 m 540 N 4.25 mAMΣ − − − 
 ( )R d= − 
( ) 5315 N m 2040 N d∴ − ⋅ = − 
 and 2.6054 md = 
 or 2.61 m to the right of d A= W 
(b) For equal reaction forces at A and B, the resultant, ,R must act at the 
 center of the span. 
 From 
2A
LM R  Σ = −    
( )( ) ( )( ) ( )( ) 800 N 1.5 m 700 N 1.5 m 540 N 1.5 m 2.5a a∴ − − + − + 
 ( )( )2040 N 3 m= − 
3060 2050 6120a+ = 
 and 1.49268 ma = 
 or 1.493 ma = W 
 
 
 
 
 
PROBLEM 3.105 
Gear C is rigidly attached to arm AB. If the forces and couple shown can 
be reduced to a single equivalent force at A, determine the equivalent 
force and the magnitude of the couple M. 
 
SOLUTION 
 
For equivalence 
( ) ( ): 90 N sin30 125 N cos40x xF RΣ − ° + ° = 
 or 50.756 NxR = 
( ) ( ): 90 N cos30 200 N 125 N sin 40y yF RΣ − ° − − ° = 
 or 358.29 NyR = − 
Then ( ) ( )2 250.756 358.29 361.87 NR = + − = 
and 358.29tan 7.0591 81.937
50.756
y
x
R
R
θ θ−= = = − ∴ = − ° 
 or 362 N=R 81.9°W 
Also 
( ) ( ) ( ) ( ) ( ) ( ): 90 N sin 35 0.6 m 200 N cos 25 0.85 m 125 N sin 65 1.25 m 0AM M      Σ − ° − ° − ° =      
 326.66 N mM∴ = ⋅ 
 or 327 N mM = ⋅ W 
 
 
 
 
 
PROBLEM 3.106 
To test the strength of a 25 20-in.× suitcase, forces are applied as shown. 
If 18 lb,P = (a) determine the resultant of the applied forces, (b) locate 
the two points where the line of action of the resultant intersects the edge 
of the suitcase. 
 
SOLUTION 
 
 
(a) 18 lbP = 
 Have ( ) ( ) ( ) ( )42 lb: 20 lb + 3 2 18 lb 36 lb
13 x y
R RΣ − − + + + = +F i i j j i i j 
( ) ( )18.9461 lb 41.297 lb x yR R∴ − + = +i j i j 
 or ( ) ( )18.95 lb 41.3 lb= − +R i j 
( ) ( )2 22 2 18.9461 41.297 45.436 lbx yR R R= + = + = 
1 1 41.297tan tan 65.355
18.9461
y
x
x
R
R
θ − −   = = = − °   −  
 
 or 45.4 lb=R 65.4°W 
(b) Have B BΣ =M M 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )42 lb4 in. 20 lb 21 in. 3 2 12 in. 36 lb 3 in. 18 lb
13B
 = − + − + + +  M j i i i j j i i j× × × × 
( ) 191.246 lb in.B∴ = ⋅M k 
 
 
 PROBLEM 3.106 CONTINUED 
Since B B=M r R× 
( ) ( )191.246 lb in. 0 41.297 18.9461
18.9461 41.297 0
x y x y∴ ⋅ = = +
−
i j k
k k 
For 191.2460, 4.6310 in.
41.297
y x= = = or 4.63 in.x = W 
For 191.2460, 10.0942 in.
18.9461
x y= = = or 10.09 in.y = W 
 
 
 
 
 
 
PROBLEM 3.107 
Solve Problem 3.106 assuming that 28 lb.P = 
Problem 3.106: To test the strength of a 25 20-in.× suitcase, forces are 
applied as shown. If 18 lb,P = (a) determine the resultant of the applied 
forces, (b) locate the two points where the line of action of the resultant 
intersects the edge of the suitcase. 
 
SOLUTION 
 
 
(a) P = 28 lb 
 Have ( ) ( ) ( ) ( )42: 20 lb 3 2 28 lb 36 lb
13 x y
R RΣ − + − + + + = +F i i j j i i j 
( ) ( )18.9461 lb 51.297 lb x yR R∴ − + = +i j i j 
 or ( ) ( )18.95 lb 51.3 lb= − +R i j 
( ) ( )2 22 2 18.9461 51.297 54.684 lbx yR R R= + = + = 
1 1 51.297tan tan 69.729
18.9461
y
x
x
R
R
θ − −   = = = − °   −  
 
 or 54.7 lb=R 69.7°W 
(b) Have B BΣ =M M 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )42 lb4 in. 20 lb 21 in. 3 2 12 in. 36 lb 3 in. 28 lb
13B
 = − + − + + +  M j i i i j j i i j× × × × 
( ) 221.246 lb in.B∴ = ⋅M k 
 
 
 PROBLEM 3.107 CONTINUED 
Since B B=M r R× 
( ) ( )221.246 lb in. 0 51.297 18.9461
18.9461 51.297 0
x y x y∴ ⋅ = = +
−
i j k
k k 
For 221.2460, 4.3130 in.
51.297
y x= = = or 4.31 in.x= W 
For 221.2460, 11.6776 in.
18.9461
x y= = = or 11.68 in.y = W