Engineering Mechanics

School of Mechanical Engineering MEE-1002: Engineering Mechanics Statics of Particle

Dr Rahul S Sikarwar Associate Professor School of Mechanical Engineering, Machine design section VIT Vellore

Determine the tension in the Cable?

Frictional force?

Motion of the Snowboarder?

Objectives 

Draw Free body diagram



Write and solve equilibrium equations for particles and rigid bodies



To analyse the structures.



To know the geometric properties of the different shapes.



To teach energy and momentum methods.

Outcomes Student will be able to



Solve the engineering problems in case of equilibrium conditions.



Calculate the reaction forces of various supports of different structures.



Solve the problems involving dry friction.



Determine the centroid, centre of gravity and moment of inertia of various surfaces and solids.



Solve the problems involving dynamics of particles and rigid bodies

Contents • Unit I Basics of Statics • Unit II Analysis of Structures • Unit III Friction • Unit IV Properties of Surfaces and Solids • Unit V Virtual Work

• Unit VI Kinematics

• Unit VII Energy and Momentum Methods

Text Books 1. Ferdinand P. Beer, E. Russell Johnston (2007), “Vector Mechanics for Engineers: Statics and Dynamics”, McGraw-Hill International Edition. 2. J. L. Meriam and L. G. Kraige (2006), “Engineering Mechanics: Statics and Dynamics (6th Edition)”, Wiley Publishers

References 1. Irving H. Shames, (2003), “Engineering Mechanics – Statics and Dynamics”, Prentice-Hall of India Private limited. 2. Russell C Hibbeler, (2009), “Engineering Mechanics: Statics and Dynamics (12th Edition)”, Prentice Hall. 3. Anthony M. Bedford and Wallace Fowler (2007), “Engineering Mechanics: Statics and Dynamics (5th Edition)”, Prentice Hall.

  

What is Mechanics? Mechanics is a physical science which deals with bodies at rest or motion under the action of forces. Mechanics is an applied science - it is not an abstract or pure science. Mechanics is the foundation of most engineering sciences.

Mechanics Mechanics of rigid bodies

Statics

Mechanics of Deformable bodies

Dynamics Kinematics

Kinetics

Mechanics of fluids

Rigid Body Mechanics Statics

Deals with bodies at rest Displacement

Kinematics

Velocity Acceleration

Dynamics

Force/Torque

Kinetics Energy/momentum

Engineering Mechanics Rigid-body Mechanics Statics: deals with equilibrium of bodies under action of forces (bodies may be either at rest or move with a constant velocity).

Engineering Mechanics Rigid-body Mechanics Dynamics: deals with motion of bodies (accelerated motion)

Fundamental concepts • Space - associated with the representation of the position of a point P given in terms of three coordinates measured from a reference point or origin. – Length :: needed to locate position of a point in space, &describe size of the physical system Distances, Geometric Properties • Time – definition of an event requires specification of the time and position at which it occurred. – measure of succession of events basic quantity in Dynamics • Mass – used to characterize and compare bodies. Two bodies of the same mass, for example, response to earth’s gravitational attraction and resistance to changes in translational motion. – quantity of matter in a body measure of inertia of a body (its resistance to change in velocity) – Mass of a body is the quantitative measure of its inertia

Force represents the action of one body on another. It changes (or) tends to change the state of rest or union motion of a body

A force is characterized by its point of application, magnitude, and direction, i.e., a force is a vector. (Free vector, Sliding Vector, Fixed Vector) In Newtonian Mechanics, space, time, and mass are absolute concepts, independent of each other. Force, however, is not independent of the other three. The force acting on a body is related to the mass of the body and the variation of its velocity with time.

Mechanics: Fundamental Concepts Newtonian Mechanics Length, Time, and Mass are absolute concepts independent of each other

Force is a derived concept not independent of the other fundamental concepts. Force acting on a body is related to the mass of the body and the variation of its velocity with time.

Force can also occur between bodies that are physically separated (Ex: gravitational, electrical, and magnetic forces) Remember: • Mass is a property of matter that does not change from one location to another. • Weight refers to the gravitational attraction of the earth on a body or quantity of mass. Its magnitude depends upon the elevation at which the mass is located • Weight of a body is the gravitational force acting on it.

Mechanics: Idealizations To simplify application of the theory Particle: A body with mass but with dimensions that can be neglected.

Size of earth is insignificant compared to the size of its orbit. Earth can be modeled as a particle when studying its orbital motion

Mechanics: Idealizations Rigid Body: A combination of large number of particles in which all particles remain at a fixed distance (practically) from one another before and after applying a load. Material properties of a rigid body are not required to be considered when analyzing the forces acting on the body. In most cases, actual deformations occurring in structures, machines, mechanisms, etc. are relatively small, and rigid body assumption is suitable for analysis

Mechanics: Idealizations Concentrated Force: Effect of a loading which is assumed to act at a point (CG) on a body. Provided the area over which the load is applied is very small compared to the overall size of the body.

Ex: Contact Force between a wheel and ground.

Mechanics: Newton’s Three Laws of Motion First Law: A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force.

First law contains the principle of the equilibrium of forces main topic of concern in Statics

Mechanics: Newton’s Three Laws of Motion Second Law: A particle of mass “m” acted upon by an unbalanced force “F” experiences an acceleration “a” that has the same direction as the force and a magnitude that is directly proportional to the force.

Second Law forms the basis for most of the analysis in Dynamics

Mechanics: Newton’s Three Laws of Motion Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear.

Third law is basic to our understanding of Force Forces always occur in pairs of equal and opposite forces.

Mechanics: Newton’s Law of Gravitation This states that two particles of mass m1 and m2 are mutually attracted with equal and opposite forces F and –F of magnitude F given by the formula.

m1m 2 F G 2 r Where r = distance between the two particles G = universal constant called the constant of gravitation Application: The force F exerted by the earth on the particle located on its surface, is defined as the weight W of the particle.

GMme W 2 r

GM g 2 r

W  mg

Mechanics: Fundamental Laws Parallelogram Law If two vector quantities are represented by two adjacent sides or a parallelogram then the diagonal of parallelogram will be equal to the resultant of these two vectors

Parallelogram Law Principle of Transmissibility The effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. Warning: This principle is not valid for deformable bodies.

Principle of Transmissibility

Systems of Units • International System of Units (SI): The basic units are length, time, and mass which are arbitrarily defined as the meter • An equation must be dimensionally (m), second (s), and kilogram (kg). Force is Homogeneous, i.e., Left and Right terms the derived unit, must be identical • Kinetic Units: length, time, mass, and force.

F  ma

 m   1 N  1 kg 1 2 

 s  • Three of the kinetic units, referred to as basic units, may be defined arbitrarily. • U.S. Customary Units: The fourth unit, referred to as a derived The basic units are length, time, and force unit, must have a definition compatible which are arbitrarily defined as the foot (ft), with Newton’s 2nd Law, second (s), and pound (lb). Mass is the  derived unit,  F  ma

F a 1lb 1slug  1ft s m

Mechanics: Units Prefixes

Method of Problem Solution • Problem Statement:

Includes given data, specification of what is to be determined, and a figure showing all quantities involved. • Free-Body Diagrams: Create separate diagrams for each of the bodies involved with a clear indication of all forces acting on each body. • Fundamental Principles:

The six fundamental principles are applied to express the conditions of rest or motion of each body. The rules of algebra are applied to solve the equations for the unknown quantities.

• Solution Check: • Test for errors in reasoning by verifying that the units of the computed results are correct • test for errors in computation by substituting given data and computed results into previously unused equations based on the six principles • always apply experience and physical intuition to assess whether results seem “reasonable”.

Numerical Accuracy • The accuracy of a solution depends on 1) accuracy of the given data, and 2) accuracy of the computations performed. The solution cannot be more accurate than the less accurate of these two. • The use of hand calculators and computers generally makes the accuracy of the computations much greater than the accuracy of the data. Hence, the solution accuracy is usually limited by the data accuracy. • As a general rule for engineering problems, the data are seldom known with an accuracy greater than 0.2%. Therefore, it is usually appropriate to record parameters beginning with “1” with four digits and with three digits in all other cases, i.e., 40.2 lb and 15.58 lb.

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Statics of Particles Many engineering problems can be solved by considering the equilibrium of a “particle.” In the case of this excavator, which is being loaded onto a ship, a relation between the tensions in the various cables involved can be obtained by considering the equilibrium of the hook to which the cables are attached.

Statics of Particles  Particle: Size and shape of the body are neglected, but not the mass.  Forces in a plane

 Vectors  Resultant of several concurrent forces (Polygon law)  Resolution of a force into components

 Rectangular components of a force: Unit vectors  Addition of forces by summing X and Y components  Equilibrium of a Particle

Forces in a plane A force represents the action of one body on another and is generally characterized by its point of application, its magnitude, and its direction. Characteristics of a force are its 1. Magnitude 2. Point of application 3. Direction Unit: Newton (N) 1 Newton is defined as the force which gives an acceleration of 1 m/s2 to a mass of 1 kg.

• • •

•

Concentrated force or a point force: it is the force acting on a very small area. Distributed force: force distributed over a length or an area or a volume Line of action of force: Direction along a straight line through its point of application, in which the force tends to move a body to which it is applied. Graphical representation of force:

• Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.

System of forces

• when several forces of various magnitude and direction act upon a body they are said to form system of forces. • Classification of system of forces: Classified as per orientation of line of action of forces. System of forces

Coplanar ( In Plane)/ 2-D

Concurrent

Non Coplanar(Space)-3-D

Non- concurrent

Parallel

Concurrent

General

Non- concurrent

Parallel

Note:

Concurrent force system: acts on a particle or rigid body Parallel and General force system: acts mainly on rigid bodies

General

Vectors Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations.

Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature Vector classifications: - Fixed or bound vectors have well defined points of application that cannot be changed without affecting an analysis. - Free vectors may be freely moved in space without changing their effect on an analysis. - Sliding vectors may be applied anywhere along their line of action without affecting an analysis. Equal vectors have the same magnitude and direction. Negative vector of a given vector has the same magnitude and the opposite direction. 2 - 31

Addition of Vectors • Trapezoid rule for vector addition

• Triangle rule for vector addition • Law of cosines, C

B C

B

R 2  P 2  Q 2  2 PQ cos B    R  PQ

• Law of sines, sin A sin B sin C   Q R A • Vector addition is commutative,     PQ  Q P • Vector subtraction     P  (Q)  P  Q

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Parallelogram law If two forces represented by vectors AB and AC acting under an angle α are applied to a body at point A. Their action is equivalent to the action of one force, represented by

vector AD, obtained as the diagonal of the parallelogram constructed on the vectors AB and AC directed as shown in the figure.

Force AD is called the resultant of AB and AC and the forces are called its components.

Lami’s Theorem If the three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces.

F1 θ3

θ2

F2

O

F3

θ1

F3 F1 F2   Sin 1 Sin 2 Sin 3

Addition of Vectors • Addition of three or more vectors through repeated application of the triangle rule

• The polygon rule for the addition of three or more vectors. • Vector addition is associative,          P  Q  S  P  Q   S  P  Q  S 

• Multiplication of a vector by a scalar

2 - 38

Resultant of Several Concurrent Forces • Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces. • Vector force components: two or more force vectors which, together, have the same effect as a single force vector.

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Example 1. SOLUTION:

The two forces act on a bolt at A. Determine their resultant.

• Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal. • Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

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• Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,

R  98 N   35 • Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,

R  98 N   35

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Trigonometric solution - Apply the triangle rule. From the Law of Cosines,

R 2  P 2  Q 2  2 PQ cos B  40 N 2  60 N 2  240 N 60 N  cos 155

R  97.73N From the Law of Sines,

sin A sin B  Q R sin A  sin B

Q R

 sin 155 A  15.04   20  A   35.04 2 - 46

60 N 97.73N

Example 2. SOLUTION: • Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 lbf.

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 lbf directed along the axis of the barge, determine

• Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions.

a) the tension in each of the ropes for  = 45o, • The angle for minimum tension in rope 2 is b) the value of  for which the tension determined by applying the Triangle Rule and observing the effect of variations in . in rope 2 is a minimum.

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• Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides.

T1  3700 lbf

T2  2600 lbf

• Trigonometric solution - Triangle Rule with Law of Sines

T1 T2 5000 lbf   sin 45 sin 30 sin 105 T1  3660 lbf

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T2  2590 lbf

• The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in . • The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.

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T2  5000 lbf  sin 30

T2  2500 lbf

T1  5000 lbf  cos 30

T1  4330 lbf

  90  30

  60

Rectangular Components of a Force: Unit Vectors • May resolve a force vector into perpendicular components so that the resulting parallelogram is a   rectangle. Fx and Fy are referred to as rectangular vector components and    F  Fx  Fy   • Define perpendicular unit vectors i and j which are parallel to the x and y axes.

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.    F  Fx i  Fy j  Fx and Fy are referred to as the scalar components of F 2 - 50

Addition of Forces by Summing Components • Wish to find the resultant of 3 or more concurrent forces,

    R  PQ S

• Resolve each force into rectangular components

        R x i  R y j  Px i  Py j  Q x i  Q y j  S x i  S y j     Px  Q x  S x i  Py  Q y  S y j





• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

Rx  Px  Qx  S x   Fx

R y  Py  Q y  S y   Fy

• To find the resultant magnitude and direction,

R 2 - 51

Rx2

 R y2

  tan

1

Ry Rx

Example 3. SOLUTION: • Resolve each force into rectangular components.

• Determine the components of the resultant by adding the corresponding force components.

Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

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• Calculate the magnitude and direction of the resultant.

SOLUTION: • Resolve each force into rectangular components.

force mag  F1 150  F2 80  F3 110  F4 100

x  comp

y  comp

 129.9

 75.0

 27.4

 75.2

0

 110.0

 96.6

 25.9

R x  199.1 R y  14.3 • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction.

R  199.12  14.32 14.3 N tan   199.1 N 2 - 53

R  199.6N

  4.1

Determine the x and y components of each of the forces and resultant of three forces shown.

Determine the x and y components of each of the forces and resultant of three forces shown.

Determine the x and y components of each of the forces and resultant of three forces shown.

Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 960-N vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

Knowing that the tension in cable BC is 145 N, determine the resultant of the three forces exerted at point B of beam AB.

Knowing that α = 65°, determine the resultant of the three forces shown.

α = 65°,

Equilibrium of a Particle • When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. • Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.

• Particle acted upon by two forces: - equal magnitude - same line of action - opposite sense 2 - 60

• Particle acted upon by three or more forces: - graphical solution yields a closed polygon - algebraic solution

  R  F  0

 Fx  0

 Fy  0

Free-Body Diagrams Free body diagram: The sketch of the isolated body which shows the external force on the body and the reactions on it by the removed elements.

General Procedure for construction of free body diagram 1. Draw the sketch of body without supporting surfaces. 2. Show all the external or applied forces on this sketch 3. Show all the reactive forces such as those caused by constraints or supports

4. Show all relevant dimensions , angles and reference axes on the sketch

External and Internal Forces • Forces acting on rigid bodies are divided into two groups: - External forces - Internal forces

• External forces are shown in a free-body diagram.

• If unopposed, each external force can impart a motion of translation or rotation, or both.

Types of support 1. Frictionless Support: Reactions are normal to the surface at the point of contact W c W A

RA

2. Hinge support or pin jointed support: RAx RAy

3. Roller Support :

4. Fixed Support :

RA

RAx RAy

MA

Draw FDB for following system WP

WQ

RA RC WP

RA

WQ

RD

RB

RC

RD

Sample Free Body Diagrams

Draw FDB for following system WP

WQ

RA RC WP

RA

WQ

RD

RB

RC

RD

Free-Body Diagrams

Space Diagram: A sketch showing the physical conditions of the problem.

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Free-Body Diagram: A sketch showing only the forces on the selected particle.

Example 4. SOLUTION:

• Construct a free-body diagram for the particle at the junction of the rope and cable. • Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle. • Apply trigonometric relations to determine the unknown force magnitudes. In a ship-unloading operation, a 3500lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope? 2 - 69

SOLUTION: • Construct a free-body diagram for the particle at A. • Apply the conditions for equilibrium. • Solve for the unknown force magnitudes.

T T AB 3500 lb  AC  sin 120 sin 2 sin 58

TAB  3570 lb T AC  144 lb

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Example 5. SOLUTION: • Choosing the hull as the free body, draw a free-body diagram.

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. It is desired to determine the drag force at a given speed on a prototype sailboat hull. • Resolve the vector equilibrium equation into two component equations. Solve for A model is placed in a test channel and the two unknown cable tensions. three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 lb in cable AB and 60 lb in cable AE. Determine the drag force exerted on the hull and the tension in cable AC.

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SOLUTION: • Choosing the hull as the free body, draw a free-body diagram.

7 ft  1.75 4 ft   60.25

tan  

1.5 ft  0.375 4 ft   20.56

tan  

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

     R  TAB  TAC  TAE  FD  0

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• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

       T AB   40 lb sin 60.26 i  40 lb cos 60.26 j    34.73 lb i  19.84 lb j    T AC  T AC sin 20.56 i  T AC cos 20.56 j    0.3512T AC i  0.9363T AC j   T  60 lb i   FD  FD i  R0



  34.73  0.3512T AC  FD  i     19.84  0.9363T AC  60 j 2 - 73

 R0

   34.73  0.3512T AC  FD  i   19.84  0.9363T AC  60 j

This equation is satisfied only if each component of the resultant is equal to zero

 Fx  0 0  34.73  0.3512TAC  FD  Fy  0 0  19.84  0.9363TAC  60 T AC  42.9 lb FD  19.66 lb

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Example 7. Determine the tension developed in each cord required for equilibrium of the 20-kg lamp.

Example 7. Determine the tension developed in each cord required for equilibrium of the 20-kg lamp. SOLUTION

Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

Ans: (a) 2.20 kN, (b) 2.43 kN

Knowing that α = 50° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.

Ans: 169.7 kN, (b) 348 kN

A boat is pulling a parasail and rider at a constant speed. Knowing that the rider weighs 550 N and that the resultant force R exerted by the parasail on the towing yoke A forms an angle of 65° with the horizontal, determine (a) the tension in the tow rope AB, (b) the magnitude of R.

Ans: (a) 405 N, (b) 830 N

A load of weight 400 N is suspended from a spring and two cords that are attached to blocks of weights 3W and W as shown. Knowing that the constant of the spring is 800 N/m, determine (a) the value of W, (b) the unstretched length of the spring.

Ans: (a) 62.8 N, (b)758 mm

Two cables tied together at C are loaded as shown. Knowing that W = 190 N, determine the tension (a) in cable AC, (b) in cable BC. Determine the range of values of W for which the tension will not exceed 240 N in either cable.

Ans: (a) 169.6 N, (b) 265 N

Two traffic signals are temporarily suspended from a cable as shown. Knowing that the signal at B weighs 300 N, determine the weight of the signal at C.

Ans: 97.7 N

Example 8 The cylinders in Figure have the indicated weights and dimensions. Assuming smooth contact surfaces, determine the reactions at A, B, C, and D on the cylinders.

Free Body Diagram

Rectangular Components in Space

 F

The vector is contained in the plane OBAC.



Resolve F into horizontal and vertical components.

Fy  F cos  y Fh  F sin  y 2 - 88

Resolve Fh into rectangular components

Fx  Fh cos  F sin  y cos Fz  Fh sin   F sin  y sin 

Rectangular Components in Space

 • With the angles between F and the axes,

Fx  F cos  x Fy  F cos  y Fz  F cos  z     F  Fx i  Fy j  Fz k     F cos  x i  cos  y j  cos  z k   F       cos  x i  cos  y j  cos  z k   •  is a unit vector along the line of action of F and cos  x , cos  y , and cos  z are the direction cosines for F



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

Rectangular Components in Space Direction of the force is defined by the location of two points,

M  x1 , y1 , z1  and N  x2 , y2 , z 2   d  vector joining M and N     d xi  d y j  d z k

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d x  x2  x1 d y  y 2  y1 d z  z 2  z1   F  F   1     d x i  d y j  d z k  d Fd y Fd x Fd z Fx  Fy  Fz  d d d

Example 9. SOLUTION: • Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. • Apply the unit vector to determine the components of the force acting on A.

The tension in the guy wire is 2500 N. Determine: a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles qx, qy, qz defining the direction of the force 2 - 91

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

SOLUTION: • Determine the unit vector pointing from A towards B.

   AB   40 m  i  80 m  j  30 m k AB 

 40 m 2  80 m 2  30 m 2

 94.3 m

  40    80    30     i    j  k 94 . 3 94 . 3 94 . 3           0.424 i  0.848 j  0.318k 

• Determine the components of the force.

  F  F

    2500 N  0.424 i  0.848 j  0.318k       1060 N i  2120 N  j  795 N k

2 - 92

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

     cos  x i  cos  y j  cos  z k     0.424 i  0.848 j  0.318k 

 x  115.1  y  32.0  z  71.5

2 - 93

Example 10

The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

Solution

Resolve Fh into rectangular components

Fx  Fh cos  F sin  y cos Fz  Fh sin   F sin  y sin 

Example 11

A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 2100 N, determine the components of the force exerted by the wire on the bolt at B.

Example 12 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 1260 N, determine the components of the force exerted by the wire on the bolt at D.

Example 13 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 degrees angles with the vertical. Knowing that the tension in wire CD is 120 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles θx , θy, and θz that the force forms with the coordinate axes. (a) the components of the force exerted by this wire on the plate,

(b) the angles θx , θy, and θz that the force forms with the coordinate axes.

Example 14 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 195 lb, determine the components of the force exerted on the plate at D.

Solution

H.W Problem: Example 14

H.W Problem: Example 15 The support assembly shown is bolted in place at B, C, and D and supports a downward force P at A. Knowing that the forces in members AB, AC, and AD are directed along the respective members and that the force in member AB is 29.2 N, determine the magnitude of P.

ANS: 55.9 N

1.

2.

A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 2100 N, determine the components of the force exerted by the wire on the bolt at B. A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 1260 N, determine the components of the force exerted by the wire on the bolt at D.

Two cables BG and BH are attached to the frame ACD as shown. Knowing that the tension in cable BG is 450 N, determine the components of the force exerted by cable BG on the frame at B. Two cables BG and BH are attached to the frame ACD as shown. Knowing that the tension in cable BH is 600 N, determine the components of the force exerted by cable BH on the frame at B.

The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension is 510 N in cable AB and 765 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

For the transmission tower shown below, determine the tensions in cables AB and AD knowing that the tension in cable AC is 1770 N and that the resultant of the forces exerted by the three cables at A must be vertical.

For the boom shown below, knowing that α = 0°, the tension in cable AB is 600 N, and the resultant of the load P and the force exerted at A by the two cables is directed along OA, determine (a) the tension in cable AC, (b) the magnitude of the load P.

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