STATICSStatics is the branch of mechanics concerned with the analysis of loads (force, torque/moment) on physical systems in static equilibrium, that is, in a state where the relative positions of subsystems do not vary over time, or where components and structures are at a constant velocity. When in static equilibrium, the system is either at rest, or its center of mass moves at constant velocity. The study of moving bodies is known as dynamics.

By Newton's first law, this situation implies that the net force and net torque (also known as moment of force) on every body in the system is zero. From this constraint, such quantities as stress or pressure can be derived. The net forces equaling zero is known as the first condition for equilibrium, and the net torque equaling zero is known as the second condition for equilibrium. See statically determinate.

Solids

Statics is used in the analysis of structures, for instance in architectural and structural engineering. Strength of materials is a related field of mechanics that relies heavily on the application of static equilibrium. A key concept is the centre of gravity of a body at rest: it represents an imaginary point at which all the mass of a body resides. The position of the point relative to the foundations on which a body lies determines its stability towards small movements. If the center of gravity falls outside the foundations, then the body is unstable because there is a torque acting: any small disturbance will cause the body to fall or topple. If the centre of gravity falls within the foundations, the body is stable since no net torque acts on the body. If the center of gravity coincides with the foundations, then the body is said to be metastable.

Composition and Resolution of Concurrent Forces by Vector Methods

THEORY: Measurable quantities may be classified as either (1) scalar quantities or (2) vector quantities. A scalar quantity has magnitude only, but a vector quantity has both magnitude and direction. For example, since to specify completely the velocity of a body it is necessary to state not only how fast it is traveling but in what direction it is going, velocity is a vector quantity. However, the mass of a body is completely specified by a magnitude, and hence mass is a scalar quantity. Since the weight of a body is the force with which it is attracted by the earth, weight has a downward direction and thus is a vector quantity. Since weight and mass are different physical concepts, they should not be measured in the same units. The gram is a unit of mass. The force with which the earth attracts a one-gram mass at a standard location sometimes is called a “gram-weight” of force. *

*Since weight is proportional to mass in any given locality, this experiment is not affected by the slight variations consequent to laboratory conditions

In order to add scalar quantities, one has merely to make the algebraic addition. When one wishes to add two vector quantities, the process is more difficult because their directions must be considered. The vector sum of two vector quantities is the single vector quantity that would produce the same result as the original pair.

The addition of vector quantities is greatly simplified by representing the vector quantity graphically. A vector is the line segment whose length represents the magnitude of a vector quantity and whose direction is that of the vector quantity. The sense along the line is indicated by an arrow. For example, a force of 100lb. acting at an angle of 30° above the horizontal may be represented by the line OA. Fig. 1, which is 5 units long and has the correct direction. Each unit of length thus represents 20lb.

When vectors do not have the same line of action, their vector sum is not their algebraic sum but a geometric sum.

This geometric sum may be determined by either graphical or analytical methods. Graphical methods are simple and direct but are limited in precision to that obtainable by drawing instruments. Analytical methods have no such inherent limitations. In this experiment both graphical and analytical methods will he applied to forces as examples of vector quantities, but the same methods apply to all vector quantities.

The vector sum, or resultant, of a set of forces is the single force that will have the same effect, insofar as motion is concerned, as the joint action of the several forces.

Vector Summation by Graphical Methods: As an example of vector addition let us consider the case of two forces acting on a body in such a direction that the forces are concurrent, that is their lines of action, if projected would intersect at a point. The vectors OA and OB representing two such forces are shown in Fig. 2. Their vector sum or resultant R, is found by constructing a parallelogram having the two vectors as sides and drawing the concurrent diagonal, as shown in Fig. 3. This diagonal vector R represents in magnitude and direction the single force that is equivalent to the origina1 pair, that is their vector sum. When the resultant of more than two vectors is to be obtained graphically a polygon method is used. This is illustrated in Fig. 4. The vector A is first constructed by the use of a chosen scale

and reference direction. Then, from the head of A, the vector B is drawn. It is clear that the vector M is the resultant of vectors A and B, since M would be the concurrent diagonal of a parallelogram if such a parallelogram had been drawn, as was done in Fig. 3. Similarly, it follows that the vector R is the resultant of M and C or of A, B, and C. When the resultant of several forces is required this method is simpler than the parallelogram method. It should be noted that when the parallelogram method is used, the arrows, with their tails together, all radiate from a common point. But in the polygon

method the tail of the second arrow coincides with the head of the first, etc.

Summation of Vectors by Analytical Methods: The resultant of two vectors may be determined analytically by the use of the trigonometric laws of sines and cosines. Consider the vectors A and B in Fig. 5. The magnitude of the resultant R can be obtained by the application of the law of cosines:

R2 = A2 + B2 + 2ABcosβ (1)

The direction of R can then be obtained from the law of sines:

Sinφ/Sinβ = B/R

Since sinβ = sinθ,

Sinφ = (B/R)sinθ (2)

Components of Vectors: Any single force may be replaced by two or more forces whose joint action will produce the same effect as the single force. These various forces are said to be components of the single force. The most useful set of components is usually a pair at right angles to each other, as shown in Fig. 6.

The force B is the resultant of Forces Bx and By. Therefore conditions are unchanged by replacing the single force B by forces Bx and By, called their X- and Y- components. It is obvious from Fig. 6 that Bx = B cosβ and By = B sinβ.

Component Method for Addition of Vectors: Fig. 7 illustrates the component method of computing the resultant of A, B,

and C. The X- axis is so chosen that it coincides with the vector A, and the vectors B and C are resolved into X- and Y-components. The three forces A, B, and C have been replaced by five forces (A has no Y- component). The slim of the component along either axis may be computed by algebraic addition. Calling the sum of the X-components Fx and the sum of the Y-components Fy, it follows that the resultant R is given by the equation

R2 = (Fx)2 + (Fy)2 (3)

and that the angle φ- the angle that R makes with the X-axis may be determined from the equation

tanφ = Fy/Fx (4)

Equilibrium: Many problems that concern the physicist and engineer involve several forces acting on a body under circumstances in which they produce no change in the motion of the body. This condition is referred to as equilibrium. The body does not necessarily have to be at rest, but its motion must retain the same velocity; hence both magnitude and direction of motion are unchanged.

First Condition for Equilibrium: Insofar as linear motion is concerned, a body is in equilibrium if there is no resultant force acting upon it, that is if the vector sum of all the forces is zero. This statement is called the first condition for equilibrium. This condition is satisfied if the vector polygon representing all the

external forces acting on the body is a closed figure. Analytically this condition is satisfied if each set of rectangular components of the forces separately adds to zero, or

Rx = ΣFx = 0 (5)Ry = ΣFy = 0 (6)

Mechanical equilibrium

A standard definition of static equilibrium is:

A system of particles is in static equilibrium when all the particles of the system are at rest and the total force on each particle is permanently zero.[1]

This is a strict definition, and often the term "static equilibrium" is used in a more relaxed manner interchangeably with "mechanical equilibrium", as defined next.[2]

A standard definition of mechanical equilibrium for a particle is:

The necessary and sufficient conditions for a particle to be in mechanical equilibrium is that the net force acting upon the particle is zero.[3]

The necessary conditions for mechanical equilibrium for a system of particles are:

(i)The vector sum of all external forces is zero; (ii) The sum of the moments of all external forces about any line is zero.[3]

As applied to a rigid body, the necessary and sufficient conditions become:

A rigid body is in mechanical equilibrium when the sum of all forces on all particles of the system is zero, and also the sum of all torques on all particles of the system is zero.[4]

[5]

A rigid body in mechanical equilibrium is undergoing neither linear nor rotational acceleration; however it could be translating or rotating at a constant velocity.

However, this definition is of little use in continuum mechanics, for which the idea of a particle is foreign. In addition, this definition gives no information as to one of the most important and interesting aspects of equilibrium states – their stability.

An alternative definition of equilibrium that applies to conservative systems and often proves more useful is:[6]

A system is in mechanical equilibrium if its position in configuration space is a point at which the gradient with respect to the generalized coordinates of the potential energy is zero.

Varignon's theorem

Varignon's theorem is a statement in Euclidean geometry that was first published by Pierre Varignon in 1731. It deals with the construction of particular parallelogram (Varignon parallelogram) from an arbitrary quadrangle.

The midpoints of the sides of an arbitrary quadrangle form a parallelogram. If the quadrangle is convex or reentrant, i.e. not a crossing quadrangle, then the area of the parallelogram is half as big as the area of the quadrangle.

Lami's theorem

Lami's theorem in statics states that

if three coplanar forces are acting on a same point and keep it stationary, then it obeys the relation where A, B and C are the magnitude of forces acting at the point (say P), and the values of α, β and γ are the angles directly opposite to the forces C, B and A respectively.

A/sinα = B/sinβ = C/ sinγ

Lami's theorem is applied in static analysis of mechanical and structural systems.

Free body diagramBlock on a ramp (top) and corresponding free body diagram of just the block (bottom).

A free body diagram is a pictorial representation often used by physicists and engineers to analyze the forces acting on a free body. A free body diagram shows all contact and non-contact forces acting on the body. Drawing such a diagram can aid in solving for the unknown forces or the equations of motion of the body. Creating a free body diagram can make it easier to understand the forces, and moments, in relation to one another and suggest the proper concepts to apply in order to find the solution to a problem. The diagrams are also used as a conceptual device to help identify the internal forces—for example, shear forces and bending moments in beams—which are developed within structures.[1][2]

Construction

A free body diagram consists primarily of a sketch of the body in question and arrows representing the forces applied to it. The selection of the body to sketch may be the first important decision in the problem solving process. For example, to find the forces on the pivot joint of a simple pair of pliers, it is helpful to draw a free body diagram of just one of the two pieces, not the entire system, replacing the second half with the forces it would apply to the first half.

What is included

The sketch of the free body need include only as much detail as necessary. Often a simple outline is sufficient. Depending on the analysis to be performed and the model being employed, just a single point may be the most appropriate.

All external contacts, constraints, and body forces are indicated by vector arrows labeled with appropriate descriptions. The arrows show the direction and magnitude of the various forces. To the extent possible or practical, the arrows should indicate the point of application of the force they represent.

Only the forces acting on the object are included. These may include forces such as friction, gravity, normal force, drag, or simply contact force due to pushing. When in a non-inertial reference frame, fictitious forces, such as centrifugal force may be appropriate.

A coordinate system is usually included, according to convenience. This may make defining the vectors simpler when writing the equations of motion. The x direction might be chosen to point down the ramp in an inclined plane problem, for example. In that case the friction force only has an x component, and the normal force only has a y component. The force of gravity will still have components in both the x and y direction: mgsin(theta) in the x and mgcos(theta) in the y, where theta is the angle between the ramp and the horizontal.

What is excluded

All external contacts and constraints are left out and replaced with force arrows as described above.

Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the ball applies a force to the table, and the table applies an equal and opposite force to the ball. The FBD of the ball only includes the force that the table causes on the ball.

Internal forces, forces between various parts that make up the system that is being treated as a single body, are omitted. For example, if an entire truss is being analyzed to find the reaction forces at the supports, the forces between the individual truss members are not included.

Any velocity or acceleration is left out. These may be indicated instead on a companion diagram, called "Kinetic diagrams", "Inertial response diagrams", or the equivalent, depending on the author.

Assumptions

The free body diagram reflects the assumption and simplifications made in order to analyze the system. If the body in question is a satellite in orbit for example, and all that is required is to find its velocity, then a single point may be the best representation. On the other hand, the brake dive of a motorcycle cannot be found from a single point, and a sketch with finite dimensions is required.

Force vectors must be carefully located and labeled to avoid assumptions that presuppose a result. For example, in the accompanying diagram of a block on a ramp, the exact location of the resulting normal force of the ramp on the block can only be found after analyzing the motion or by assuming equilibrium.

Other simplifying assumptions that may be considered include two-force members and three-force members.

Example

A simple free body diagram, shown above, of a block on a ramp illustrates this.

All external supports and structures have been replaced by the forces they generate. These include:

mg: the product of the mass of the block and the constant of gravitation acceleration: its weight.

N: the normal force of the ramp.

Ff: the friction force of the ramp.

The force vectors show direction and point of application and are labeled with their magnitude.

It contains a coordinate system that can be used when describing the vectors.

Numerical Example 1. Two loads 400N and 500N are suspended in a vertical plane by three springs as shown in Fig. /Exmpl.1. Find the tension in the strings

Fig. /Exmpl.1

Solution: Obviously tension in OB = TOB = 500N

Resolving the forces at O vertically POA.sin300 + POB.sin30o = 400Nor, POA = 300N

Resolving the forces at O horizontallyPOC + POA.cos30o = POB.cos30O

or, POC + 300N(cos30o) = (500N)cos30o

400N

500N

A

B

CO

30°30°

POC = 100√3 N

Tension in OA = 300NTension in OB = 500NTension in OC = 100√3 N

Problem. 1 An electric light fixture weighing 15 Newtons hangs from a point C, by two strings AC and BC. AC is inclined at 60° to the horizontal and BC at 45° to the vertical as shown in Fig. /Prob.1.Using Lami’s theorem or otherwise determine the forces in the strings AC and BC.

15N

Ans- TCB =7.76N ; TCA =10.98N

B

A

45O

60O

C