Post on 08-Feb-2022
Problem 2.130 The magnitudes jUj D 10 and jVj D20.
(a) Use the definition of the cross product to determineU ð V.
(b) Use the definition of the cross product to determineV ð U.
(c) Use Eq. (2.34) to determine U ð V.
(d) Use Eq. (2.34) to determine V ð U.
U
V
x
y
45°30°
Solution: From Eq. (228) U ð V D jUjjVj sin �e. From the sketch,the positive z-axis is out of the paper. For U ð V, e D �1k (points intothe paper); for V ð U, e D C1k (points out of the paper). The angle� D 15°, hence (a) U ð V D �10��20��0.2588��e� D 51.8e D �51.8k.Similarly, (b) V ð U D 51.8e D 51.8k (c) The two vectors are:
U D 10�i cos 45° C j sin 45� D 7.07i C 0.707j,
V D 20�i cos 30° C j sin 30°� D 17.32i C 10j
U ð V D∣∣∣∣∣∣
i j k7.07 7.07 017.32 10 0
∣∣∣∣∣∣ D i�0� � j�0� C k�70.7 � 122.45�
D �51.8k
(d) V ð U D∣∣∣∣∣∣
i j k17.32 10 07.07 7.07 0
∣∣∣∣∣∣ D i�0� � j�0� C k�122.45 � 70.7�
D 51.8k
Problem 2.131 The force F D 10i � 4j (N). Deter-mine the cross product rAB ð F.
y
x
B
A
rAB
(6, 3, 0) m
(6, 0, 4) m
F
z
Solution: The position vector is
rAB D �6 � 6�i C �0 � 3�j C �4 � 0�k D 0i � 3j C 4k
The cross product:
rAB ð F D∣∣∣∣∣∣
i j k0 �3 410 �4 0
∣∣∣∣∣∣ D i�16� � j��40� C k�30�
D 16i C 40j C 30k (N-m)
y
x
B
A
rAB
(6, 3, 0)
(6, 0, 4) Fz
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c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 3.25 A traffic engineer wants to suspend a200-lb traffic light above the center of the two rightlanes of a four-lane thoroughfare as shown. Determinethe tensions in the cables AB and BC.
30 ft
CA
B10 ft
20 ft80 ft
Solution:∑
Fx : � 6p37
TAB C 2p5
TBC D 0
∑Fy :
1p37
TAB C 1p5
TBC � 200 lb D 0
Solving: TAB D 304 lb, TBC D 335 lb
61
2
1
TBC
TAB
200 lb
Problem 3.26 Cable AB is 3 m long and cable BC is4 m long. The mass of the suspended object is 350 kg.Determine the tensions in cables AB and BC.
C
B
A
5m
Solution:∑
Fx : � 3
5TAB C 4
5TBC D 0
∑Fy :
4
5TAB C 3
5TBC � 3.43 kN D 0
TAB D 2.75 kN, TBC D 2.06 kN
4
4
33
TAB
TAC
3.43 kN
98
c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 3.27 In Problem 3.26, the length of cable ABis adjustable. If you don’t want the tension in either cableAB or cable BC to exceed 3 kN, what is the minimumacceptable length of cable AB?
Solution: Consider the geometry:
We have the constraints
LAB2 D x2 C y2, ,4 m+2 D ,5 m % x+2 C y2
These constraint imply
y D√
,10 m+x % x2 % 9 m2
L D√
,10 m+x % 9 m2
Now draw the FBD and write the equations in terms of x
∑Fx : % xp
10x % 9TAB C 5 % x
4TBC D 0
∑Fy :
p10x % x2 % 9p
10x % 9TAB C
p10x % x2 % 9
4TBC % 3.43 kN D 0
If we set TAB D 3 kN and solve for x we find x D 1.535, TBC D2.11 kN < 3 kN
Using this value for x we find that LAB D 2.52 m
x
y
5−x
4 mLAB
TAB
TBC
x5−x
yy
4
3.43 kN
c# 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Problem 4.142 The vector sum of the forces acting onthe truss is zero, and the sum of the moments about theorigin O is zero.
(a) Determine the forces Ax , Ay , and B.(b) If you represent the 2-kip, 4-kip, and 6-kip forces
by a force F, what is F, and where does its line ofaction intersect the y axis?
(c) If you replace the 2-kip, 4-kip, and 6-kip forces bythe force you determined in (b), what are the vectorsum of the forces acting on the truss and the sumof the moments about O?
2 kip
4 kip
6 kip
x
6 ft
3 ft
3 ft
3 ft
AxO
Ay B
y
Solution: (a) The sum of the forces is∑FX D -AX % 2 % 4 % 6,i D 0,
from which AX D 12 kip
∑FY D -AY C B,j D 0.
The sum of the moments about the origin is
∑MO D -3,-6, C -6,-4, C -9,-2, C 6-B, D 0,
from which B D %10j kip. (b) Substitute into the force balance eq-uation to obtain AY D %B D 10 kip. (b) The force in the new systemwill replace the 2, 4, and 6 kip forces, F D -%2 % 4 % 6,i D %12i kip.The force must match the moment due to these forces: FD D 3-6, C-6,-4, C -9,-2, D 60 kip ft, from which D D 60
12D 5 ft, or the action
line intersects the y axis 5 ft above the origin. (c) The new system isequivalent to the old one, hence the sum of the forces vanish and thesum of the moments about O are zero.
c# 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Problem 4.149 Consider the system shown in Problem4.148. The tension in each of the cables AB and CD is400 N. If you represent the forces exerted on the rightpost by the cables by a force F, what is F, and wheredoes its line of action intersect the y axis?
Solution: From the solution of Problem 4.148, the tensions are
TAB D %400*i cos*%26.6°)Cj sin*%26.6°)) D %357.77i C 178.89j,
and
TCD D %400*i cos*%20.6°)Cj sin*%20.6°)) D %374.42i C 140.74j.
The equivalent force is equal to the sum of these forces:
F D *%357.77 % 374.42)i C *178.77 C 140.74)j
D %732.19i C 319.5j (N).
The sum of the moments about O is∑M D 0.3*357.77) C 0.8*140.74 C 178.89)k D 363k (N-m).
The intersection is D D 363
732.19D 0.496 m on the positive y axis.
c# 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Problem 5.36 This structure, called a truss, has a pinsupport at A and a roller support at B and is loaded bytwo forces. Determine the reactions at the supports.
Strategy: Draw a free-body diagram, treating the entiretruss as a single object.
b b b b
4 kN 2 kN
A
45°
B
b
30°
Solution:∑MA : %*4 kN)
p2b % *2 kN cos 30°)3 b
C *2 kN sin 30°)b C B*4 b) D 0∑Fx : Ax C 4 kN sin 45° % 2 kN sin 30° D 0
∑Fy : Ay % 4 kN cos 45° % 2 kN cos 30° C B D 0
Solving:
Ax D %1.828 kN, Ay D 2.10 kN, B D 2.46 kN
B
45° 30°
Ay
Ax
4 kN 2 kN
c# 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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