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349
SOLUTION
4–121.
Replace the loading on the frame by a single resultant force.
Specify where its line of action intersects a horizontal line
along member CB, measured from end C.
1 m
B
y
0.5 m
1 m
0.5 m
400 N
600 N
54
3
400 N
900 N
1.5 m
5
4
3
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–122.
SOLUTION
a
Ans.d=0.8274 mm =827 mm
+(MR)A=©MA; 533.01(d)=500 cos 30°(2) -500 sin 30°(0.2) -250a3
5b(0.5) -250a4
5b(3) +300(1)
The magnitude of the resultant force FRis given by
Ans.
The angle of FRis
Ans.
Location of the Resultant Force: Applying the principle of moments,Figs. aand b,
and summing the moments of the force components algebraically about point A,
u=tan -1
B
(F
R)y
(F
R)x
R
=tan -1c100
533.01 d=10.63° =10.6° b
u
F
R=2(F
R)x2+(F
R)y2=2533.012+1002=542.31 N =542 N
Replace the force system acting on the post by a resultant
force, and specify where its line of action intersects the post
AB measured from point A.
250 N
500 N
0.2 m
0.5 m
3
4
5
300 N
1 m
30
1 m
B
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
FR=-10 kN
*4–124.
Replace the parallel force system acting on the plate by a
resultant force and specify its location on the x–z plane.
SOLUTION
Ans.
The negative sign indicates that acts along the negative y axis.
Resultant Moment: Using the right-hand rule, and equating the moment of to
the sum of the moments of the force system about the xand zaxes,
FR
FR
=-10 kN
1m
1m
0.5 m
0.5 m
5kN
3kN
x
y
z
2kN
4–127.
750mm
z
xy
650 mm
100 mm
150 mm
600 mm
700 mm
100 mm
150 mm
8kN
6kN
F
A
F
B
O
metsys ecrof eht tneserper, fI
acting on the corbels by a resultant force, and specify its
location on the x–y plane.
FA
=7 kN and
FB
=5kN
SOLUTION
*4–128.
Determine the magnitudes of and so that the
resultant force passes through point Oof the column.
F
B
F
A
750 mm
z
xy
650 mm
100 mm
150 mm
600 mm
700 mm
100 mm
150 mm
8kN
6kN
F
A
F
B
O
SOLUTION
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
FC=600
N
FD=500
N
4–129.
SOLUTION
(1)
(2)
Solving Eqs. (1) and (2) yields:
Ans.F
C=600 N
FD=500 N
F
C+F
D=1100
©My=0; 500(0.2) +600(0.2) -F
C(0.2) -F
D(0.2) =0
F
C-F
D=100
The tube supports the four parallel forces. Determine the
magnitudes of forces and acting at Cand Dso that
the equivalent resultant force of the force system acts
through the midpoint Oof the tube.
FD
FC
z
A
D
C
O
400 mm
500 N
600 N
F
C
F
D
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
FA=30
kN
FB=20
kN
FR=190
kN
4–133.
If the resultant force is required to act at the center of the
slab, determine the magnitude of the column loadings
and and the magnitude of the resultant force.FB
FA
2.5 m
2.5 m
0.75 m
0.75 m
3m
3m
0.75 m 90 kN
30 kN
20 kN
x
y
z
F
A
F
B
SOLUTION
(2)
(3)
Solving Eqs.(1) through (3) yields
Ans.
F
A
=
30 kN
F
B
=
20 kN
F
R
=
190 kN
F
R=1.769F
A+1.769F
B+101.54
(MR)y=©My;F
R(3.25) =30(0.75) +20(0.75) +90(3.25) +F
A(5.75) +F
B(5.75)
F
R=0.2F
B+1.8F
A+132
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
FR=5141i+100j+159k6
N
M
RO
=5122i-183k6
N#m
4–134.
Replace the two wrenches and the force, acting on the pipe
assembly, by an equivalent resultant force and couple
moment at point O.
SOLUTION
Equivalent Force and Couple Moment At Point O:
Ans.
The position vectors are and
Ans.=122i-183kN#m
+100k+127.28i-127.28k
+
ij k
0 1.1 0
141.42 0 -141.42
=3ij k
0 0.5 0
00300 3
MRO=r1*F1+r2*F2+M1+M2
MRO=©MO;
r2=51.1j6m.r1=50.5j6m
=5141i+100j+159k6N
=141.42i+100.0j+1300 -141.422k
FR=F1+F2+F3
FR=©F;
=5127.28i-127.28k6N#m
M2=1805cos 45°i-sin 45°k6N#m
M1=5100k6N#m
=5141.42i-141.42k6N
AB
O
z
Cy
100 N
100N·m
300 N
SOLUTION
4–135.
Replace the force system by a wrench and specify the
magnitude of the force and couple moment of the wrench
and the point where the wrench intersects the x–z plane.
z
O
y
0.5 m
3 m
200 N
400 N
2 m
200 N
5
3
4
365
Ans:
FR=379 N
MR=590 N #m
z=2.68 m
x=-2.76 m
SOLUTION
*4–136.
Replace the five forces acting on the plate by a wrench.
Specify the magnitude of the force and couple moment for
the wrench and the point P(x, z) where the wrench intersects
the x–z plane.
z
4 m
400 N
800 N
200 N
4 m
2 m
2 m
367
Ans:
MR=-1.37 kN #m
x=2.68 m
z=-0.233 m
