HW_2_solutions.pdf

(1)

Position Vectors

Description: ± Includes Math Remediation. In this tutorial, students will find the position vectors for two points that have a common origin, find the position vector between the two points, and determine the force acting along a given direction.

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 relative positions. From previous tutorials, we know that Cartesian coordinates give both magnitude and direction of vectors; therefore, position vectors can be expressed as Cartesian vectors with three-dimensional

components. To best illustrate the procedure for finding position vectors, picture two points lying along a number line; the position vector between these two points is their difference. The position vector’s direction is such that it points from the starting point toward the ending point. The position vector’s sign is then given by the direction in which it points; thus if the value increases from the start to the end the sign is positive. If the 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

C

_z

= 1.90 ft

A

9.36 ft

A C

= 27.0

B

8.50 ft

C

B

x

B

y

7.70 ft

4.70 ft

A

B

(

x

1

,

y

1

)

(

x

2

,

y

2

)

1

2 {

x

2

−

x

1

,

y

2

−

y

1

}

2

1

2

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

(2)

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 the distances 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 the end of to end of , .

Using vector addition, we can write . Algebraically solve for and evaluate using Cartesian components. ANSWER:

A

B

= 6.60

=

r

A B z

ft

r

A B

A

B

A

B

A

B r

B = A + r

r

(3)

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 answer in component form.

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

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 unit vectors, its magnitude is one. The unit direction vector is found by dividing the position vector components by 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 Pythagorean theorem. 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) k

r

A B

A

x

B

x

A

y

B

y

A

z

= (−

−

) i+ (

−

) j+ (l−

−

) k

r

A B

B

x

A

x

B

y

A

y

C

z

A

z

= (

−

i+ (

−

j+ (

− l+

k

r

A B

A

x

B

x

)

2

A

y

B

y

)

2

A

z

C

z

)

2

= (−

−

i+ (

−

j+ (l−

k

r

_{A B}

B

x

A

x

)

2

B

y

A

y

)

2

A

z

)

2 , , = -7.70, 4.70, 6.60

=

r

A B

i ft,j ft,k ft

=

u

r r r

A

B

v =

v

x2

+

v

y2

+

v

z2

−

−−−−−−−−−−−

−

= 11.2

=

r

A B

ft

(4)

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 information and the dimensions provided in the figure, find the position vector from to . Express the position vector in component form.

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

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 angle with the y axis. Using right-triangle trigonometry and the Pythagorean theorem the x and y components can be 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 y axis, , 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

=

u

A B

i,j,k

A C

9.36 ft

= 27.0

A

C

A C

C

9.36 ft

A C

27.0

= 3.80

=

r

A C x , , = 3.80, 8.34, -1.90

=

r

_{A C}

i ft,j ft,k ft

(5)

Part D - Position vector from B to C

Find the position vector from to . Express your answer in component form. Express your answers, separated by commas, to three significant figures.

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 common third point, then the position vector between the two arbitrary points can be found from the known position vectors using vector addition. Recall that vectors add tip-to-tail. To add vectors tip-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 at the origin. The position vector is the difference between points and .

2.

Lastly, a third method is to determine the displacement for each coordinate direction between the two points, then determine the correct sign of the component vector. For example, if the first point is and the second point is , then the displacement in the x direction 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 vectors found 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 x component of position vector . Recall the equation ; this equation can be evaluated by summing the Cartesian components for each direction.

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 its length. The tensile force will attempt to pull toward . The force exerted on has a z component with a magnitude of . What is the magnitude of the tension in the cable, ?

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

Hints (2)

B

C

A

C

B

(5,7,− 1)

(7,4,2)

2 A

B

C

B

C

=

−

r

B C

r

A B

r

A C

r

A B

r

A C

r

B C

r

B C

=

r

A C

−

r

A B = 11.5

=

r

B C x

ft

, , = 11.5, 3.64, -8.50

=

r

B C

i ft,j ft,k ft

B

C

T

B C

B

C

B

10 lb

T

B C

(6)

Hint 1. Geometry of the angles

The geometry of the cable will determine the components of force for a given magnitude. Notice that the angle between the position vector and its components is the same for the force and its components. By finding the angle between the position vector and one of its components, the tensile force can be found using 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. The first step in solving for the tension is to solve for the cosine of the angle. What is the cosine of the angle between 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 z component of the tension acts in the negative direction, it has a negative value when calculating

. ANSWER: ANSWER:

z

= -0.576

cos( )=

= 17.4

=

T

B C

(7)

2–94.

The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles of the resultant force. Take x = 20 m,y = 15 m.

a, b, g

SOLUTION

Ans. Ans. Ans. Ans. g = cos-1_a-1466.71 1501.66 b = 168° b = cos-1_a-16.82 1501.66b = 90.6° a = cos-1_a321.66 1501.66b = 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 F_DC= 600a16 34 i -18 34j -24 34kb N F_DB = 800a -6 25.06 i + 4 25.06 j -24 25.06kb N F_DA = 400a 20 34.66 i + 15 34.66 j -24 34.66kb N x z y x y 6 m 4 m 18 m C A D 400 N 800 N 600 N 24 m O 16 m B

(8)

2–111.

The window is held open by chain AB. Determine the length of the chain, and express the 50-lb force acting at A along the chain as a Cartesian vector and determine its coordinate 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 = r_rAB AB = -3.830i - 3.00j + 8.786k 10.043 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 y 5 ft 12 ft 8 ft = 50 lb A B 5 ft z 5 ft F

(9)

2–119.

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

u

SOLUTION

Position Vector:

The magnitudes of the position vectors are

The Angles Between Two Vectors The dot product of two vectors must be

determined first. Then, Ans. u = cos-1 rAO

#

rAB rAOrAB = cos -1 3 3.00 3.00 = 70.5° = 3 = 0122 + 1-321-12 + 01-22 rAC

#

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 C 3 ft 2 ft 2 ft 2 ft z

(10)

2–139. z A O x y 300 mm 300 mm 300 mm F 300 N 30 30

Determine the magnitude of the projected component of the force F = 300 Nacting along line OA.

SOLUTION

Force and Unit Vector: The force vector F and unit vector u_OAmust be determined

first. 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 + 129.90kB

#

A-0.75i + 0.5j + 0.4330kB 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

(11)

Coplanar Force Systems

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

Learning Goal:

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

As shown, a frictionless pulley hangs from a system of springs and a cable. The pulley is equidistant between the two supports attaching the springs to the ceiling. The distance between the supports is . The cable cannot stretch and 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 to hang from the ceiling. Assume that the pulley has no mass. What is the weight of the mass?

Express your answer to three significant figures 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 a

two-dimensional plane (x–y, y–z, x–z). Coplanar force systems allow for a simplified analysis of equilibrium

d = 1.50 m

16.0 N

h = 130.0 cm

(12)

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 the direction of action of the force. For unknown forces, a direction or sign can be assumed; if the answer yields 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 be written in component form. Before the components of the tensile force in the cable can be written, the angle between 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 between the 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. All angles are measured from the positive x axis and are positive in the counterclockwise direction. ANSWER:

F = 0

i+

j= 0

F

x

F

y

= 0

F

x

F

y

= 0

d

h

=

tan

− 1

(

d

),

=

(

)

2h

tan

− 1 2h d

=

tan

− 1

(

2d

),

=

(

)

h

tan

− 1 h 2d

=

tan

− 1

(

2h

),

=

(

)

d

tan

− 1 d 2h

=

tan

− 1

( ),

d

=

( )

h

tan

− 1 h d

(13)

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. Express your answer to three significant figures and include the appropriate units. 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, the tension 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 the mass of the pulley.

Express your answer to three significant = 13.9 Also accepted: = 13.9

=

T

y = 27.7 Also accepted: = 27.7

W =

m = 5.1 kg

36.9 N

h = 147.2 cm

(14)

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 two dimensional. The equation of equilibrium for a coplanar system is:

which can be evaluated with the magnitudes as and .

g = 9.81 m /s

2

i+

j= 0,

F

x

F

y

= 0

F

x

F

y

= 0

(15)

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. All angles 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 be the 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 or horizontal. ANSWER: ANSWER:

T

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

(16)

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 pulley are 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 ; after hanging the mass and pulley, the spring’s length is . Determine the spring constant of Spring 3, . Express your answer to three significant

figures 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 in length 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 this

problem, 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 constant and is the displacement of the spring. To solve for the unknown spring constant of Spring 3, you must know the force, , and the displacement of the spring. Given the unloaded and loaded lengths of Spring 3 find the displacement in meters.

k

3

m = 5.1 kg

m

p

= 1.6 kg

h = 115.0 cm

₃

= 34.5 cm

= 42.5 cm

3

k

3

= ks,

F

s

k

s

h = 115.0 cm

m

p

= 1.6 kg

m = 5.1 kg

g = 9.81 m /s

2 = 39.2

T =

= ks

F

s

k

s

F

s

3

(17)

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 find it, draw the free-body diagram of the spring and cable (on the side of the pulley containing Spring 3). From the free-body diagram, notice that the tension in the cable is transferred entirely to the spring. ANSWER: ANSWER:

F

s = 8.00×10−2

=

s

3

m

= 490

=

k

3

(18)

3–3.

SOLUTION

Free-Body Diagram: By observation, the force has to support the entire weight

of 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; FACcos u - FABcos 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 of 500 kg. Determine the force in each of the cables AB and AC as a function of If the maximum tension allowed in each cable is 5 kN, determine the shortest lengths of cables AB and AC that can be used for the lift. The center of gravity of the container is located at G.

u. _A C B 1.5 m 1.5 m G F θ θ

(19)

■_3–15.

The spring has a stiffness of and an unstretched length of 200 mm. Determine the force in cables BC and BD when the spring is held in the position shown.

k = 800 N>m A k 800 N/m B D 500 mm 400 mm 400 mm 300 mm C

SOLUTION

The Force in The Spring:Thespringstretches

Applying 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° - FBDa 3 5b = 0 0.7071FBC + 0.8FBD= 240 :+ ©Fx = 0; FBC cos 45° + FBDa 4 5b - 240 = 0 Fsp = ks = 800(0.3) = 240 N s = l - l0 = 0.5 - 0.2 = 0.3 m.

(20)

3–30.

SOLUTION

Geometry: The angle u which the surface makes with the horizontal is to be

determined 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 8.4957m_B + 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 the mass of block B needed to hold it in the equilibrium position shown. mB B A y x 0.4 m 0.4 m 60 y 2.5x2

(21)

3–31.

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

A B E C D 4 3 5 30 30

SOLUTION

Equations 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 the free-body diagram of joint B shown in Fig. b,

Ans. Ans. FBA = 86.6 lb :+ ©Fx = 0; 69.78cos 30° + 43.61a3₅b - FBA = 0 FBC = 69.78 lb = 69.8 lb + c ©Fy = 0; FBC sin 30° - 43.61a4₅b = 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; FEDcos 30° - FEBa3₅b = 0

(22)

*3–56.

SOLUTION

Cartesian Vector Notation:

Determine the force in each cable needed to support the 3500-lb platform. Set d = 2 ft. 3 ft d y x 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 _{+ 1}2 _{+ (-10)}2≤ = -0.3698FADi+ 0.09245FADj - 0.9245FADk FAC = FAC¢ 2i + 3j - 10k 222 _{+ 3}2 _{+ (-10)}2≤ = 0.1881FACi + 0.2822FACj - 0.9407FACk FAB = FAB¢ 4i - 3j - 10k 242 _{+ (-3)}2 _{+ (-10)}2≤ = 0.3578FABi - 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 0.3578F_AB + 0.1881F_AC - 0.3698F_AD = 0 + (-0.8944FAB - 0.9407FAC - 0.9245FAD + 3500)k = 0 Equations of Equilibrium: ©F = 0; FAB+ FAC + FAD + F = 0

(23)

SOLUTION

Force Vectors: We can express each of the forces on the free-body diagram shown in

Fig. 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 lb FAB = 808 lb -6 7 FAB -6 7 FAC + FAE - 1200 = 0 -3 7 FAB -2 7 FAC + 500 = 0 2 7 FAB - 3 7 FAC = 0 ¢2₇ FAB -3 7 FAC≤i + ¢ -3 7 FAB -2 7 FAC + 500≤j + ¢ -6 7 FAB -6 7 FAC + FAE - 1200≤k = 0 ¢2 7 FAB i -3 7 FAB j -6 7 FAB k≤ + ¢-3 7 FAC i -2 7 FAC j -6 7 FAC k≤ + (500j - 1200k) + F AE k = 0 gF= 0; F AB+ F AC + F AD + F AE = 0 FAE = F AE k FAD = F_ADC (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 = F AC C ( - 15 - 0)i + ( - 10 - 0)j + ( - 30 - 0)k 2(-15 - 0)2 ₊ _{( - 10 - 0)}2 ₊ _{( - 30 - 0)}2S = -3 7 FAC i -2 7 FAC j -6 7 FAC k FAB = F_ABC (10 - 0)i + ( - 15 - 0)j + ( - 30 - 0)k 2(10 - 0)2 ₊ _{( - 15 - 0)}2₊ _{( - 30 - 0)}2S = 2 7 FAB i -3 7 FAB j -6 7 FAB k 3–61.

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

15 ft 15 ft 10 ft 10 ft z x B _E D C A y 30 ft 12.5 ft