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MasteringEngineering: Lecture 3

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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. C is below A by a distance C z = 1.90 ftand connected to A by a cable 9.36 ftlong. Cable A C forms an angle = 27.0 with the positive y axis. B is 8.50 ftabove C and the distances B x and B y are 7.70 ftand 4.70 ft, respectively.

Part A - Position vector from A to B Using the dimensions in the figure, find the position vector from A to B 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 (x1,y1)and (x2 ,y2), the position vector from 1 to 2 is {x2 − x1,y2 − y1}. Naturally, the position vector from 2 to 1 is the negative of the position vector from 1 to 2. Care should be taken to determine the correct sign for each component.

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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 A to B ? Express your answer to three significant figures. ANSWER:

rA B z =

= 6.60 ft

Hint 3. Identify the equations used to calculate the components Which of the following equations expresses the position vector, rA B , from A to B 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 A and B and the vector connecting the end of A to end of B , r.

Using vector addition, we can write B = A + r. Algebraically solve for r and evaluate using Cartesian components.

ANSWER:

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rA B = (A x + B x)i+ (A y − B y)j+ (A z − l)k rA B = (− B x − A x)i+ (B y − A y)j+ (l− C z − A z)k rA B = (A x − B x)2 i+ (A y − B y)2 j+ (A z − l+ C z)2 k rA B = (− B x − A x)2 i+ (B y − A y)2 j+ (l− A z)2 k

ANSWER: ,

rA B =

,

= -7.70, 4.70, 6.60 ift,jft,k ft

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: ur = rr.

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 A to B . Express your answer to three significant figures.

Hints (1)

Hint 1. Equation for the magnitude of a vector The formula for the magnitude of a vector is

v=

−−−−−−−−−−−−− vx2 + vy2 + vz2.

ANSWER:

rA B =

= 11.2 ft

ANSWER:

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,

uA B =

,

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

Part C - Position vector from A to C The length of cable A C is 9.36 ft, and the cable forms the angle = 27.0 with the y axis. Given this information and the dimensions provided in the figure, find the position vector from A to C . 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 A C and other information given.

From this diagram, you can see that the side of the triangle from point C 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, 9.36 ft, the angle cable A C makes with the positive y axis, 27.0 , and the z component, determine the x component. Express your answer to three significant figures and include the appropriate units. ANSWER: = 3.80

rA C x =

ANSWER:

rA C =

,

,

= 3.80, 8.34, -1.90 ift,jft,k ft

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Part D - Position vector from B to C Find the position vector from B to C . 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. 1. 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. 2. A second, simpler method is to write the Cartesian coordinates of the points with point A at the origin. The position vector is the difference between points C and B . 3. 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 (5,7,− 1)and the second point is (7,4,2), then the displacement in the x direction is positive 2, etc. For the first method, recognize that point A is the common point for which B and C are already defined. The position vector between the B and C then can be written in terms of the previous position vectors found in Part A and Part C: rB C = rA B − rA C .

Hint 2. Calculate the x component of the position vector from B to C Using the vector components for position vectors rA B and rA C found in Parts A and B, determine the x component of position vector rB C . Recall the equation rB C = rA C − rA B ; this equation can be evaluated by summing the Cartesian components for each direction. Express your answer to three significant figures. ANSWER: = 11.5 ft

rB C x =

ANSWER:

rB C =

,

,

= 11.5, 3.64, -8.50 ift,jft,k ft

Part E - Tension in the cable A cable is attached between B and C . The cable is attached in such a way that it has a tension of T B C along its length. The tensile force will attempt to pull B toward C . The force exerted on B has a z component with a magnitude of 10 lb . What is the magnitude of the tension in the cable, TB C ? Express your answer to three significant figures and include the appropriate units.

Hints (2)

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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 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? Express your answer to three significant figures.

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 z direction, it has a negative value when calculating .

ANSWER:

cos( )=

= -0.576

ANSWER:

TB C =

= 17.4

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2–94. z

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 a, b, g of the resultant force. Take x = 20 m, y = 15 m.

D 600 N 400 N

800 N

SOLUTION FDA = 400 a

20 15 24 i + j kb N 34.66 34.66 34.66

FDB = 800 a

4 24 -6 i + j kb N 25.06 25.06 25.06

FDC = 600a

24 m

16 m

18 24 16 i j kb N 34 34 34

FR = FDA + FDB + FDC

18 m

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

a = cos-1 a

321.66 b = 77.6° 1501.66

Ans.

b = cos-1 a

-16.82 b = 90.6° 1501.66

Ans.

g = cos-1 a

-1466.71 b = 168° 1501.66

Ans.

4m

B

6m x

y A

x

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

= 1501.66 N = 1.50 kN

O

C

y

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.

z

5 ft

B

F = 50 lb

SOLUTION

12 ft A

Unit Vector: The coordinates of point A are

5 ft 40˚

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

y 5 ft

Then

8 ft x

rAB = 510 - 3.8302i + 15 - 8.002j + 112 - 3.2142k6 ft = 5-3.830i - 3.00j + 8.786k6 ft rAB = 21-3.83022 + 1-3.0022 + 8.7862 = 10.043 ft = 10.0 ft uAB =

Ans.

-3.830i - 3.00j + 8.786k rAB = rAB 10.043 = - 0.3814i - 0.2987j + 0.8748k

Force Vector: F = FuAB = 505 -0.3814i - 0.2987j + 0.8748k6 lb = 5-19.1i - 14.9j + 43.7k6 lb

Ans.

Coordinate Direction Angles: From the unit vector uAB obtained above, we have cos a = - 0.3814

a = 112°

Ans.

cos b = - 0.2987

b = 107°

Ans.

cos g = 0.8748

g = 29.0°

Ans.

2–119. Determine the angle u between the y axis of the pole and the wire AB.

z

3 ft C

θ

2 ft

SOLUTION Position Vector:

2 ft x

rAC = 5 -3j6 ft

2 ft

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

B

= 52i - 1j - 2k6 ft The magnitudes of the position vectors are rAC = 3.00 ft

rAB = 22 2 + 1-122 + 1-222 = 3.00 ft

The Angles Between Two Vectors U: The dot product of two vectors must be determined first. rAC # rAB = 1-3j2 # 12i - 1j - 2k2 = 0122 + 1-321-12 + 01-22 = 3 Then, u = cos-1

rAO # rAB rAOrAB

= cos-1

3 3.00 3.00

= 70.5°

Ans.

A

y

2–139. Determine the magnitude of the projected component of the force F = 300 N acting along line OA.

30

F

z A

30 300 mm

O 300 mm

SOLUTION

x

300 mm

Force and Unit Vector: The force vector F and unit vector uOA must be determined first. From Fig. a F = (- 300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k) = { - 75i + 259.81j + 129.90k} N uOA =

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

Vector Dot Product: The magnitude of the projected component of F along line OA is FOA = F # uOA =

A -75i + 259.81j + 129.90k B # A - 0.75i + 0.5j + 0.4330k B

= (- 75)( -0.75) + 259.81(0.5) + 129.90(0.4330) = 242 N

Ans.

y

300 N

MasteringEngineering: Lecture 4

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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 d = 1.50 m . 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 16.0 N in the cable and the pulley to hang h = 130.0 cm 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

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because only two directions need to be considered. So

for the x–y plane and can be evaluated using

F = 0 simplifies to

F x i+ F y j= 0 F x = 0 and F y = 0 .

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 d and h. ANSWER: d = tan− 1 ( 2h ),

= tan− 1 ( 2h ) d

= tan− 1 ( 2d ), h

h = tan− 1 ( 2d )

= tan− 1 ( 2h ), d

d = tan− 1 ( 2h )

= tan− 1 ( dh ),

= tan− 1 ( hd )

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:

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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. Express your answer to three significant figures and include the appropriate units. ANSWER: = 13.9

Ty = Also accepted:

= 13.9

ANSWER:

W = Also accepted:

= 27.7 = 27.7

Part B - Finding the mass of the pulley As shown, an object with mass m = 5.1 kgis hung from a pulley and spring system. When the object is hung, the tension in the cable is 36.9 N and the pulley is h = 147.2 cm 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

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figures and include the appropriate units. For this problem, use g = 9.81 m /s2 .

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:

F x i+ F y j= 0, F x = 0 and F y = 0.

which can be evaluated with the magnitudes as

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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. 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 T be the tension in the cable, m be the mass of the hanging weight, and m p 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:

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

ANSWER: 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 k3, replaces Spring 2. The mass of the weight and pulley are unchanged: m = 5.1 kgand m p = 1.6 kg. However, because of the different spring constant the distance the pulley hangs below the ceiling is now, h = 115.0 cm . The unloaded length of Spring 3 is 3 = 34.5 cm ; after hanging the mass and pulley, the spring’s length is 3 = 42.5 cm . Determine the spring constant of Spring 3, k3. 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

F s = ks, where k is the spring constant, which has units of force over length, and s is the deformation or change in length of the spring.

Hint 2. Find the magnitude of the tension Given h = 115.0 cm , the mass of the pulley, m p = 1.6 kg, and the mass of the hanging object, m = 5.1 kg, 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 g = 9.81 m /s2 . ANSWER:

T=

= 39.2

Hint 3. Find the displacement in the spring, s The spring force equation for a linear elastic spring is F s = ks . In this equation k is the spring constant and s is the displacement of the spring. To solve for the unknown spring constant of Spring 3, you must know the force, F s, and the displacement of the spring. Given the unloaded and loaded lengths of Spring 3 find the displacement s3 in meters. Express your answer to three significant figures.

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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, F s, 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:

s3 =

= 8.00×10−2 m

ANSWER:

k3 =

= 490

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3–3. 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 u. 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.

F A

SOLUTION

B

Free-Body Diagram: By observation, the force F1 has to support the entire weight of the container. Thus, F1 = 50019.812 = 4905 N.

θ

θ

1.5 m

1.5 m

Equations of Equilibrium: + ©F = 0; : x

FAC cos u - FAB cos u = 0

+ c ©Fy = 0;

4905 - 2F sin u = 0

FAC = FAB = F

F = 52452.5 cos u6 N

G

Thus, FAC = FAB = F = 52.45 cos u6 kN

Ans.

If the maximum allowable tension in the cable is 5 kN, then 2452.5 cos u = 5000 u = 29.37° From the geometry, l =

1.5 and u = 29.37°. Therefore cos u l =

1.5 = 1.72 m cos 29.37°

Ans.

C

■3–15.

The spring has a stiffness of k = 800 N>m and an unstretched length of 200 mm. Determine the force in cables BC and BD when the spring is held in the position shown.

C

400 mm A

k

800 N/m

B

SOLUTION

300 mm

The Force in The Spring: The spring stretches s = l - l0 = 0.5 - 0.2 = 0.3 m. Applying Eq. 3–2, we have

D

Fsp = ks = 800(0.3) = 240 N

500 mm

Equations of Equilibrium: + ©F = 0; : x

4 FBC cos 45° + FBD a b - 240 = 0 5 0.7071FBC + 0.8FBD = 240

+ c ©Fy = 0;

(1)

3 FBC sin 45° - FBD a b = 0 5 FBC = 0.8485FBD

(2)

Solving Eqs. (1) and (2) yields, FBD = 171 N

FBC = 145 N

Ans.

400 mm

3–30. A 4-kg sphere rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown.

y

B

60

SOLUTION A

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

y

2.5x2

0.4 m

tan u `

x = 0.4 m

=

dy = 5.0x ` = 2.00 ` dx x = 0.4 m x = 0.4 m

x 0.4 m

u = 63.43° Free-Body Diagram: The tension in the cord is the same throughout the cord and is equal to the weight of block B, WB = mB (9.81). Equations of Equilibrium: + ©F = 0; : x

mB (9.81) cos 60° - Nsin 63.43° = 0 N = 5.4840mB

+ c ©Fy = 0;

(1)

mB (9.81) sin 60° + Ncos 63.43° - 39.24 = 0 8.4957mB + 0.4472N = 39.24

(2)

Solving Eqs. (1) and (2) yields mB = 3.58 kg

N = 19.7 N

Ans.

3–31. If the bucket weighs 50 lb, determine the tension developed in each of the wires.

C B

30

A 5

4

D

3

SOLUTION

E

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. + ©F = 0; : x

3 FED cos 30° - FEB a b = 0 5

(1)

+ c ©Fy = 0;

4 FED sin 30° + FEB a b - 50 = 0 5

(2)

Solving Eqs. (1) and (2), yields FED = 30.2 lb

FEB = 43.61 lb = 43.6 lb

Ans.

Using the result FEB = 43.61 lb and applying the equations of equilibrium to the free-body diagram of joint B shown in Fig. b, + c ©Fy = 0;

4 FBC sin 30° - 43.61a b = 0 5 FBC = 69.78 lb = 69.8 lb

+ ©F = 0; : x

30

Ans.

3 69.78 cos 30° + 43.61a b - FBA = 0 5 FBA = 86.6 lb

Ans.

*3–56. Determine the force in each cable needed to support the 3500-lb platform. Set d = 2 ft.

z 3500 lb A

10 ft

SOLUTION Cartesian Vector Notation: FAB = FAB ¢ FAC = FAC ¢ FAD = FAD ¢

4i - 3j - 10k 24 + ( -3)2 + (- 10)2 2

2i + 3j - 10k 222 + 32 + (- 10)

≤ = 0.3578FAB i - 0.2683FAB j - 0.8944FAB k

D B

≤ = 0.1881FAC i + 0.2822FAC j - 0.9407FACk 2

- 4i + 1j - 10k 2(- 4)2 + 12 + ( -10)2

≤ = - 0.3698FAD i + 0.09245FAD j - 0.9245FAD k x

Equations of Equilibrium: FAB + FAC + FAD + F = 0

(0.3578FAB + 0.1881FAC - 0.3698FAD) i + (- 0.2683FAB + 0.2822FAC + 0.09245FAD)j + ( - 0.8944FAB - 0.9407FAC - 0.9245FAD + 3500)k = 0 Equating i, j, and k components, we have 0.3578FAB + 0.1881FAC - 0.3698FAD = 0

(1)

-0.2683FAB + 0.2822FAC + 0.09245FAD = 0

(2)

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

(3)

Solving Eqs. (1), (2) and (3) yields FAB = 1369.59 lb = 1.37 kip FAD = 1703.62 lb = 1.70 kip

FAC = 744.11 lb = 0.744 kip

4 ft

3 ft

F = {3500k} lb

©F = 0;

2 ft

Ans. Ans.

C 3 ft

d

4 ft

y

z

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.

A 30 ft

C

10 ft

10 ft B

E D

15 ft 12.5 ft x

SOLUTION

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

FAB = FAB C

FAC = FAC C

FAD = FAD C

(10 - 0)i + (-15 - 0)j + (-30 - 0)k 2(10 - 0) + (-15 - 0) + (-30 - 0) 2

2

2

S =

(-15 - 0)i + (-10 - 0)j + ( -30 - 0)k 2(-15 - 0)2 + (-10 - 0)2 + (-30 - 0)2 (0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 2(0 - 0)2 + (12.5 - 0)2 + (-30 - 0)2

2 3 6 FAB i - FAB j - FAB k 7 7 7

3 2 6 S = - FAC i - FAC j - FAC k 7 7 7

S = {500j - 1200k} lb

FAE = FAE k Equations of Equilibrium: Equilibrium requires gF = 0; 2 7

¢ FAB i 2 7

¢ FAB -

FAB + FAC + FAD + FAE = 0 3 6 3 2 6 FAB j - FAB k ≤ + ¢- FAC i - FAC j - FAC k ≤ + (500j - 1200k) + FAE k = 0 7 7 7 7 7 3 3 2 6 6 F ≤ i + ¢- FAB - FAC + 500 ≤ j + ¢ - FAB - FAC + FAE - 1200 ≤ k = 0 7 AC 7 7 7 7

Equating the i, j, and k components yields 3 2 F - FAC = 0 7 AB 7 2 3 - FAB - FAC + 500 = 0 7 7 6 6 - FAB - FAC + FAE - 1200 = 0 7 7

(1) (2) (3)

Solving Eqs. (1) through (3) yields FAB = 808 lb FAC = 538 lb FAE = 2354 lb = 2.35 kip

Ans. Ans. Ans.

15 ft y