Chapter 2: Force and Force Systems

A: FORCES

2.1 Preface

2.2 Actions and Effects Of Forces

2.3 Force Distributions

2.4 Force As A Vector Quantity

2.5 Principle Of Transmisibility

2.6 Addition Of Forces

2.7 Cartesian Force Vector

2.8 Resolution Of Forces

B: MOMENTS

2.9 Basic Concept Of Moments

2.10 Formulation Of Moments Using Vectors

2.11 Moments About An Inclined Axis

2.12 Varignon’s Theorem

2.13 Couples

2.14 Torques

C: EQUIVALENT FORCE SYSEMS

2.15 Definition Of Equivalent Force Systems

2.16 Equivalent Force System 1: Single Resultant Force

2.17 Equivalent Force System 2: Combination of a Resultant Force And a Couple

2.18 Distributed Forces

2.19 Summary

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2.1. PREFACE

Statics, as well as whole study of mechanics, is the study about the actions of forces and force systems on bodies and the effects of these actions.  An understanding of the characteristics of force systems and specific methods to analyse them, forms the basis to master the study of mechanics.

In this chapter, we will study the action of forces and force systems commonly encountered in engineering analyses.  To facilitate the study, the chapter is divided into three parts: Part A deals with forces, Part B deals with moments, and Part C deals with equivalent force systems.

A: FORCES

2.2. ACTIONS AND EFFECTS OF FORCES

A force is defined as the action of a body on another body.  A force is applied either through a direct contact or through a remote action.  Forces applied through a remote action are gravitational, electrical, and magnetic forces.  All other forces are applied through direct contacts.

A force acting on a body produces effects that can be divided into two types, namely external effect and internal effect.

As an illustration, consider the structure portrayed in Figure 2.1.  Assume that the structure is of negligible mass.  The applied force F acting on the structure causes the reactive forces R1 and R2 which are applied by the support surfaces onto the structure so as to balance F.  The reactions R1 and R2 are effects external to the body which are caused by F.  Hence, external forces acting on a body are of two types, namely applied external forces (also called active forces) and reactive forces.

 

 

Figure 2.1

Beside causing the reactions R1 and R2, the applied force F also causes effects internal to the members of the structure.  These effects are in the forms of stresses and strains that appear in the material of the sturcture. The effects exist to balance the respective external actions and effects.

In the study of mechanics, only the external effects of a force are considered.  The internal effects are studied in disciplines of study specific to them, for example Mechanics Of Materials (also known as Strength Of Materials).

2.3 FORCE DISTRIBUTION

Any force applied to a body will act on a finite area of application.  The force is distributed over the area of application.  In many cases, the area of application is extremely small compared to the dimensions of the body being upon acted by the force.  In such cases, the dimension of the area of distribution can be neglected and the area of application can be considered as a point.  A force acting on such a point of action is called a concentrated force.  The tension of the cable of the crane in Figure 2. 2(a) is a concentrated force.

 

 

 

Figure 2.2

If the dimension of the area of application of a force cannot be neglected, we obtain a distributed force which acts either a line, an area, or a volume.  The push of the wind acting on the sign board in Figure 2.2(b) ia a distributed force.

The effect of a distributed force depends on the nature of distribution of the force.  This will be studied in more details in Section 2.18 and in Chapter 6.

2.4 FORCE AS A VECTOR QUANTITY

A concentrated force is a vector quantity.  This has been determined through experiments which shows that concentrated forces add-up according to the parallelogram law and when a concentrated is applied to a body, the effect depends on the magnitude, direction, and point of application of the force.  Hence, mathematical analyses of a concentrated force is based on the mathematical rules of vectors (see Appendix).

MAGNITUDE AND DIRECTION OF A FORCE.  The magnitude of a force is a scalar quantity.  In the SI system, it is measured in the unit newton (N).  A force acting on a body is illustrated graphically by a vector arrow at the point of application of the force, where the orientation of the arrow represents the direction of the line of action of the force whilst the arrow head represents the sense of the force, Figure 2.3.  A force is fully defined once its magnitude, direction, sense, and point of application is specified.

 

 

 

 

Figure 2.3

The magnitude, direction, and sense of a force F can be represented fully by a mathematical expression using a unit vector s which corresponds to line of action and sense of the force.  For example, the force F in Figure 2.4 has a magnitude F.  Its direction of action corresponds to that of the unit vector s.  Hence F can be written as F=Fs.

 

 

 

 

Figure 2.4

2.5 PRINCIPLE OF TRANSMISSIBILITY

In Mechanics, we consider only the external effects of a force on the body that it acts on. This consideration enables us to handle an applied force as a sliding vector because the external effects of a force depend on its magnitude, sense, and line of action only.  The position of the point of application of the force does not influence the effects involved in the analyses of mechanics.  As an example, the rigid body in Figure 2.1 mentioned earlier, is now shown again Figure 2.5(a).  Line a-a represents the line of action of an applied force F acting on point A.   If the force F is made to slide along its line of action from A to a new point of application B, the values of the reactions exerted by the support surface on the structure do not change.  The same applies to the trolley in Figure 2.5(b).  The same motion is obtained whether the trolley is pulled from the front or or pushed from the back with the same effort acting one the same line of action.  This fact is formulated from experience into what is called the principle of transmissibility which states that:

The overall effect of the action of an external force on a rigid body is dependent only on the magnitude, direction, and line of action of the force.

Noe that this principle cannot be used if the what to be analysed is an effect of the force on any part internal to the body.  In such a case, the force needs to be handled as a fixed vector.  In Mechanics, all concentrated forces are taken as sliding vectors and the principle of transmissibility can be used for all analyses.

2.6 ADDITION OF FORCES

In general, forces are added according to the parallelogram laws.  Here, we will discuss special cases often met in our analyses.

2.6.1 Concurrent Forces

Parallelogram law.  Two concurrent forces, F1 and F2, are added graphically according to the parallelogram law on their common plane to produce the resultant force P.  For addition by this method, the tails of the vectors are placed at a common point, Figure 2.6(a).  If necessary, the principle of transmissibility can be applied as shown in Figure 2.6(b).  The forces F1 and F2 in Figure 2.6(a) and 2.6(b) are replaced by a single force P without changing the external effects on the body that they acted on.  This addition is written mathematically by using the vector expression

P = F1 + F2

From mathematics, it is known that every vector equation can solve for two unknown scalar quantities.  Hence, in the solution of this equation, two unknowns can be determined.  The two unknowns are normally the magnitude and direction of the resultant force P.

Triangle law.  The addition of vectors illustrated by the equation given above can also be carried out graphically by using the triangle rule, Figure 2.7.  The application of this rule requires the shifting of the line of action of one of the forces F1, F2 or P.  If the lines of action of F1 and F2 are maintained, then the resultant force P has the correct magnitude and direction but its line of action is not obtained because the line  of action does not pass through the point of intersection O, Figure 2.7(d).

Often, more than two concurrent forces are to be added.  The resultant is obtained by adding, in turn, two forces by two forces, using the parallelogram , Figure 2.8(a), or the triangle law, Figure 2.8(b).

Polygon law.  In the case of planar forces, the resultant can also be obtained by drawing the vectors of the forces head-to-tail to form the respective force polygon, Figure 2.9.  Note that a closed polygon is obtained if the resultant force is zero.

Calculation of values.  When the three laws are applied, the unknown values are obtained as follows:

  1. if a scaled drawing is made, the unknown values are measured directly from the drawing; and
  2. if a sketch is made, the sine rule or cosine rule is used to get the unknowns.

2.6.2 Parallel Forces

Parallel forces P and Q are normally added by adding the magnitudes of the forces, |P+Q|=|P|+|Q|.  Use the negative sign for opposing direction.  The direction of the resultant force is the same as the direction of the original forces, Figure 2.10(a).  The position of the line of action of the resultant is normally determined by using the principle of moments, as later discussed in Section 2.16.3.

The parallelogram law can also be used but not directly.  The forces are added using the parallelogram law by first adding two equal , opposite, and collinear forces to the original system.  One of the added forces is applied to the point of action of the first original force whilst the other is applied to the point of application of the second original force.  The added forces, F and -F, do not give rise to any external effects on the body which the forces act on.

The process of adding of two parallel forces F1 and F2 is carried out according to the order shown in Figure 2.10(b).  F is added to F1 to produce the resultant P1.  Then -F is added to F2 to produce the resultant P2.  The resultant P, the addition of F1 and F2, is  obtained by adding P1 and P2.  Hence, for two parallel forces, the resultant P is obtained through

P = F1+ F2 = (F1+F) + (F2-F)

The force P obtained has the correct magnitude, direction, and line of action.  Note that the magnitude of P is the same as the algebraic sum of the magnitudes of F1 and F2 and its direction is the same as the direction of the forces being summed up.  I ought to be stated here that the position of the line of action of the resultant force produced by a parallel-force system is determined by using the principle of moments, as stated above.

The actual solution for the addition of forces is obtained either:

(a) by drawing, using a suitable scale,  the polygon of the forces concerned and measuring the required quantities from the polygon; or

(b) by sketching the polygon of forces and then calculating the values required using geometrical calculations.

Note that adding two known forces means determining the magnitude and direction of their resultant.

In general, this method enables the determination of two scalar unknowns in a system consisting of six quantities, namely the magnitude and direction of the two original forces and the magnitude and direction of the resultant force.  Any two of these quantities can be determined when the rest are known.  Please note that, in the system described, only four quantities can be chosen arbitrarily; the other two are fixed by them.

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EXAMPLES

EXA2.1.

EXA2.2

EXA2.3

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EXERCISES

EXE2.1-EXE2.15

(Pls refer to Tutorial Manual)

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2.7. CARTESIAN FORCE VECTORS

Any force F can be written in the form of a Cartesian vector.  Such an expression uses the addition of the force’s component vectors in the x, y, and z directions of the axes of the right-hand coordinate.  If the magnitude of its component in the x direction is Fx, in the y direction Fy, an in the z direction Fz, then F is expressed as

F = Fx i + Fy j + Fz k

where i, j, and k are the Cartesian unit vectors.

Expressing A Force Vector By Using Its Magnitude And Two Points On Its Line Of Action.  The Cartesian vector expression for a force is often obtained by using the Cartesian position vector that gives the Cartesian unit vectors corresponding to the direction concerned.  For example, the force F in Figure 2.11(b) has a line of action that passes through points P(x1,y1,z1) and Q(x2,y2,z2).

 

 

 

Figure 2.11

Let the position vector r that points from P to Q describes the relative position between P and Q.  From the diagram, we obtain the relationship

r = -r1 + r2

However, from Figure 2.11(b), we can formulate the relationships  r1 = x1i + y1j + z1k and r2 = x2i + y2j + z2k.  By inserting these relationships into the terms for  r1 and r2 in the equation above, we get

 r = (x2x1)i + (y2y1)j + (z2z1)k          …..(2.2)

Hence the components of r in the x, y, and z directions are obtained by subtracting the coordinates of the tail from those of the heads.   Note that if the coordinates of Q are subtracted from those of P, we will get the vector -r.

From Equation 2.22, we will get the unit vector s (corresponding to the line of action and the direction of F) as

s = r

=                                                                    …..(2.3)

In this expression, as usual, r is the magnitude of the position vector r and is given by the relationship

r =

A force of magnitude F and having its line of action, direction, and sense corresponding to the unit vector s in Figure 2.3 can thus be expressed in the Cartesian vector form by writing F = F s and then expanding the expression on the right of the equal sign.

The coordinate direction angles for the line of action of F are obtained from the cosine direction relationship for s, i.e.

 

…..(2.4)

ADDITION OF FORCES.  The concept of force resultant can be applied to a concurrent force system which is written in the Cartesian vector form.  If P is the resultant of F1, F2, F3, … hence

PFi

=  (F1x + F2x + …)i + (F1y + F2y + …)j + (F1z + F2z + …)k

= ΣFixi + ΣFiyj + ΣFizk                                                                                                                 …..(2.5)

where ΣFix, ΣFiy, and ΣFiz are the sums of the magnitudes of the forces in the corresponding direction.

2.8 RESOLUTION OF FORCES

Any force can be resolved into its components. There are a number of methods that can be used to resolve a force. The method use depends on the problem at hand.  The different methods are described below.

2.8.1. Parallelogram Law

A force acting at any point can be resolved into components that act in two desired directions through the parallelogram law.  Force F in Figure 2.12(a), for example, can be replaced by two components acting in directions 1 and 2.  The resolution is implemented by drawing a parallelogram with F as the diagonal and its two non-parallel sides along directions 1 and 2, Figure 2.12(b).

 

 

 

 

Figure 2.12

Note that resolution is the reverse process of adding two forces into their resultant.  Hence the resolution involves six quantities, i.e. the magnitude and direction of F and of the two components, where two of them can be determined when the other four are known.

Rectangular components.  When force F is resolved into perpendicular directions, we obtain the rectangular components.  This is shown in Figure 2.13 for a 2D case, where the respective perpendicular directions are represented by the x-axis and the y-axis.  In this case, we get

Fx = F cos θ

Fy = F sin θ

where  θ = tan-1 (Fy/Fx)

Addition Of Forces.   Forces can be added by using their components, normally the rectangular components in the Cartesian directions. Consider forces F1 and F2 that are concurrent at O being added using the force polygon, Figure 2.14(a).  The resultant P is

P = F1 + F2

where

 

 

2.8.2. Direction Cosines

2.8.3. Dot Product

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EXAMPLES

EXA2.4.

EXA2.5.

EXA2.6.

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B: MOMENTS

2.9. THE GENERAL CONCEPT FOR A MOMENT

2.10. FORMULATION OF A MOMENT USING VECTORS

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EXAMPLES

EXA2.7.

EXA2.8.

EXA2.9.

EXA2.10.

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2.11. MOMENTS ABOUT AN INCLINED AXIS

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EXAMPLES

EX2.11.

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2.12. VARIGNON’S THEOREM

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EXAMPLES

EX2.12.

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2.13. COUPLE

2.14. TORQUE

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EXAMPLE

EX2.13.

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C: EQUIVALENT FORCE SYSEMS

2.15. DEFINITION OF EQUIVALENT FORCE SYSTEMS

2.16. EQUIVALENT FORCE SYSTEM 1: SINGLE RESULTANT FORCE

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EXAMPLES

EXA2.14.

EXA2.15.

EXA2.16.

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2.17. EQUIVALENT FORCE SYSTEM 2: COMBINATION OF A RESULTANT FORCE AND A COUPLE

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EXAMPLE

EXA2.17.

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2.18. DISTRIBUTED FORCES

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EXAMPLE

EXA2.18.

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2.19. SUMMARY

This chapter discusses forces and their systems.  To facilitate the study, this chapter is divided into three parts that cover three areas, namely forces, moments, and equivalent force systems.

Part A introduces the concept of concentrated forces and explains the methods of handling the forces mathematically.  It is stressed that a concentrated force is a vector quantity and the force obeys the laws of vector algebra.

Part B discusses a moment, i.e. the tendency of a force to rotate a body that it acts on.  The methods to formulate a moment using scalar and vector approaches are explained. Thereafter, the concept of  a moment about an inclined axis is explained. The way to handle moments mathematically is also explained.  At the end of this part, moments of types couples and torques are introduced.

Part C discusses equivalent force systems, i.e. the simplest forms of a force system that can replace a given force system without changing the external effects of the original force system.  Two equivalent force systems are discussed: a single resultant force and combination of a resultant force and a resultant couple. Force systems that can be replaced by one of the equivalent force systems and the method to carry out the replacement have been described.   At the end of this part, the concept of distributed forces was introduced and it is stressed that a distributed force can be replaced by a single resultant force that acts through the centroid of the original given force. [END]

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