Unit - 1
Introduction to Engineering Mechanics
Force is basically a vector quantity which may be either push or pull exerted during interaction of one object with other. Force is sort of interaction, when the interaction stops between objects, force is considered to be absent.
Types of Forces:
They can be divided as Contact Forces and Action at a distance type forces.
These forces act when there exists contact between surfaces or objects and are subdivided as:
Frictional Force: This type of force exists between the two surfaces making contact and having irregularities or roughness between the mating surfaces. E.g., Frictional force between the tyre of car and road causing it to wear and tear.
Tension Force: This type of force gets transmitted through ropes, wires or cables when some object is suspended or held using them. E.g., A block of material freely suspended using some rope and pulley arrangement.
Normal Force: This type of force exists when a particular object is placed on some other object, thus the weight of one object acts as a kind of force onto the other stable object. E.g., Book placed on the table.
Air resistance Force: This kind of force comes into action when motion of one object in presence of air, such that the air particle opposes the motion of that object, creating a kind of resistance force acting onto the body. E.g., An object falling from sky gets heated due to air resistance on the object.
Applied Force: This type of force is acted on an object externally, either to create a motion or retard the ongoing motion. E.g., Force applied by brakes to stop a moving vehicle.
Spring Force: This type of force exists when the spring which is compressed or stretched and gives a reaction force causing compression or stretching.
b. Action-at-a-Distance type Forces
These types of forces do not need to have physical contact between the two surfaces to exist. Some of these types of forces are
Gravitational Force: This force is considered to exist between any two bodies and is governed using Law of Gravitational Force given by Newton.
Electrical Force: This type of force is considered to be existing between two charged particles placed near each other causing pull or push between them.
Magnetic Force: This type of force is present between magnetic poles which may cause either pull or push of the magnets.
Effects of a force:
It may produce the effects on body on which it is acted as:
1. Motion of the body may be changed.
2. It may retard the motion of a body.
3. It may reduce the acting forces to make it stop or be in equilibrium
4. It may even create internal stresses in the body.
Characteristics of a force:
Following are the characteristics of a force:
1. Magnitude of the force (i.e., 100 N, 50 N, 20 kN, 5 kN, etc.)
2. The direction of the line, along which the force acts.
3. Nature of the force. This is denoted by placing an arrow head on the line of action of the force.
4. The point at which (or through which) the force acts on the body.
A particle is said to be in equilibrium if it remains in rest condition or in a particular constant velocity if it is already in motion. If we consider in Static Equilibrium, then essentially a particle is considered to be in rest condition. Thus, for a particle to be in rest condition, it needs to satisfy the Newton’s First Law of motion that states that a particle remains in state of rest or motion, until an external unbalanced force act on it. Thus, the generalized equation for a particle to be in equilibrium states that resultant force acting on any particle should be zero for the particle to be in equilibrium,
i.e., 
.
Some generalized steps for solving particle equilibrium problem in 2D and 3D systems are
Drawing a free-body diagram which is discussed in next section.
Using equations of equilibrium for solving the unknowns and desired quantities.
Thus, the equations of equilibrium for two-dimensional coplanar force system are given by considering that the particle is present in x-y plane as shown in fig. below, then the force can be resolved into i and j components.
For equilibrium, the forces must add on to produce a zero-force output, i.e.,
Σ F = 0
Σ Fx i + Σ Fy j = 0
Thus, for satisfying this vector equation, the x and y component must be zero as mentioned in earlier points. Hence,
Σ Fx = 0
Σ Fy = 0
Thus, for solving two-dimensional coplanar force problems using conditions of equilibrium, we can follow the below-mentioned procedure.
Free-Body Diagram.
• In any orientation, establish x and y axes.
• Then mark all the known and unknown forces in figure.
• Unknown forces are also assumed and plotted.
Equilibrium equation.
• Apply the equations of equilibrium, Σ Fx = 0 and Σ Fy = 0.
• The forces are directed as positive for those along positive axis and vice-versa.
• If the resultant comes out to be negative value, it means that the direction needs to be reversed for that is already considered.
As we know, the sufficient case for stating equilibrium for a body under combination of forces is Σ F = 0.
For any three-dimensional force system, as shown in Fig. below, we can resolve the forces into their respective i, j, k components, so that
Σ Fx i + Σ Fy j + Σ Fz k = 0
To satisfy this equation we require,
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
These three equations state that the algebraic sum of the components of all the forces acting on the particle along each of the coordinate axes must be zero. Using these equations mentioned we can solve for maximum three of the unknowns, shown by coordinate direction angles or magnitudes of forces shown on the particle’s free-body diagram.
For solving three-dimensional problem, we can use the following procedure.
Free Body Diagram:
• In any orientation, establish x and y axes.
• Then mark all the known and unknown forces in figure.
• Unknown forces are also assumed and plotted.
Equations of Equilibrium.
• Apply the equations of equilibrium, Σ Fx = 0, Σ Fy = 0 and Σ Fz = 0.
• Initially we can display as a Cartesian vector if it becomes to solve the problem regularly, substitute these vectors into Σ F = 0 and then set the i, j, k components equal to zero.
• If the resultant comes out to be negative value, it means that the direction needs to be reversed for that is already considered.
Problems in two and three dimensions:
The above-mentioned procedure (2.2) is used for finding solutions to two- and three-dimensional problems. This procedure can be better understood through some of the examples mentioned below.
e.g., 1: Determine the tension that will be there in cables BA and BC necessary to support the 60-kg cylinder in Fig. below.
Solution:
The first step involves drawing the free-body diagram of the given example. As this is two-dimensional structure, we will follow the method mentioned for two-dimensional coplanar structure.
2. Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes, we have
Σ Fx = 0 i.e., TC cos 45° - (4/5) TA = 0
Σ Fy = 0 i.e., TC sin 45° + (3/5) TA - 60(9.81) N = 0
Solving the above equations, we get, TC = 475.66 N = 476 N
Then, substituting this value in above equation,
TA = 420 N
In this way, using the equilibrium equations, we can solve for the unknown force in the given problem.
e.g., 2: The 200-kg crate in Fig. below is suspended using the ropes AB and AC. Each rope can withstand a maximum force of 10 kN before it breaks. If AB always remains horizontal, determine the smallest angle to which the crate can be suspended before one of the ropes breaks.
Solution:
Free-Body Diagram: We will initially understand the equilibrium of ring. There are three forces acting on it, as shown in fig. (b). The magnitude of is equal to the weight of the crate, i.e.,
FD = 200 (9.81) N = 1962 N < 10 kN
Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes,
i.e. - FC cos θ + FB = 0
Therefore FC = (FB / cos θ)
Σ Fy = 0
i.e., FC sin θ - 1962 N = 0
As seen from the above equations, FC is always greater than FB as cos θ is less than or equal to 1. So, rope AC will reach the max value of tensile force of 10 kN before that of rope AB. Thus, substituting FC = 10 kN in the above equations, we get
[10(103) N] sin θ - 1962 N = 0
Thus, θ = 11.30
The force developed in rope AB can be obtained by substituting the values for θ and FC into Equations mentioned above.
We get, FB = 9.81 kN.
In this way, using equations of equilibrium and free-body diagrams, we can calculate unknown quantities of the given problem.
Problems in Three-dimensions:
e.g., 3: The 10-kg lamp in Fig. below is suspended from the three equal-length cords. Find the smallest vertical distance s from the ceiling if force built in cord is not to go beyond 50 kN.
Solution:
For solving these three-dimensional numerical, we will follow the same procedure mentioned above.
Free-Body Diagram: Because of symmetry, the distance are as 
mm. It follows that from Σ Fx = 0 and Σ Fy = 0, the tension T in each cord will be the same. Also, the angle between z axis and every cord is 
.
Equation of Equilibrium: Applying the equilibrium equation along the z-axis, with T = 50 N, we have
Σ Fz = 0;
3[(50 N) cos 
] - 10(9.81) N = 0
= cos-1 (98.1 / 150)
= 49.160
Now,
By geometry shown in free-body diagram,
tan 49.160 = (600 / s)
s = 519 mm
In this way, three-dimensional problems are solved using equations of equilibrium.
e.g., 4: A 90-lb load is being suspended from a hook as shown in Fig. below. If the load is being supported by two cables and a spring having a stiffness of k = 500 lb/ft, find the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x–y plane and cable AC lies in the x–z plane.
Solution:
Initially, force in spring is determined to determine stretching of spring.
Free-Body Diagram. A point is chosen for all analysis as all the points are concurrent here. The free-body diagram for the problem is shown in Fig. below.
Equations of Equilibrium. We can resolve each force in x, y and z components, and so we can use three scalar equations of equilibrium. Considering components directed along each positive axis as “positive,” we have
Σ Fx = 0
FD sin 30° - (4/5) FC = 0
Σ Fy = 0
-FD cos 30° + FB = 0
Σ Fz = 0
(3/5) FC – 90 lb = 0
Thus, solving the above-mentioned equations, we get,
FC = 150 lb
FD = 240 lb
FB = 207.8 lb
The stretch of the spring is therefore
FB = ksAB
207.8 lb = (500 lb/ft) (sAB)
sAB = 0.416 ft
Thus, the method discussed earlier can be followed as shown in the examples for finding out the unknown quantities in two- and three-dimensional problems using equations of equilibrium.
Consider a rigid body acted upon by various kinds of forces as shown in fig. like F1, F2, F3 etc. Thus, for a body to be in equilibrium, the resultant force, which is sum of all the forces and the resultant couple moment, both should be equal to zero. If these conditions are satisfied, then the body is said to be in equilibrium.
Thus, mathematically, the rigid body can be expressed using the following equations.
Thus, the method of solving the rigid body equilibrium problems involves the same steps as mentioned above.
Drawing a free-body diagram.
Using equations of equilibrium for solving the unknowns and desired quantities.
Thus, the only change takes place here is the equations of equilibrium that are used.
The procedure for solving is same as that of the particle equilibrium, just we need to consider the additional degree of freedom to be locked.
When two or more forces act, then we called it as the system of forces, like:
1. Coplanar forces: Those having line of action in same plane.
2. Collinear forces. Those whose line of action is in the same line.
3. Concurrent forces. Those which meet at one particular point in space.
4. Coplanar concurrent forces. Those having same plane and meet at one point.
5. Coplanar non-concurrent forces. Those who are in same plane but do not meet at a point.
6. Non-coplanar concurrent forces. Those, which meet at a point in space but are not in same plane.
7. Non-coplanar non-concurrent forces. Those, which do not meet at a point in space nor in the same plane.
These are the forces that are present in the same plane and they all meet at one particular point, thus being called coplanar concurrent force.
As shown in the figure above, the forces lie in the same plane and also meet at one particular point.
When many forces are acting, the overall effect can be considered in a single force. This single force is called resultant force and the given individual forces are called component forces.
Methods for the resultant force
Though there are many methods for finding out the resultant force of a number of given forces, yet the following are important from the subject point of view:
Analytical method includes the Parallelogram law of Vectors which is described as “If two forces, acting simultaneously on a particle, be shown by two sides of parallelogram, then the resultant will be the diagonal between these sides of parallelogram.”
Mathematically, resultant force,
Where R is the resultant force, A and B are component vectors and 
is the angle between the two vectors, as shown below.
For finding the direction of resultant vector, we need to find the angle between A and R i.e., 
.
Method of resolution for the resultant force:
1. Horizontal resolution of forces is done and sum is taken (i.e., ∑H).
2. Vertical resolution of forces is done and sum is taken (i.e., ∑V).
3. The resultant R of the given forces will be given by the equation:
4. When the resultant is being inclined by an angle of θ, with the horizontal, we have
Thus, value of the angle θ will vary depending upon the values of ∑V and ∑H as discussed below:
1. When ∑V is +ve, the resultant makes an angle between 0° and 180°. But when ∑V is –ve, the resultant makes an angle between 180° and 360°.
2. When ∑H is +ve, the resultant makes an angle between 0° to 90° or 270° to 360°. But when ∑H is –ve, the resultant makes an angle between 90° to 270°.
Moment of a Force about a point:
When a force is applied such that it creates rotation at the point of application along line of application, this tendency to rotate is called as torque, or the moment of a force. For example, consider a wrench used to unscrew the bolt in Fig. below.
Note that if the force F is applied at an angle less than 900, Fig. (bas moment arm is smaller, it will be more difficult to turn.
If F is applied along the length of wrench, Fig. (c), the moment arm will be zero. Hence moment will also be zero.
Thus, using the above observations, we can generalize accordingly.
The magnitude of the moment is given by,
where d is the moment arm or perpendicular distance from the axis at point O to the line of action of the force. Units are. N*m or lb*ft.
Resultant Moment: For two-dimensional problems, where all the forces lie within the x–y plane, Fig. below, the resultant moment about point O (the z axis) can be determined by finding the algebraic sum of the moments caused by all the forces in the system. As a convention, we will generally consider positive moments as counterclockwise since they are directed along the positive z axis (out of the page). Clockwise moments will be considered to be negative. Doing this, the directional sense of each moment can be represented by a plus or minus sign. Thus, fig. below shows the sign conventions.
As we know that product of Scalar and vector quantity is always a vector quantity. Hence, the Moment of a force or torque is a vector quantity, and can be calculated using cross-product.
The moment of a force F about point O, or actually about the moment axis passing through O and perpendicular to the plane containing O and F, Fig. below, can be expressed using the vector cross product, namely,
MO = r X F
Here r represents a position vector directed from O to any point on the line of action of F. We will now show that indeed the moment when determined by this cross product, has the proper magnitude and direction.
The magnitude of the cross product is defined as MO = rFsin
where the angle 
is measured between the tails of r and F. To establish this angle, r must be treated as a sliding vector so that 
can be constructed properly. Since the moment arm d = rsin
therefore,
MO = rFsin
MO = Fd
Cartesian Vector Formulation: If we establish x, y, z coordinate axes, then the position vector r and force F can be expressed as Cartesian vectors, Fig. (a) below.
Formulating the equation in matrix format we have,
Where,
rx, ry, rz represent the x, y, z components of the position vector drawn from point O to any point on the line of action of the force.
Fx, Fy, Fz represent the x, y, z components of the force vector.
If the determinant is expanded, we will get the value of MO as,
MO = (ryFz – rzFy) i – (rxFz – rzFx) j + (rxFy – ryFx) k
Moment of a Force about a Specified Axis:
Sometimes, the moment produced by a force about a specified axis must be determined. For example, suppose the lug nut at O on the car tire in Fig. above needs to be loosened. The force applied to the wrench will create a tendency for the wrench and the nut to rotate about the moment axis passing through O; however, the nut can only rotate about the y axis. Therefore, to determine the turning effect, only the y component of the moment is needed, and the total moment produced is not important. To determine this component, we can use either a scalar or vector analysis.
Scalar Analysis. To use a scalar analysis in the case of the lug nut in Fig. above, the moment arm perpendicular distance from the axis to the line of action of the force is dy = d cos 
. Thus, the moment will be
MOy = F (d cos
)
Vector Analysis: To find the moment of force F in above figure about the y axis using a vector analysis, we must first determine the moment of the force about any point O on the y axis by applying MO = r x F. The component My along the y axis is the projection of MO onto the y axis. It can be found using the dot product discussed above, so that My = j*MO = j*(r x F) where j is the unit vector for the y axis.
We can generalize this approach by letting ua be the unit vector that specifies the direction of the axis shown in Fig. 4–21. Then the moment of F about the axis is
This combination is referred to as the scalar triple product. If the vectors are written in Cartesian form, we have
Equivalent Moment:
The equivalent moment acting on a system of forces can be calculated as illustrated in the following examples:
e.g., 5: A force of 15 N is applied perpendicular to the edge of a door 0.8 m wide as shown in Fig. below. Find the moment of the force about the hinge.
Solution:
Given: Force applied (P) = 15 N and width of the door (l) = 0.8 m
Moment when the force acts perpendicular to the door as shown in fig.
We know that the moment of the force about the hinge is given as,
MO = P × l = 15 × 0.8 = 12.0 N-m
If the force is to be applied at a particular angle according to fig. below,
Then the calculations are done in following procedure.
As the force applied in the question above is inclined at an angle of 600 with the horizontal, we can calculate the moment of force by two methods, either by finding out the perpendicular distance between the hinge and the line of action of the force as shown in Fig. (a) or by finding out the vertical component of the force as shown in Fig. (b) below.
From the geometry of Fig. (a), we find that the perpendicular distance between the line of action of the force and hinge,
OC = OB sin 60° = 0.8 × 0.866 = 0.693 m
∴ Moment = 15 × 0.693 = 10.4 N-m
In the second case, we know that the vertical component of the force
= 15 sin 60° = 15 × 0.866 = 13.0 N
∴ Moment = 13 × 0.8 = 10.4 N-m
Calculating the resultant of Force System:
As discussed above, the equivalent force can be calculated using two methods, either by Analytical method like simple laws of vector addition, subtraction or Parallelogram law of vector addition or Method of Resolution.
e.g.,1: Two forces of 100 N and 150 N are acting simultaneously at a point. What is the resultant of these two forces, if the angle between them is 45°?
Solution:
Here, the two forces are First force (F1) = 100 N; Second force (F2) = 150 N and angle between F1 and F2 (θ) = 45°.
We know that using parallelogram law of vector addition, we get,
R = 232 N.
In this way, we can calculate the resultant between two force vectors.
e.g.,2: Two forces act at an angle of 120°. The bigger force is of 40 N and the resultant is perpendicular to the smaller one. Find the smaller force.
Solution:
In this problem, one of the forces is given as 40 N and the angle between the two forces is given as 1200, as depicted in fig below. Also, the angle between the resultant and smaller force is given as 900. Thus, the resultant parallelogram is shown below.
Angle between the forces ∠AOC = 120°, Bigger force (F1) = 40 N and angle between the resultant and F2 (∠BOC) = 90°;
Let F2 = Smaller force in N
From the geometry of the figure, we find that ∠AOB,
α = 120° – 90° = 30°
We know that
Now,
By putting the values as,
α = 300
= 1200
F1 = 40 N, we get,
Thus,
F2 = 20 N
In this way, we can calculate the unknown quantities by analytical method of force resolution.
Now, we will understand the procedure of finding the resultant or unknown quantity using the method of resolution of forces vertically and horizontally, as discussed earlier.
e.g., 3: A machine component 1.5 m long and weight 1000 N is supported by two ropes AB and CD as shown in Fig. given below. Calculate the tensions T1 and T2 in the ropes AB and CD.
Solution:
Given: Here, weight of the component is 1000 N. Thus, for solving this example we will resolve the forces vertically as well as horizontally. Initially we will resolve the forces horizontally. As the component is in equilibrium horizontally, (i.e., along BC) we will get:
T1 cos 60° = T2 cos 45°
Thus, T1 = 1.414 T2
Now, we will resolve the tension and self-weight vertically, we will get,
T1 sin 60° + T2 sin 45° = 1000
(1.414 T2) 0.866 + T2 × 0.707 = 1000
1.93 T2 = 1000
Thus, T2 = 518.1 N
Therefore, T1 = 1.414*518.1
T1 = 732.6 N.
In this way, using the resolution of forces vertically and horizontally, we can get the required quantities.
If we wanted to find the equivalent force of many forces acting on a body, we can use the resolution of forces method. It can be explained using the example below.
e.g., 4: The forces 20 N, 30 N, 40 N, 50 N and 60 N are acting at one of the angular points of a regular hexagon, towards the other five angular points, taken in order. Find the magnitude and direction of the resultant force.
Solution:
The system of forces is as shown in the fig. above. So, for finding the equivalent force, we will resolve the forces horizontally and vertically, and then calculate the resultant force.
Now, resolve the above-mentioned force system horizontally, (i.e., Along AB) we will get,
ΣH = 20 cos 0° + 30 cos 30° + 40 cos 60° + 50 cos 90° + 60 cos 120° N
= (20 × 1) + (30 × 0.866) + (40 × 0.5) + (50 × 0) + 60 (– 0.5) N
= 36 N
Also, resolving the force system vertically, we get,
ΣV = 20 sin 0° + 30 sin 30° + 40 sin 60° + 50 sin 90° + 60 sin 120° N
= (20 × 0) + (30 × 0.5) + (40 × 0.866) + (50 × 1) + (60 × 0.866) N
= 151.6 N
Now, for calculating the magnitude of resultant force of this system,
R = 155.8 N
For finding the direction of resultant force vector,
Let, θ = Angle, which the resultant force makes with the horizontal (i.e., AB).
We will use the formula,
Thus, θ = 76.6°
In this way, using the resolution of forces method, we can find the equivalent force of the system.
Couple:
Couple can be defined as combination of two forces which are of equal magnitude but are opposite in direction, and are separated by a perpendicular distance d as shown in figure below. This can be illustrated by considering that when we are driving a car, to take a right turn, the right hand imparts a pull downwards on the steering wheel and the left hand imparts a push on steering wheel upwards, thus creating a couple. This moment created by couple is known as Couple Moment.
This, in scalar formulation, Formula for couple is given as,
Where F is magnitude of force and d is perpendicular distance between the forces. For finding the direction, the same method is used which is used for moment about a point, i.e. Right Hand rule.
If more than two moments are created, their resultant can be determined using vector addition. i.e., MR = M1 + M2.
Generalized equation for more than one couple moments is given as,
A system of forces is said to be in equilibrium if the resultant force which is summation of all the forces acting on the body is zero. If we consider in Static Equilibrium, then the system is considered to be in rest condition. Thus, for a system to be in rest condition, it needs to satisfy the Newton’s First Law of motion. Thus, the generalized equation for a system to be in equilibrium states that resultant force acting on the system should be zero for the system to be in equilibrium,
i.e., 
.
This is further explained in the topic of equations of equilibrium for the system.
To apply the equation of equilibrium, we have to consider all types of forces (ΣF) which are acting on the particle. A drawing in which all the forces are depicted is called free body diagram (FBD).
Process to Draw a Free-Body Diagram
To construct a free body diagram, the following three steps are necessary.
Imagine that the particle is to be isolated from its environments by drawing its outlined shape.
b. Show All the forces.
Then on this sketch, show all the types of forces. Identify Each Force that are known and unknown and display them.
c. Identify each force
Known forces are displayed with magnitude and direction. Letters are used to represent the magnitudes and directions of forces that are unknown.
For understanding the free-body diagram, we will consider the following example:
Suppose, in this example there are two strings (AC and BC) attached to an object at point C. Thus, the first step involves drawing a simple diagram representing the outline of all components.
Then, we need to show all the forces that are contributing in the given diagram. Here, there is 15 N force acting at point C, and there is tension in strings BC and AC due to 15 N force acting at C. Thus, identifying and displaying these forces onto the outline diagram is the second step.
The third step is finding the magnitude and direction of the forces and displaying them on the outline diagram. Using these steps, we can draw the free-body diagram as shown below.
1. If the body moves in any direction, it means that there is a resultant force acting on it. In other words, the horizontal component of all the forces (Σ H) and vertical component of all the forces (ΣV) must be zero. Mathematically,
Σ H = 0 and Σ V = 0
2. If there is rotation of body, without moving, that means there is couple acting. In other words, the resultant moment of all the forces (Σ M) must be zero. Mathematically,
Σ M = 0
3. If the body moves in any direction and at the same time it rotates about itself, if means that there is a resultant force and also a resultant couple acting on it. In other words, horizontal component of all the forces (Σ H), vertical component of all the forces (Σ V) and resultant moment of all the forces (Σ M) must be zero. Mathematically,
Σ H = 0 Σ V = 0 and Σ M = 0
4. If the body is completely at rest, it means there is no force nor any couple acting on it. A little consideration will show, that in this case the following conditions are already satisfied:
Σ H = 0 Σ V = 0 and Σ M = 0
The above-mentioned equations are known as the conditions/equations of equilibrium.
When the static equilibrium equations, which may be in 2D or 3D, that we have discussed in above sections, are not enough for solving the forces present which are unknown, then the structure may be considered as Statically Indeterminate structure. For e.g., if we consider a 2D rigid body having 4 unknown quantities, in that case, we have 3 equilibrium equations. Thus, in this case, we cannot get all the unknown quantities using the equilibrium equations. Hence, this type of situation is called as Static Indeterminacy.
1. Irving H. Shames (2006), Engineering Mechanics, 4th Edition, Prentice Hall
2. F. P. Beer and E. R. Johnston (2011), Vector Mechanics for Engineers, Vol I - Statics, Vol II, – Dynamics, 9th Ed, Tata McGraw Hill
3. R. C. Hibbler (2006), Engineering Mechanics: Principles of Statics and Dynamics, Pearson Press.
4. Andy Ruina and Rudra Pratap (2011), Introduction to Statics and Dynamics, Oxford University Press
5. Shanes and Rao (2006), Engineering Mechanics, Pearson Education
