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

68

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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

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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

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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.

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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.

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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

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