STATICS UNIT I FORCES INTRODUCTION

Transcription

1 STATICS UNIT I FORCES INTRODUCTION Mechanics: Mechanics is the science which deals with the effects of forces on material bodies. Under the influence of forces, a body may be in motion or in rest. Statics: Statics is the science which deals with the conditions for lack of motion under given forces. Force: Force as any cause which produces or tends to produces a change in the exisisting state of rest of a body or of its uniform motion in the straight line. Types of forces: Attraction & Repulsion Tension or thrust Action & Reaction Equilibrium: i. When a numbers of forces act on the body and keep it at rest, the forces are said to be in equilibrium. ii. If two forces be equal and opposite then these two forces are in equilibrium. Resultant of two forces on a particle: If a particle is acted on by two forces the particle. is said to be the resultant force on 1

2 Bookwork (1): To find the magnitude and direction of the resultant of Expression for the resultant of two forces acting at a point. Proof: Let be any two forces then the resultant is To find the magnitude of the resultant: Magnitude of (or) Analytical To find the direction of resultant: Let be the angle between and the resultant Here Corollary 1: If are of equal 2

3 Corollary 2: If are of perpendicular to each offer then choosing in their directions Book work 2: To express the force acting in a plane in terms of its components in two perpendicular directions in the plane. Proof: Let be the give force and let be the unit vectors alloy the perpendicular directions. Then 3

4 Problems: 1. The magnitude of the resultant of two given forces P,Q is R, If Q is doubled then R is doubled. If Q is reversed, then also R is doubled. Show that Solution: Given be two forces and be the resultant of two forces. Give Q=2Q then R=2R Give Q= -Q then R=2R From From From 4

5 The Square ratio 2. The resultant of the forces acting at O is R. if any transversal meets the lines of action of R at Proof: Let be the two forces and the transversal meets these forces at respectively. Drawn ON, the perpendicular to the transversal. Let be the unit vector along ON The angles between and the forces are 3. Two forces of magnitudes act at a point. They are inclined at an angle. If the forces are interchanged, show that their resultant is turned through the angle. Proof: Let the two forces be then the resultant is -----(1) If these two forces are interchanged the new resultant is 5

6 Let be the triangle between where is the angle between. Then Sing componendo and dividend rule, 6

7 4. Two forces of magnitudes P+Q, P Q acting at a point include an angle. Show that if their resultant makes with the bisector of the angle between the forces an angle Then. Proof: Let the two forces be and the angle between the two forces are The resultant of the two forces Direction of the resultant is Here, 7

8 RESULTANT OF THE FORCES RELATED TO A TRIANGLE ACTING A POINT. 1. Forces of magnitudes, act on a particle. If their directions are parallel to where ABC is a triangle show that the magnitude of their resultant is Solution: The given forces are. 8

9 2. Find the magnitude and direction of the resultant of three co-planer forces P, 2P, 3P acting at a point and inclined mutually at on equal angle of 120 Proof: Let P, 2P, 3P be the given three forces. Considering P, P, P from the given 3 forces. The system of 3 forces is reduced to the system of 2forces The magnitude of resultant of 3 forces Direction of resultant of 2 forces 9

10 RESULTANT OF SEVERAL FORCES ACTING ON A PARTICLE: 1. E is the midpoint of the side CD of a square ABCD. Forces act along Show that they are in equilibrium. Solution: Let ABCD be a square and E is a midpoint of the side CD. All the forces act through A. let be the unit vectors along AB, AD Then ir AB = a, Consider So the unit vectors along are 10

11 EQUILIBRIUM OF A PARTICLE: TRIANGLE OF FORCES: If three forces acting on a particle can be represented in magnitude and direction by the sides of a triangle, taken in order, then the forces keep the particle in equilibrium. Converse of triangle of forces: If a particle is kept in equilibrium by three forces, then the forces can be represented in magnitude and direction by the sides of a triangle, taken in order Proof: Let the three forces be in equilibrium (ie) Let AB and AD denote The diagonal AC is the resultant in magnitude and direction. Complete the parallelogram ABCD. Hence CA represented. LAMI S theorem: If a particle is in equilibrium under the action of three forces forces then to show that Proof: Let the three forces keep the particle in equilibrium Multiplying by =0 11

12 Multiply by From 2 & 3 12

13 Problems: 1. I is the incentre of a triangle ABC. If forces of magnitudes P,Q,R acting along the bisectors IA, IB, IC are ub equilibrium show that. Proof: Let P, Q, R be the magnitude of three forces acting along IA, IB, IC such that ABC which are the equilibrium The opposite angles of P, Q, R are Now from Similarly. 13

14 LIMITING EQUILIBRIUM OF A PARTICLE ON AN INCLINED PLANE. Bookwork: Suppose a particle of weight W lying on a rough plane inclined at an angle to the horizontal is subjected to a force P along the plane in the upward direction. If the equilibrium is limiting to find P. Proof: Now the following two cases of limiting equilibrium arise: (i) The particle has a tendency to slide up the Plane. (ii) The particle has a tendency to slide down the plane. Case (i): Now it is enough to consider the forces acting on the particle along the plane. They are (i) (ii) (iii) component of weight P up the plane Frictional force F down the plane. Where F has its maximum value so that Where is the co-efficient of friction. Thus in this case, We have Case (ii): In the case in which the particle has a tendency to move down word the frictional force is up the plane and hence we ve 14

15 UNIT II FORCES ON A RIGID BODY Moment of a force: Let be a force and A, a point on its line of action. Let O be a point in space. Then the vector is called the moment of about O. 1. Forces of magnitudes 3P, 4P, 5P act along the slides BC, CA, AB of an equilateral triangle of slide a. find the moment of the resultant about A. Sol: Let the moment of the resultant about A equals the sum of the moment of the individual forces about A. But the forces 4P, 5P Passes through A. So their moment about A are Zero. 15

16 EQUIVALENT (OR EQUIPOLENT) SYSTEMS OF FORCES: Definition: Two systems of forces which produce the same motion on a given rigid body are equilvalent or equipollent. Parallel forces: Forces whose lines of action are parallel are called parallel forces. If their directions are in the same sense, then they are called like parallel forces. Otherwise they are called unlike parallel forces. Bookwork: Find the resultant of two parallel forces acting on a rigid body. Proof: Case (i) : Like parallel forces: Let the like parallel force acting at respectively. Let be the unit vector in the direction of. Introduce a force. Since these two forces are equal in magnitude and opposite in direction and act along the same line. Their introduction will not affect the system Let 16

17 Complete the parallelograms. Now the resultant of The resultant of These two resultant they meet at 0. The resultant of these two resultant is Note that this resultant is parallel to the original forces. To find the point of inter section of. From similar triangles, From similar triangles Which implies x divides. Case ii: Let the given unlike parallel force be acting at respectively. Let be the unit vector in the direction of. Introduce a force. Since these two forces are equal in magnitude and opposite in direction their introduction will not affect the system 17

18 Let Complete the parallelograms. Now the resultant of The resultant of The resultant of these two resultant of these two resultant is Note that the resultant is not parallel to the given forces. To find the point of intersection of. From similar triangles, 18

19 From similar triangles Which implies x divides. Problems: 1. If two like parallel forces of magnitude P, Q(P>Q)acting on a rigid body at A, B are interchanged in position. Show that the line of action of the resultant is displaces through distance Solution: Let AB = a and let the resultant intersect AB at a distance If the forces are interchanged let the distance between the resultant A be at 19

20 Equation - we get Which is the displacement of the resultant. 2. A uniform plank AB of length 2a and weight W is supported horizontally on two horizontal pegs C and D at a distance d part. The greatest weight that can be placed at the two ends in succession with out upsetting the plank are Solution: Let G be the mass centre and CG = x, GD = y. let When the greatest weight is placed at A. In this situation taking moment about C. Similarly when the weight is zero. Taking moments about D. 20

21 Equation + we get, VARIGNON S THEOREM: The sum of the moment of two intersecting or parallel forces about any point is equal to the moment of the resultant of the forces about the same point. Proof: Case (i): Intersecting forces: Let the lines of action of the forces intersect at A. Then the moments are are Then moment of moment of Case ii: Parallel forces: Let the parallel forces be acting. Let Moment of Moment of 21

22 The sum of these moments = moment of the resultant. COUPLES Couple : A couple, acting on a right body, is just a pair of unlike parallel forces of equal magnitude acting on the body along two different lines. Moment of a couple: The sum of the moment of the constituent forces of a couple about any point is called the moment of the couple. Arm and axis of couple: If is the unit vector perpendicular to the plane of a couple such from a right handed trial, then the moment of the couple is Where is the angle between is the perpendicular distance between the lines of action of the forces. P is called the arm of the couple and the direction of is called the axis of the couple. 22

23 Book work: A system of coplanar forces reduce to a single force or to a couple. Proof: Let n coplanar forces act on a rigid body. Then, choosing two of them which do not form a couple, we can reduce them to a single force. This reduction will reduce the given system to n-1 forces. If this process is repeated successively then the system will reduce to just two forces. If these two forces are unlike parallel forces of equal magnitude, then they form a couple. Otherwise, they further reduce to a single force. Therefore, a system of coplanar forces reduce either to a couple or to a single force. EQUILIBRIUM OF A RIGID BODY UNDER THREE COPLANAR FORCES Cotangent formulae or trigonometrically formula: If D is any point on the base of triangle ABC, is the ratio m:n and angle Solution: Case i: Consider 23

24 Case ii: Consider 24

25 UNIT III FRICTION When two bodies are in contact with one another a force is exerted between then at their point of contact to prevain one body from slidiny on the other is called friction. A force exerted is called force of friction. Types of friction: 1. Statical Friction 2. Limiting Friction 3. Dynamical Friction Laws of friction: Law 1: The friction acts opposite to the direction in which the body moves or has a tendency to move. Law 2: When the bodies are at rest, the friction called in to play is just sufficient to prevent the motion of any of the bodies. Law 3: There is limit to the amount of friction that can be called in to play. This limit bears a constant ratio to the normal reaction. This constant depends upon the nature and material of the surfaces in contact and not on the measure of the areas in contact. Law 4: When a body slides on another, the friction called into play is slightly less then the limiting friction. Law 5: When a body slides on another, the friction called into play is independent of its relative speed. 25

26 Co-efficient of Friction: The ratio of the limiting friction to the normal reaction is called the Co-efficient of friction. It is denoted by. Problems: 1. A ladder which stands on a horizontal ground learning against a vertical wall, has its centre of gravity at distances a and b from its lower and upper ends respectively. Show that, if the ladder is in limiting equilibrium, and if at the lower and upper contacts, its inclination to the vertical is given by are the co-efficient of friction Proof: A has a tendency to move away from the wall and has the tendency to move down wards. Therefore, the frictional force at A is towards the wall and the frictional force at B is opposite direction to the wall. The forces acting on the ladder are 26

27 Now the weight and the reaction at B and the reaction at A meet at a point 0. Then AO is inclined to the normal AD to the floor at an angle is inclined to the normal BD to the wall at an angle. Consider the Then from the co-tangent formula. 2. A solid hemisphere of weight W rests in limiting equilibrium with its curved surface on a rough inclined plane, its plane force being kept horizontal by a weight, P attached to a point A in its rim prove that the co-efficient of friction is Solution: Let OB be the centre and a, the radius of the sphere and B, the point of contact of the inclined plane. The forces acting on the hemisphere are Since w and p are vertical, R should be inclined to BO, the normal to the plane at B. Taking moments about B. 27

28 Consider, 3. A rod is in limiting equilibrium resting horizontally with its ends on two inclined planes which are at right angles and one of which makes an angle with horizontal. If the co-efficient of friction is the same for both the ends, show that Solution: A has a tendency to slide downwards and B has a tendency to slide upwards. The forces acting on the rod are So that OG is vertical but, AB is in horizontal. So OA=OB. 28

29 Hence the 4. A rod AB rests within a fixed hemispherical bowl whose radius is equal to the length of the rod. If is the co-efficient of friction, show that in limiting equilibrium, the inclination of the rod to the horizontal is given by. Solution: The radii AO, BO and a rod AB form an equilateral triangle. The reactions of the bowl at A and B and the weight of the rod meet at l. Then AL, BL are inclined to AO, BO and the same angle of friction 29

30 By cotangent formula, 30

31 5. A uniform ladder AB rests in a limiting equilibrium with the end A on a rough floor, the co-efficient of friction being and with the other end B against a smooth vertical wall. Show that, If is the inclination of the ladder to the vertical then (i) (ii) If, find Solution: A has a tendency to move away from the wall. Therefore the frictional force F at A towards the wall. The forces acting on the ladder are Let the weight and the reaction at B intersect at O. Also the reaction at A passes through O. The reaction is inclined through the normal to the floor. (ie) to the vertical at an angle. Now consider, the Then by cotangent formula, 31

32 LIMITING EQUILIBRIUM OF A PARTICLE ON AN INCLINED PLANE Bookwork: Suppose a particle of weight W lying on a rough plane inclined at an angle to the horizontal is subjected to a force P along the plane in the upward direction. If the equilibrium is limiting to and P. Sol: Case i: The particle has tendency to slide up the plane. The forces acting on the system are (i) The force P up the plane. (ii) The frictional forces F down the plane. (iii) The normal reaction R acting at the particle. (iv) The weight of the particle w acting downwards. Resolving horizontally. Resolving Vertically. Substitute Equation 2 in 1. Case (ii): The particle has tendency to slide down the plane. The forces acting on the system are (i) The force P up the plane. (ii) The frictional forces F up the plane. (iii) The normal reaction R acting at the particle. (iv) The weight of the particle w acting downwards. (v) Resolving horizontally. 32

33 Resolving horizontally. Substitute Equation 2 in 1. PROBLEMS: 1. A ladder is in limiting equilibrium having contact with a rough horizontal floor and a rough vertical wall whose co-efficient of friction are. If is the inclination of the ladder to the vertical then show that, Solution: A has a tendency to move away from the wall and B has a tendency to move downwards. The frictional forces at A is towards the wall and the frictional force at B is the frictional force at B is opposite direction to the wall. The forces acting on the ladder are, Now the weight and the reaction at B and the reaction at A meet at a point O. Then AO is inclined to the normal AD to the floor at angle and BO is inclined to the normal BD to the wall at angle. Consider the Then from the cotangent formula. 33

34 . 2. A body weight 4kg rest in limiting equilibrium on a rough plane whose slope is if the plane is raised to a slope find the force along the plane required to support the body. Solution: Case (i) The forces acting on the system are, (i) The frictional forces F up the plane. (ii) The normal reaction R acting upward. (iii) The weight 4kg of the particle acting downwards. Resolving horizontally. 34

35 Resolving Vertically Substitute 2 in 1 we get, Case (ii) The forces acting on the system are, (i) The frictional forces F up the plane. (ii) The normal reaction R acting upward. (iii) The weight 4kg of the particle acting downwards. (iv) P acting upward. Resolving horizontally Resolving horizontally Substitute 2 in 1 we get, 35

36 UNIT IV VIRTUAL WORK Work: A force is said to do work, when its point of application moves. Virtual work: The work done by the forces in a virtual work done by the forces. Virtual displacement: A change in the configuration of a dynamical system, as a result of any arbitrary infinitesimal change in the coordinates of the system, is called a virtual displacement of the system. Bookwork: Principle of virtual work: If a dynamical system is in equilibrium, then the work done by the applied forces in a virtual displacement is zero, provided the virtual displacement is such that the forces, due to the constraints imposed on the system, do no work. Proof: Let the system be constituted by n particles of masses and their position vectors with reference to a fixed point O, be the forces acting an each particle may be classified as (i) Applied forces (ii) Internal forces due to its contact with other particles. (iii) Forces due to the external constraints imposed on the system. Let the forces on a general particle of mass be (i) (ii) (iii) 36

37 Equal in magnitude but opposite in direction Let be the infinite simal displacement of in a virtual displacement in which the forces due to the constraints imposed on the system do no work. If are particles having contact with each other, then and consequently, The equation of motion of is Multiply this scalarly by Adding such equation for all of we get, Of which vanishes by (2) and vanishes by the imposed condition that the virtual displacement is such that the forces due to the constraints do no work. By the system is in equilibrium. So the accelerations are zero. 37

38 1. Four equal uniform rods are linked together to form a rhombus ABCD and the frame is suspended from A. The corners A and C are connected by an inelastic string to prevent the frame from collapsing. Show that the tension in the string is 2w, where w is the weight of each rod. Solution: We shall use the principle of virtual work. Let us have the following assumption, P, Q, R, S: Mid points AB, BC, CD, DA 2a: length of each rod. Q: inclination of each rod to the vertical. Then the forces acting on the system, their point of application and their position vectors with respect to the fixed point A are as follows, Point Force Position Vectors Displacement P Q C R S A 38

39 2. A square frame work formed by four rods of equal weights w joined together, is lung up by one corner. A weight w is suspended from each of the three lower corners. If the shape of the square is preserved by a light rod along the horizontal diagonal diagonal, show that its thrust T on the frame is 4 W. Solution: We shall use the principle of virtual work, let us have the following P, Q, R, S: Mid points AB, BC, CD, DA 2a: length of each rod. Q: inclination of each rod to the vertical. The inclination of each rod to the vector. 39

40 Then the forces acting on the system, their point of application and their position vectors with respect to the fixed point A are as follows, Position Vectors Displacement Point Force P B Q C R D S A 40

41 3. A ladder of weight w rests at an angle to the horizon, with its ends resting on a smooth floor and against a smooth vertical wall, the lower end being attached by string to the junction O of the wall and the floor. Find the tension of the string when a man whose weight is one-half of the ladder, has ascended the ladder two-thirds of its length Solution: Lets us have the following assumption 2a: length of the rod. G: Mass centre of the ladder. There are five forces act on the ladder. The weight of the ladder and the man, the tension and normal of two reaction. In a displacement due to a small change in, normal reaction R and S do no work. The forces which do work are as follows with respect to the origin O. Position Vectors Displacement Point Force M G A 41

42 4. A uniform beam of length 2a, rests in equilibrium against a smooth vertical wall and upon a smooth peg O at a distance d from the wall. Show that in the position of equilibrium, the beam is inclined to the wall at an angle given by Solution: In a displacement due to a smell change in, the normal reaction S,R of the peg and the wall do no work. The only force which does work is the weight of the rod, namely where is the unit vector. Let Ol, MN be the perpendiculars drawn from O,M to the wall. Then with the represent to the fixed point O. 42

43 To find LN In 43

44 UNIT V HANGING STRINGS Bookwork: Derive the equation of catenary (or) derive the intrinsic equation of catenary (or) show that the shape of a uniform string henging under gravity is a catenary. Proof: Let A and B be the points of suspension not lying in the same vertical line. Let C be the lowest point of the string and Q be the Point intersection of the tangents at C and P. Let be the angle between these tangents. Let S be the length of the arc CP. Then the applied forces acting on the portion CP of the String are: (i) The tension T at P along the tangent (ii) The tension T o at C along the tangent (iii) The weight sw acting through Q. Let the length of the string whose weight equals T o be C. then T o =CW. Now the portion CP is in equilibrium. Therefore, the sum of the horizontal components of the applied forces on CP is Zero. And the sum of vertical components equal to Zero. Resolving horizontally, 44

45 Resolving Vertically, Bookmark: To obtain the equation the curve formed by the string in the parameter in form as And in the Cartesian form Proof: We know that, Consider 45

46 Integrating with respect to Initial condition are A=0 Consider, 46

47 Next to prove Cartesian form: W.K.T Adding 1 & 2 47

48 Problems: 1. A uniform string of length l is suspended from the points A,B in the same horizontal line. i) If the tension at A is n times the tension at the lowest point C, then show that the span is ii) If the tension at A is twice that at the lowest point C, then show that the span is Solution: The given tension at Tension at I given We know that 48

49 WKT Sub 1 in * ii) n= 2 49

50 2. A string of length 2 l hangs hangs over two small pegs in the same horizontal level. Show that, if h is the sag in the middle, the length of either part of the string that hangs vertically is Solution: Let A and B be the pegs and C be the vertex of the catenary. Form by the string. Let O be the point blow C at a distance C and ox, oy, be the x & y. Let a be the length of the string hanging on either side, WKT 50

51 3. Show that the length of a chain whose ends are tied together and which is hanging over a circular pulley of radius a. so as to be in contact with two-thirds of the circumference of the pulley, is Sol Let o be the centre of the circle. Let ACB be the hanging portion of the chain in the form of a centenary. Now Let the length of arc CA be l. Then the intrinsic Co-ordinates. Here AL = x 51

52 Comparing 1 & 2 Definition Span & Sag If A, B are the points of suspension in the sama horizontal line, AB is called span and the depth of C below AB is called sag. 52

53 Sag: Bookwork: To calculate approximately the sag of a telephone wire in terms of its length and span when the wire is tightly pulled and tied and to find the tension if w is the weight of the wire per unit length. Proof: Let the length of the wire be 2l, its span AB = 2a and its sag CD = b. let the tension at C be denoted by CW. Take O below C at the depth C and horizontal line ox. X=a; y =b+c; S=C To find the parameter C: WKT 53

54 To find sag: Consider Tension at 54

55 Tension at 1. Prove that the difference between the length and span Proof: WKT Sag Multiply & Divided by 2 55

56 Suspension Bridge A suspension bridge is a bridge hung horizontally from two parallel hanging cables by means of vertical strings. At equal horizontal distances. Bookmark: A heterogeneous string, the weight of any portion of which varies as its projection on the horizontal, hangs under gravity. To show that the curve formed by the string is a parabola. Proof: Let A, B be points of suspension and C, the lowest point of the string. Choose the horizontal and vertical lines through C in the plane of the string as the x, y axes. Let p(x, y) be any point on the string. Let the weight of the string vertically above a unit length be w. then the weight of the portion CP of the string is wx. The other forces acting on the portion CP are the tension T at P and the tension to say wc, at C. Since the portion CP is in equilibrium, from forces at CP are, Tension T at p(x, y) Tension Resolving horizontally, 56

57 Resolving vertically, 2 & 1 To find the value of A. the initial condition are x=0 & y=0 A = 0 Which represent to the parabola. 57

58 1. The span of a suspension bridge is 100m. and the sag at the middle of each cable is 10m. if the total load on each cable is 750 quintals, find the greatest tension in each cable and the tension at the lowest point. Solution: Let ACB be one cable with C as its middle point. Its form is parabolic. Choosing the horizontal and vertical lines cx and cy as the x, y axis, the equation of the parabola is of the form. Now terminal A is (50, 10) This satisfies The weight w per unit length along the bridge is The greatest tension T at A is, The lowest tension 58