Problemas de Equivalência de Força-Casal

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PROBLEM 3.94 
A 25-lb force acting in a vertical plane parallel to the yz plane is applied 
to the 8-in.-long horizontal handle AB of a socket wrench. Replace the 
force with an equivalent force-couple system at the origin O of the 
coordinate system. 
 
SOLUTION 
 
 
 
Have 
: BΣ =F P F 
where 
( ) ( )25 lb sin 20 cos20B  = − ° + ° P j k 
 ( ) ( )8.5505 lb 23.492 lb= − +j k 
 ( ) ( )or 8.55 lb 23.5 lb= − +F j k W 
Have 
/:O B O B OΣ =M r P M× 
where 
( ) ( ) ( )/ 8cos30 15 8sin 30 in.B O  = ° + − ° r i j k 
 ( ) ( ) ( )6.9282 in. 15 in. 4 in.= + −i j k 
 6.9282 15 4 lb in.
0 8.5505 23.492
O∴ − ⋅ =
−
i j k
M 
( ) ( ) ( )318.18 162.757 59.240 lb in.O  = − − ⋅ M i j k 
 ( ) ( ) ( )or 318 lb in. 162.8 lb in. 59.2 lb in.O = ⋅ − ⋅ − ⋅M i j kW 
 
 
 
 
 
PROBLEM 3.95 
A 315-N force F and 70-N · m couple M are applied to corner A of the 
block shown. Replace the given force-couple system with an equivalent 
force-couple system at corner D. 
 
SOLUTION 
 
 
 
 
Have 
 : DΣ =F F F 
 AI F= λ 
 ( ) ( ) ( ) ( )0.360 m 0.120 m 0.180 m 315 N
0.420 m
− += i j k 
 ( )( )750 N 0.360 0.120 0.180= − +i j k 
 ( ) ( ) ( )or 270 N 90.0 N 135.0 ND = − +F i j kW 
Have 
/:D I D DΣ + =M M r F M× 
where 
 ACM=M λ 
 ( ) ( ) ( )0.240 m 0.180 m 70.0 N m
0.300 m
−= ⋅i k 
 ( )( )70.0 N m 0.800 0.600= ⋅ −i k 
( )/ 0.360 mI D =r k 
( )( ) ( ) 70.0 N m 0.8 0.6 0 0 0.36 750 N m
0.36 0.12 0.18
D∴ = ⋅ − + ⋅
−
i j k
M i k 
 ( ) ( )56.0 N m 42.0 N m= ⋅ − ⋅i k ( ) ( )32.4 N m 97.2 N m + ⋅ + ⋅ i j 
 ( ) ( ) ( )or 88.4 N m 97.2 N m 42.0 N mD = ⋅ + ⋅ − ⋅M i j k W 
 
 
 
 
 
PROBLEM 3.96 
The handpiece of a miniature industrial grinder weighs 2.4 N, and its 
center of gravity is located on the y axis. The head of the handpiece is 
offset in the xz plane in such a way that line BC forms an angle of 25° 
with the x direction. Show that the weight of the handpiece and the two 
couples 1M and 2M can be replaced with a single equivalent force. 
Further assuming that 1 0.068 N mM = ⋅ and 2 0.065 N m,M = ⋅ 
determine (a) the magnitude and the direction of the equivalent force, 
(b) the point where its line of action intersects the xz plane. 
 
SOLUTION 
 
 
 
 
First assume that the given force W and couples 1M and 2M act at the 
origin. 
Now W= − jW 
and ( ) ( )1 2 2 1 2cos 25 sin 25M M M= + = − ° + − °M M M i k 
Note that since W and M are perpendicular, it follows that they can be 
replaced with a single equivalent force. 
(a) Have ( ) or 2.4 NF W= = − = −jW F j 
( )or 2.40 N= −F jW 
(b) Assume that the line of action of F passes through point P (x, 0, z). 
 Then for equivalence 
/P O= ×M r F 
 where /P O x z= +r i k 
 ( ) ( )2 1 2 cos25 sin 25M M M∴ − ° + − °i k 
 ( ) ( )0
0 0
x z Wz Wx
W
= = −
−
i j k
i k 
 
 
 
 
PROBLEM 3.96 CONTINUED 
 Equating the i and k coefficients, 
1 2cos25 sin 25 and zM M Mz x
W W
− ° − ° = = −   
(b) For 1 22.4 N, 0.068 N m, 0.065 N mW M M= = ⋅ = ⋅ 
0.068 0.065sin 25 0.0168874 m
2.4
x − °= = −− 
 or 16.89 mmx = − W 
0.065cos25 0.024546 m
2.4
z − °= = − 
 or 24.5 mmz = − W 
 
 
 
 
 
PROBLEM 3.97 
A 20-lb force 1F and a 40- lb ft⋅ couple 1M are applied to corner E of the 
bent plate shown. If 1F and 1M are to be replaced with an equivalent 
force-couple system ( )2 2, F M at corner B and if ( )2 0,zM = determine 
(a) the distance d, (b) 2F and 2.M 
 
SOLUTION 
 
(a) Have 2: 0Bz zM MΣ = 
 ( )/ 1 1 0H B zM+ =k r F⋅ × (1) 
 where ( ) ( )/ 31 in. 2 in.H B = −r i j 
1 1EH F=F λ 
 ( ) ( ) ( ) ( )6 in. 6 in. 7 in. 20 lb
11.0 in.
+ −= i j k 
 ( )20 lb 6 6 7
11.0
= + −i j k 
1 1zM = k M⋅ 
 1 1EJ M=M λ 
 ( ) ( ) ( )
2
3 in. 7 in.
480 lb in.
58 in.
d
d
− + −= ⋅
+
i j k
 
 Then from Equation (1), 
( )( )
2
0 0 1
7 480 lb in.20 lb in.31 2 0 0
11.0 586 6 7 d
− ⋅⋅− + =
+−
 
 
 PROBLEM 3.97 CONTINUED 
 Solving for d, Equation (1) reduces to 
( )
2
20 lb in. 3360 lb in.186 12 0
11.0 58d
⋅ ⋅+ − =
+
 
 From which 5.3955 in.d = 
or 5.40 in.d = W 
(b) ( )2 1 20 lb 6 6 711.0= = + −F F i j k 
 ( )10.9091 10.9091 12.7273 lb= + −i j k 
( ) ( ) ( )2or 10.91 lb 10.91 lb 12.73 lb= + −F i j k W 
 2 / 1 1H B= +M r F M× 
 ( ) ( )5.3955 3 720 lb in.31 2 0 480 lb in.
11.0 9.3333
6 6 7
− + −⋅= − + ⋅
−
i j k
i j k
 
 ( )25.455 394.55 360 lb in.= + + ⋅i j k 
 ( )277.48 154.285 360 lb in.+ − + − ⋅i j k 
( ) ( )2 252.03 lb in. 548.84 lb in.= − ⋅ + ⋅M i j 
( ) ( )2or 21.0 lb ft 45.7 lb ft= − ⋅ + ⋅M i jW 
 
 
 
 
 
PROBLEM 3.98 
A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each 
loading with an equivalent force-couple system at end A of the beam. 
(b) Which of the loadings are equivalent? 
 
SOLUTION 
 (a) 
 
 
 
 (a) Have : 200 lb 100 lby aF RΣ − − = 
 or 300 lba =R W 
 and ( )( ): 900 lb ft 100 lb 4 ftA aM MΣ ⋅ − = 
 or 500 lb fta = ⋅M W 
(b) Have : 300 lby bF RΣ − = 
 or 300 lbb =R W 
 and : 450 lb ftA bM MΣ − ⋅ = 
 or 450 lb ftb = ⋅M W 
(c) Have : 150 lb 450 lby cF RΣ − = 
 or 300 lbc =R W 
 and ( )( ): 2250 lb ft 450 lb 4 ftA cM MΣ ⋅ − = 
 or 450 lb ftc = ⋅M W 
(d) Have : 200 lb 400 lby dF RΣ − + = 
 or 200 lbd =R W 
 and ( )( ): 400 lb 4 ft 1150 lb ftA dM MΣ − ⋅ = 
 or 450 lb ftd = ⋅M W 
(e) Have : 200 lb 100 lby eF RΣ − − = 
 or 300 lbe =R W 
 and ( )( ): 100 lb ft 200 lb ft 100 lb 4 ftA eM MΣ ⋅ + ⋅ − = 
 or 100 lb fte = ⋅M W 
 
 
 
 
 
 
(b) 
PROBLEM 3.98 CONTINUED 
(f) Have : 400 lb 100 lby fF RΣ − + = 
 or 300 lbf =R W 
 and ( )( ): 150 lb ft 150 lb ft 100 lb 4 ftA fM MΣ − ⋅ + ⋅ + = 
 or 400 lb ftf = ⋅M W 
(g) Have : 100 lb 400 lby gF RΣ − − = 
 or 500 lbg =R W 
 and ( )( ): 100 lb ft 2000 lb ft 400 lb 4 ftA gM MΣ ⋅ + ⋅ − = 
 or 500 lb ftg = ⋅M W 
(h) Have : 150 lb 150 lby hF RΣ − − = 
 or 300 lbh =R W 
 and ( )( ): 1200 lb ft 150 lb ft 150 lb 4 ftA hM MΣ ⋅ − ⋅ − = 
 or 450 lb fth = ⋅M W 
 Therefore, loadings (c) and (h) are equivalent W 
 
 
 
 
 
 
PROBLEM 3.99 
A 4-ft-long beam is loaded as shown. Determine the loading of Problem 
3.98 which is equivalent to this loading. 
 
SOLUTION 
 
 
 
Have : 100 lb 200 lby RΣ − − =F 
 or 300 lb=R 
and ( )( ): 200 lb ft 1400 lb ft 200 lb 4 ftAM MΣ − ⋅ + ⋅ − = 
 or 400 lb ft= ⋅M 
 Equivalent to case (f) of Problem 3.98 W 
Problem 3.98 Equivalent force-couples at A 
 case R M 
 ( )a 300 lb 500 lb ft⋅ 
 ( )b 300 lb 450 lb ft⋅ 
 ( )c 300 lb 450 lb ft⋅ 
 ( )d 200 lb 450 lb ft⋅ 
 ( )e 300 lb 100 lb ft⋅ 
 ( )f 300 lb 400 lb ft⋅ W 
 ( )g 500 lb 500 lb ft⋅ 
 ( )h 300 lb 450 lb ft⋅ 
 
 
 
 
 
 
PROBLEM 3.100 
Determine the single equivalent force and the distance from point A to its 
line of action for the beam and loading of (a) Problem 3.98b, 
(b) Problem 3.98d, (c) Problem 3.98e. 
Problem 3.98: A 4-ft-long beam is subjected to a variety of loadings. 
(a) Replace each loading with an equivalent force-couple system at end A 
of the beam. (b) Which of the loadings are equivalent? 
 
SOLUTION 
 (a) 
 
(b) 
 
(c) 
 
 
For equivalent single force at distance d from A 
Have : 300 lbyF RΣ − = 
 or 300 lb=R W 
and ( )( ): 300 lb 450 lb ft 0CM dΣ − ⋅ = 
 or 1.500 ftd = W 
Have : 200 lb 400 lbyF RΣ − + = 
 or 200 lb=R W 
and ( )( ) ( )( ): 200 lb 400 lb 4 1150 lb ft 0CM d dΣ + − − ⋅ = 
 or 2.25 ftd = W 
Have : 200 lb 100 lbyF RΣ − − = 
 or 300 lb=R W 
and ( )() ( )( ): 100 lb ft 200 lb 100 lb 4 200 lb ft 0CM d dΣ ⋅ + − − + ⋅ = 
 or 0.333 ftd = W