hw_2_solutions.pdf

Position VectorsDescription: ± Includes Math Remediation. In this tutorial, students will find the position vectors for two points thathave a common origin, find the position vector between the two points, and determine the force acting along a givendirection.

Learning Goal:

To find a position vector between two arbitrary points.

As shown, two cables connect three points. is below by a distance and connected to by a cable long. Cable forms an angle with the positive y axis. is above and the distances

and are and , respectively.

Part A - Position vector from A to B

Using the dimensions in the figure, find the position vector from to in component form.

Express your answers, separated by commas, to three significant figures.

Hints (3)

Hint 1. Position vectors

Position vectors give information about the distance between two points and the direction of their relativepositions. From previous tutorials, we know that Cartesian coordinates give both magnitude and direction ofvectors; therefore, position vectors can be expressed as Cartesian vectors with three-dimensionalcomponents. To best illustrate the procedure for finding position vectors, picture two points lying along anumber line; the position vector between these two points is their difference. The position vector’s directionis such that it points from the starting point toward the ending point. The position vector’s sign is then givenby the direction in which it points; thus if the value increases from the start to the end the sign is positive. Ifthe value decreases from the start to the end the sign is negative.

So if two points lying in the x–y plane have coordinates and , the position vector from to is . Naturally, the position vector from to is the negative of the position vector

from to . Care should be taken to determine the correct sign for each component.

C A = 1.90 ftCz A9.36 ft AC = 27.0 B 8.50ft C Bx

B y 7.70 ft 4.70 ft

A B

( , )x1 y1 ( , )x2 y2 12 { − , − }x2 x1 y2 y1 2 1

1 2

MasteringEngineering: Lecture 3 https://session.masteringengineering.com/myct/itemView?view=solution&...

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Hint 2. Calculate the z component of the position vector from A to B

What is the z component of the position vector from to ?

Express your answer to three significant figures.

ANSWER:

Hint 3. Identify the equations used to calculate the components

Which of the following equations expresses the position vector, , from to correctly in terms of thedistances indicated in the picture?

Hints (1)

Hint 1. A useful diagram

The diagram below shows a sketch of arbitrary vectors and and the vector connecting theend of to end of , .

Using vector addition, we can write . Algebraically solve for and evaluate usingCartesian components.

ANSWER:

A B

= 6.60=rABz ft

rAB A B

A BA B r

B = A + r r


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ANSWER:

Part B - Unit direction vector for line AB

For the position vector found in Part A, find the unit direction vector acting in the same direction. Express your answerin component form.


Hints (2)

Hint 1. Unit direction vector

A unit direction vector is a unit vector with the same direction and sign as the position vector. Like all unitvectors, its magnitude is one. The unit direction vector is found by dividing the position vector componentsby the magnitude of the position vector:

.

Hint 2. Calculate the magnitude of the position vector

The magnitude of a vector with known components is found through the application of the Pythagoreantheorem. Find the magnitude of the position vector from to .


Hints (1)

Hint 1. Equation for the magnitude of a vector

The formula for the magnitude of a vector is

.

ANSWER:

ANSWER:

= ( + ) i+ ( − ) j+ ( − l) krAB A x B x A y B y A z

= (− − ) i+ ( − ) j+ (l− − ) krAB B x A x B y A y Cz A z

= ( − i+ ( − j+ ( − l+ krAB A x B x)2 A y B y)

2 A z Cz)2

= (− − i+ ( − j+ (l− krAB B x A x)2 B y A y)

2 A z)2

, , = -7.70, 4.70, 6.60=rAB i ft,j ft,k ft

=urrr

A B

v = + +vx2 vy2 vz2− −−−−−−−−−−−−

= 11.2=rAB ft


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Part C - Position vector from A to C

The length of cable is , and the cable forms the angle with the y axis. Given this informationand the dimensions provided in the figure, find the position vector from to . Express the position vector incomponent form.


Hints (2)

Hint 1. Finding the x component of the position vector AC

The x component of the position vector is not explicitly given. It must be calculated from the length of cable and other information given.

From this diagram, you can see that the side of the triangle from point to the y axis forms a right anglewith the y axis. Using right-triangle trigonometry and the Pythagorean theorem the x and y components canbe determined.

Hint 2. Calculate the x component of the position vector

Using the known values for the length of the cable, , the angle cable makes with the positive yaxis, , and the z component, determine the x component.

Express your answer to three significant figures and include the appropriate units.

ANSWER:

ANSWER:

, , = -0.689, 0.420, 0.590=uAB i,j,k

AC 9.36 ft = 27.0A C

AC

C

9.36ft AC27.0

= 3.80=rACx

, , = 3.80, 8.34, -1.90=rAC i ft,j ft,k ft


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Part D - Position vector from B to C

Find the position vector from to . Express your answer in component form.


Hints (2)

Hint 1. Determining the position vector between two points

The position between two arbitrary points can be found using several methods.

If the two points are already defined (position vectors are known) with respect to a commonthird point, then the position vector between the two arbitrary points can be found from theknown position vectors using vector addition. Recall that vectors add tip-to-tail. To add vectorstip-to-tip or tail-to-tail requires a sign change.

1.

A second, simpler method is to write the Cartesian coordinates of the points with point atthe origin. The position vector is the difference between points and .

2.

Lastly, a third method is to determine the displacement for each coordinate direction betweenthe two points, then determine the correct sign of the component vector. For example, if thefirst point is and the second point is , then the displacement in the xdirection is positive , etc.

3.

For the first method, recognize that point is the common point for which and are already defined.The position vector between the and then can be written in terms of the previous position vectorsfound in Part A and Part C: .

Hint 2. Calculate the x component of the position vector from B to C

Using the vector components for position vectors and found in Parts A and B, determine the xcomponent of position vector . Recall the equation ; this equation can be evaluatedby summing the Cartesian components for each direction.


ANSWER:

ANSWER:

Part E - Tension in the cable

A cable is attached between and . The cable is attached in such a way that it has a tension of along itslength. The tensile force will attempt to pull toward . The force exerted on has a z component with a magnitudeof . What is the magnitude of the tension in the cable, ?


Hints (2)

B C

AC B

(5,7,− 1) (7,4,2)2

A B CB C= −rBC rAB rAC

rAB rACrBC = −rBC rAC rAB

= 11.5=rBCx ft

, , = 11.5, 3.64, -8.50=rBC i ft,j ft,k ft

B C T B CB C B

10 lb TBC


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Hint 1. Geometry of the angles

The geometry of the cable will determine the components of force for a given magnitude. Notice that theangle between the position vector and its components is the same for the force and its components. Byfinding the angle between the position vector and one of its components, the tensile force can be foundusing the known force component.

Hint 2. Find the cosine of the angle between the cable and the positive z axis

The angle, , is the same between the z components of both the position vector and the tensile force. Thefirst step in solving for the tension is to solve for the cosine of the angle. What is the cosine of the anglebetween the z axis and the cable?


Hints (1)

Hint 1. Direction of the z component

Recall that is defined between the cable and the positive z axis. However, because the zcomponent of the tension acts in the negative direction, it has a negative value when calculating

.

ANSWER:

ANSWER:

z

= -0.576cos( )=

= 17.4=TBC


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2–94.

The tower is held in place by three cables. If the force ofeach cable acting on the tower is shown, determine themagnitude and coordinate direction angles of theresultant force. Take , .y = 15 mx = 20 m

a, b, g

SOLUTION

Ans.

Ans.

Ans.

Ans.g = cos-1a -1466.711501.66

b = 168°

b = cos-1a -16.821501.66

b = 90.6°

a = cos-1a 321.661501.66

b = 77.6°

= 1501.66 N = 1.50 kN

FR = 2(321.66)2 + (-16.82)2 + (-1466.71)2

= {321.66i - 16.82j - 1466.71k} N

FR = FDA + FDB + FDC

FDC = 600a1634

i -1834

j -2434

kb N

FDB = 800a -625.06

i +4

25.06 j -

2425.06

kb N

FDA = 400a 2034.66

i +15

34.66 j -

2434.66

kb N

x

z

yxy6 m

4 m

18 m

C

A

D

400 N 800 N600 N

24 m

O16 m

B

2–111.

The window is held open by chain AB. Determine thelength of the chain, and express the 50-lb force acting at Aalong the chain as a Cartesian vector and determine itscoordinate direction angles.

SOLUTION

Unit Vector: The coordinates of point A are

Then

Ans.

Force Vector:

Ans.

Coordinate Direction Angles: From the unit vector obtained above, we have

Ans.

Ans.

Ans.cos g = 0.8748 g = 29.0°

cos b = -0.2987 b = 107°

cos a = -0.3814 a = 112°

uAB

= 5-19.1i - 14.9j + 43.7k6 lb

F = FuAB = 505-0.3814i - 0.2987j + 0.8748k6 lb

= -0.3814i - 0.2987j + 0.8748k

uAB =rAB

rAB=

-3.830i - 3.00j + 8.786k10.043

rAB = 21-3.83022 + 1-3.0022 + 8.7862 = 10.043 ft = 10.0 ft

= 5-3.830i - 3.00j + 8.786k6 ft

rAB = 510 - 3.8302i + 15 - 8.002j + 112 - 3.2142k6 ft

A15 cos 40°, 8, 5 sin 40°2 ft = A13.830, 8.00, 3.2142 ft 40˚

x

y5 ft

12 ft

8 ft

= 50 lb

A

B5 ft

z

5 ft

F

2–119.

Determine the angle between the y axis of the pole andthe wire AB.

u

SOLUTIONPosition Vector:

The magnitudes of the position vectors are

The Angles Between Two Vectors The dot product of two vectors must bedetermined first.

Then,

Ans.u = cos-1 rAO# rAB

rAO rAB= cos-1 3

3.00 3.00= 70.5°

= 3

= 0122 + 1-321-12 + 01-22rAC

# rAB = 1-3j2 # 12i - 1j - 2k2

U:

rAC = 3.00 ft rAB = 222 + 1-122 + 1-222 = 3.00 ft

= 52i - 1j - 2k6 ft

rAB = 512 - 02i + 12 - 32j + 1-2 - 02k6 ft

rAC = 5-3j6 ft

θ

x

y

B

A

C3 ft

2 ft

2 ft

2 ft

z

2–139.

zA

O

x y

300 mm

300 mm

300 mm

F 300 N

30

30Determine the magnitude of the projected component ofthe force acting along line OA.F = 300 N

SOLUTIONForce and Unit Vector: The force vector F and unit vector uOA must be determinedfirst. From Fig. a

Vector Dot Product: The magnitude of the projected component of F along line OA is

Ans.= 242 N

= (-75)(-0.75) + 259.81(0.5) + 129.90(0.4330)

FOA = F # uOA = A -75i + 259.81j + 129.90k B # A -0.75i + 0.5j + 0.4330k B

uOA =rOA

rOA=

(-0.45 - 0)i + (0.3 - 0)j + (0.2598 - 0)k

2(-0.45 - 0)2 + (0.3 - 0)2 + (0.2598 - 0)2= -0.75i + 0.5j + 0.4330k

= {-75i + 259.81j + 129.90k} N

F = (-300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k)

Coplanar Force SystemsDescription: In this tutorial, students will use the spring equation to solve for unknown forces in two dimensions andfind the new length of the springs and will also find an unknown spring constant from an applied force and the system’sgeometry.

Learning Goal:

To use the equations of equilibrium to find unknown forces in two dimensions; understand the relationship between aspring’s unloaded length, its displacement, and its loaded length; and use the spring equation to solve problems involvingmultiple springs.

As shown, a frictionless pulley hangs from a system of springs and a cable. The pulley is equidistant between the twosupports attaching the springs to the ceiling. The distance between the supports is . The cable cannot stretchand its length between the two springs is 1.8 m.

Part A - Finding an unknown weight

As shown, a mass is hung from the pulley. This mass causes a tensile force of in the cable and the pulley tohang from the ceiling. Assume that the pulley has no mass. What is the weight of the mass?

Express your answer to three significantfigures and include the appropriate units.

Hints (4)

Hint 1. Coplanar force system

A coplanar force system is a collection of forces that act entirely within a particular plane of interest.Therefore, all the forces acting on a point of interest are two dimensional and act only within atwo-dimensional plane (x–y, y–z, x–z). Coplanar force systems allow for a simplified analysis of equilibrium

d = 1.50 m

16.0Nh = 130.0 cm


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because only two directions need to be considered. So simplifies to

for the x–y plane and can be evaluated using and .

It is important to continue using a consistent sign convention for all forces, since the sign accounts for thedirection of action of the force. For unknown forces, a direction or sign can be assumed; if the answeryields a negative value the force acts in the opposite direction from what was assumed.

Hint 2. Identify the equation for the angle of action of the tension

To use the equations of equilibrium for the component directions, the forces acting on a particle must bewritten in component form. Before the components of the tensile force in the cable can be written, the anglebetween the force and a coordinate axis needs to be determined from the geometry of the system.In this problem, two separate angles can be used to solve the problem. The first, , is the angle betweenthe positive y axis and the cable. The second, , is the angle between the x axis and the cable.

Identify the correct trigonometric relationships defining the angles in terms of the dimensions and .

ANSWER:

Hint 3. Draw the free-body diagram of the pulley

Draw the free body diagram of the pulley. Recall that the pulley has no mass.

Start all forces in the center of the pulley. Draw all forces as tensile forces, or pulling forces. Allangles are measured from the positive x axis and are positive in the counterclockwise direction.

ANSWER:

F = 0

i+ j= 0Fx F y

= 0Fx = 0Fy

d h

= ( ), = ( )tan−1 d2h

tan−1 2hd

= ( ), = ( )tan−1 2dh

tan−1 h2d

= ( ), = ( )tan−1 2hd

tan−1 d2h

= ( ), = ( )tan−1 dh

tan−1 hd


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Hint 4. Find the y component of the tension

Find the magnitude of the y component of the tensile force in one side of the cable.


ANSWER:

ANSWER:

Part B - Finding the mass of the pulley

As shown, an object with mass is hung from a pulley and spring system. When the object is hung, thetension in the cable is and the pulley is below the ceiling.

Because the tensile force is greater than the object’s weight, the pulley cannot be massless as assumed. Find themass of the pulley.

Express your answer to three significant

= 13.9

Also accepted: = 13.9

=Ty

= 27.7

Also accepted: = 27.7

W =

m = 5.1 kg36.9 N h = 147.2cm


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figures and include the appropriate units.For this problem, use .

Hints (3)

Hint 1. Coplanar forces

Coplanar force systems have forces that act entirely within a plane. In other words, all forces are twodimensional. The equation of equilibrium for a coplanar system is:

which can be evaluated with the magnitudes as and .

g= 9.81 m /s2

i+ j= 0,Fx F y

= 0Fx = 0Fy


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Hint 2. Draw the free-body diagram of the pulley

Draw the free-body diagram of the pulley. Recall that the pulley has a mass.

Start all forces in the center of the pulley. Draw all forces as tensile forces, or pulling forces. Allangles are measured from the positive x axis and are positive in the counterclockwise direction.

ANSWER:

Hint 3. Determine the equation of equilibrium

Determine the equation of equilibrium needed to solve for the object of the pulley. For this equation let bethe tension in the cable, be the mass of the hanging weight, and be the mass of the pulley. Let be

the angle between the cable and the positive y axis and be the angle between the cable and the ceiling orhorizontal.

ANSWER:

ANSWER:

Tm m p

2T cos( )− 9.81m − 9.81m p

2T sin( )− 9.81m − 9.81m p

2T sin( )+ 9.81m + 9.81m p

2T cos( )+ 9.81m + 9.81m p

mass of pulley = = 1.60


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Part C - Finding the spring constant

As shown, Spring 3, which has an unknown spring constant , replaces Spring 2. The mass of the weight and pulleyare unchanged: and . However, because of the different spring constant the distance the

pulley hangs below the ceiling is now, . The unloaded length of Spring 3 is ; afterhanging the mass and pulley, the spring’s length is . Determine the spring constant of Spring 3, .

Express your answer to three significantfigures and include the appropriate units.

Hints (3)

Hint 1. Spring force equation

For linear elastic springs, the equation defining the force developed as they stretch is

where is the spring constant, which has units of force over length, and is the deformation or change inlength of the spring.

Hint 2. Find the magnitude of the tension

Given , the mass of the pulley, , and the mass of the hanging object,

, find the magnitude of the tension in the cable.

Express your answer to three significant figures and include the appropriate units. For thisproblem, use .

ANSWER:

Hint 3. Find the displacement in the spring, s

The spring force equation for a linear elastic spring is . In this equation is the spring constantand is the displacement of the spring. To solve for the unknown spring constant of Spring 3, you mustknow the force, , and the displacement of the spring. Given the unloaded and loaded lengths of Spring 3find the displacement in meters.


k3m = 5.1 kg = 1.6kgm p

h = 115.0cm = 34.5 cm3= 42.5 cm3 k3

= ks,Fsk s

h = 115.0 cm = 1.6kgm p

m = 5.1 kg

g= 9.81 m /s2

= 39.2T =

= ksFs ks

Fss3


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Hints (1)

Hint 1. Force in Spring 3

To use the spring force equation the force in or applied to the spring, , must be known. To findit, draw the free-body diagram of the spring and cable (on the side of the pulley containing Spring3). From the free-body diagram, notice that the tension in the cable is transferred entirely to thespring.

ANSWER:

ANSWER:

Fs

= 8.00×10−2=s3 m

= 490=k3


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3–3.

SOLUTIONFree-Body Diagram: By observation, the force has to support the entire weightof the container. Thus,

Equations of Equilibrium:

Thus,

Ans.

If the maximum allowable tension in the cable is 5 kN, then

From the geometry, and Therefore

Ans.l =1.5

cos 29.37°= 1.72 m

u = 29.37°.l =1.5

cos u

u = 29.37°

2452.5 cos u = 5000

FAC = FAB = F = 52.45 cos u6 kN

4905 - 2F sin u = 0 F = 52452.5 cos u6 N+ c ©Fy = 0;

FAC cos u - FAB cos u = 0 FAC = FAB = F:+ ©Fx = 0;

F1 = 50019.812 = 4905 N.F1

The lift sling is used to hoist a container having a mass of500 kg. Determine the force in each of the cables AB andAC as a function of If the maximum tension allowed ineach cable is 5 kN, determine the shortest lengths of cablesAB and AC that can be used for the lift. The center ofgravity of the container is located at G.

u. A

CB

1.5 m 1.5 m

G

F

θ θ

■3–15.

The spring has a stiffness of and an unstretchedlength of 200 mm. Determine the force in cables BC and BDwhen the spring is held in the position shown.

k = 800 N>m

A Bk 800 N/m

D

500 mm 400 mm

400 mm

300 mm

C

SOLUTIONThe Force in The Spring: sehcterts gnirps ehTApplying Eq. 3–2, we have


(1)

(2)

Solving Eqs. (1) and (2) yields,

Ans.FBD = 171 N FBC = 145 N

FBC = 0.8485FBD

+ c ©Fy = 0; FBC sin 45° - FBDa35b = 0

0.7071FBC + 0.8FBD = 240

:+ ©Fx = 0; FBC cos 45° + FBDa45b - 240 = 0

Fsp = ks = 800(0.3) = 240 N

s = l - l0 = 0.5 - 0.2 = 0.3 m.

3–30.

SOLUTIONGeometry: The angle u which the surface makes with the horizontal is to bedetermined first.

Free-Body Diagram: The tension in the cord is the same throughout the cord and is equal to the weight of block B, .


(1)

(2)

Solving Eqs. (1) and (2) yields

Ans.mB = 3.58 kg N = 19.7 N

8.4957mB + 0.4472N = 39.24

+ c ©Fy = 0; mB (9.81) sin 60° + Ncos 63.43° - 39.24 = 0

N = 5.4840mB

:+ ©Fx = 0; mB (9.81) cos 60° - Nsin 63.43° = 0

WB = mB (9.81)

u = 63.43°

tan u `x = 0.4 m

=dy

dx`x = 0.4 m

= 5.0x `x = 0.4 m

= 2.00

A 4-kg sphere rests on the smooth parabolic surface.Determine the normal force it exerts on the surface and themass of block B needed to hold it in the equilibriumposition shown.

mB

B

A

y

x0.4 m

0.4 m

60

y 2.5x2

3–31.

If the bucket weighs 50 lb, determine the tension developedin each of the wires.

A

B

E

C

D4

3

5

30

30

SOLUTIONEquations of Equilibrium: First, we will apply the equations of equilibrium along the x and y axes to the free-body diagram of joint E shown in Fig. a.

(1)

(2)

Solving Eqs. (1) and (2), yields

Ans.

Using the result and applying the equations of equilibrium to thefree-body diagram of joint B shown in Fig. b,

Ans.

Ans.FBA = 86.6 lb

:+ ©Fx = 0; 69.78 cos 30° + 43.61a 35b - FBA = 0

FBC = 69.78 lb = 69.8 lb

+ c ©Fy = 0; FBC sin 30° - 43.61a 45b = 0

FEB = 43.61 lb

FED = 30.2 lb FEB = 43.61 lb = 43.6 lb

+ c ©Fy = 0; FED sin 30° + FEBa45b - 50 = 0

:+ ©Fx = 0; FED cos 30° - FEBa 35b = 0

*3–56.

SOLUTIONCartesian Vector Notation:

Determine the force in each cable needed to support the3500-lb platform. Set .d = 2 ft

3 ft d yx

C

D

B

A

3500 lb

4 ft

3 ft

10 ft

4 ft

2 ft

z

F = {3500k} lb

FAD = FAD¢ -4i + 1j - 10k

2(-4)2 + 12 + (-10)2≤ = -0.3698FAD i + 0.09245FAD j - 0.9245FAD k

FAC = FAC¢ 2i + 3j - 10k

222 + 32 + (-10)2≤ = 0.1881FAC i + 0.2822FAC j - 0.9407FACk

FAB = FAB¢ 4i - 3j - 10k

242 + (-3)2 + (-10)2≤ = 0.3578FAB i - 0.2683FABj - 0.8944FABk

(0.3578FAB + 0.1881FAC - 0.3698FAD) i + (-0.2683FAB + 0.2822FAC + 0.09245FAD)j

Equating i, j, and k components, we have

(1)

(2)

(3)

Solving Eqs. (1), (2) and (3) yields

Ans.

Ans.FAD = 1703.62 lb = 1.70 kip

FAB = 1369.59 lb = 1.37 kip FAC = 744.11 lb = 0.744 kip

-0.8944FAB - 0.9407FAC - 0.9245FAD + 3500 = 0

-0.2683FAB + 0.2822FAC + 0.09245FAD = 0

0.3578FAB + 0.1881FAC - 0.3698FAD = 0

+ (-0.8944FAB - 0.9407FAC - 0.9245FAD + 3500)k = 0


©F = 0; FAB + FAC + FAD + F = 0

SOLUTION

Force Vectors: We can express each of the forces on the free-body diagram shown inFig. a in Cartesian vector form as

Equations of Equilibrium: Equilibrium requires

Equating the i, j, and k components yields

(1)

(2)

(3)

Solving Eqs. (1) through (3) yields

Ans.Ans.Ans.FAE = 2354 lb = 2.35 kip

FAC = 538 lbFAB = 808 lb

-

67

FAB -

67

FAC + FAE - 1200 = 0

-

37

FAB -

27

FAC + 500 = 0

27

FAB -

37

FAC = 0

¢27

FAB -

37

FAC≤ i + ¢- 37

FAB -

27

FAC + 500≤j + ¢- 67

FAB -

67

FAC + FAE - 1200≤k = 0

¢27

FAB i -

37

FAB j -

67

FAB k≤ + ¢- 37

FAC i -

27

FAC j -

67

FAC k≤ + (500j - 1200k) + FAE k = 0

gF = 0; FAB + FAC + FAD + FAE = 0

FAE = FAE k

FAD = FAD C (0 - 0)i + (12.5 - 0)j + (-30 - 0)k

2(0 - 0)2+ (12.5 - 0)2

+ (-30 - 0)2S = {500j - 1200k} lb

FAC = FAC C (-15 - 0)i + (-10 - 0)j + (-30 - 0)k

2(-15 - 0)2+ (-10 - 0)2

+ (-30 - 0)2S = -

37

FAC i -

27

FAC j -

67

FAC k

FAB = FAB C (10 - 0)i + (-15 - 0)j + (-30 - 0)k

2(10 - 0)2+ (-15 - 0)2

+ (-30 - 0)2S =

27

FAB i -

37

FAB j -

67

FAB k

3–61.

If cable AD is tightened by a turnbuckle and develops atension of 1300 lb, determine the tension developed incables AB and AC and the force developed along theantenna tower AE at point A.

15 ft 15 ft

10 ft10 ft

z

x

B ED

C

A

y

30 ft

12.5 ft

hw_2_solutions.pdf

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