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Space Forces

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Space Forces

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  1. www.ekeeda.com Contact : 9029006464 Space Forces Space Forces Email : care@ekeeda.com P INTRODUCTION We have so far discussed and worked with a coplanar all system of forces, wherein the forces in a system would lie in a single plane i.e. it was a two dimensional force system. In this chapter we deal with" a system of forces lying in different planes forming a three dimensional force system. Such a system is also referred as a space Force System and a vector approach is required to deal with these problems. The chapter begins with the study of basic operations with forces using a vector approach. Later on we will learn to find the resultant of a space force system and finally we shall deal with equilibrium problems. BASIC OPERATIONS USING VECTOR APPROACH Since the forces have three dimensions, we employ a vector approach, which simplifies the working. Here we will learn to represent a force vectorially, to find vectorially moment of a force about a point and about a line, to find magnitude and direction of a force given in vector form and other basic operations useful in solution of a space force problem. Force in vector form Fig. shows a force of magnitude F in space passing through A (x1, y1, z1) and. B (x2,y2, z2). The force in vector form is  ˆ F F.e AB                    x x i y y j z z k  2 1 2 1 2 1 F F x       2 2 2      x y y z z 2 1 2 1 2 1    F F i F j F k Note:i, j and k printed in bold type denote unit x y z vectors along the x, y and z axis respectively. 1

  2. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com MAGNITUDE AND DIRECTION OF FORCE  Fig. shows a force making angles F F i+F j+Fk x y z  xy and z, with the x, y and z axis respectively. Here Fx is the component of force in the x direction. Similarly Fy and Fz are the force components in the y and z direction. The magnitude of the force is    2 2 2 F F F F x y z Also,    F F F cos θ Fcosθ x x y y F Fcos θ z z Here xy and  z are known as the force directions, the value of which lies between 0 and 180. There is an important identity which relates them.       2 2 2 cos cos cos 1 x y z 2

  3. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com MOMENT OF A FORCE ABOUT A POINT This is a very important operation while dealing with forces. For coplanar forces, moment about g point was the product of the force and the  distance. Here if the force is in space the moment calculation requires a vector approach. Fig. shows a force F in space passing through points A (x1, y1, z1) and B (x2, y2, z2) on its line of action. Let C (x3, y3, z3) be the moment centre i.e. the point about which we have to find the moment. The procedure of finding the moment of the force about the point is as follows. Step 1: Put the force in vector form i.e.  F F i+F j+Fk x y z Step 2: Find the position vector extending from the moment centre to any  point on the force i.e. . r r i+r j+r k x y z Step 3: Perform the cross product of the position vector and the force vector to get the moment vector i.e. F r F   M point i j k r F  r F r F x y z x y z 3

  4. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 1 1.A force of 50 N acts parallel to the y axis in the -ve direction. Put the force in vector form. 2.A 130 KN force acts at B (12, O, O) and passes through C (0, 3, 4). Put the force in vector form. 3.A force F = (3i -4j 12 + k) N acts at a point A (1,-2, 3) m. Find a.moment of the force about origin. b.moment of the force about point B (2, l,2)m. 4.A force F = 80 i + 50 j - 60 k passes through a point A (6, 2, 6). Compute its moment about a point B (8, 1, 4) 5.A 700 N force passes through two points A (-5,-1, 4) towards B (1, 2, 6) m. Find moment of the force about a point C (2,-2, 1)m. 6.A force of magnitude 1200 N passes from A (2, -4, 6) to B (-3, 2, 3). Calculate the moment of this force about the origin. Also determine the component of this force along the line AC where C has the position vector defined by i 2j 3kN     . 7.A rectangular plate is supported by brackets to the wall at A and B by wire CD as shown in figure. Knowing that tension in wire is 2OO N determine the moment about point A, of the force exerted by wire on point C. All dimensions are in mm. 8.Find moment of force P = 500 N about point H. (M. U. Dec12) c r 9.Three forces act on a rectangular box as shown in figure. Determine the moment of each force about the origin. 4

  5. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com RESULTANT OF CONCURRENT SPACE FORCE SYSTEM Resultant of a concurrent space force system is a single force R , which acts through the point of concurrence. Fig. (a) shows a concurrent system at point P. The resultant of the system is shown 1n Fig. (b) and calculated as, RESULTANT OF PARALLEL SPACE FORCE SYSTEM The resultant of a parallel space force system is a single force R which acts 0 parallel to the force system. The location of the resultant can be found out using Varignon's theorem. Figure (a) shows a parallel system of three forces F1, F2 and F3. The resultant of the system is shown in figure (b) and is calculated as,    R F F F 1 2 3 The resultant acts at P. The coordinates (x, y, z) of the point P can be calculated by using Varignon's theorem, the moments for which can be taken about any convenient point like point O. The equation of Varignon's theorem  F O R O  M M for space forces is 5

  6. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com RESULTANT OF GENERAL SPACE FORCE SYSTEM A general space force system is neither a concurrent nor a parallel system. The resultant of such a system is a single force R and a moment M at any desired point. Since the resultant contains one force and one moment, it is also known as a Force Couple System. Fig. (a) shows a general system of four forces F1, F2, F3 and F4. If it is desired to have the resultant at point O, then as per figure (b). The single force 1 2 R F F    and the single moment O M M M    F F 4 3   F O F F O F O M M 3 1 2 4 6

  7. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 2 1.A force P1 = 10 N in magnitude acts along direction AB whose co-ordinates of points A and B are (3, 2, -l) and (8, 5, 3) respectively. Another force P2 = 5 N in magnitude acts along BC where C has co-ordinates (-2, 11, -5). Determine a.The resultant of Pr and Pz. b.The moment of the resultant about a point D (1, 1, 1). 2.Knowing that the tension in AC = 20 kN, determine the required values of TAB and TAD so that the resultant of the three forces applied at A is vertical. Also find the resultant. 3.A plate foundation is subjected to five vertical forces as shown. Replace these five forces by means of a single vertical force and find the point of replacement. 4.Determine the loads to be applied at A and F, if the resultant of all six loads is to pass through the centre of the foundation of hexagonal shape of side 3 m. 7

  8. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 5.Knowing that the resultant of the three parallel forces is 500 N acting in positive y direction and passing through the centre of the rectangular plate, determine the forces F1 , F2 and F3. 6.A square foundation mat supports four columns as shown in figure. Determine magnitude and point of application of resultant of four loads. 7.A rectangular parallelepiped carries four forces as shown in the figure. Reduce the force system to a resultant force applied at the origin and moment around the origin. OA = 5 m, OB = 2m, OC = 4 m. (M. U Dec 12) 8.Figure shows a rectangular parallelepiped subjected to four forces in the direction shown. Reduce them to a resultant force at the origin and a moment. 8

  9. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 9.A tetrahedron A B C D is loaded by forces F1 = 100 N at A along DA, F2 = 2OO N at B along CB and F3 = 300 N at C along DC as shown in the figure. Replace the three force system by a single resultant force R at B and a single resultant moment vector M at B. Take the co-ordinates in metre units. 10.The following forces act on the block shown in figure. F1 = 40 kN at point C along CD, F2 = 3O kN at point D along DB and F3 = 1OO kN at point D along DE. Find the resultant force and resultant moment of these forces acting at O. (SPCE Nov 12) 9

  10. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EQUILIBRIUM OF SPACE FORCES When the resultant of a system is zero, the system is said to be in equilibrium. For the resultant to be zero, the resultant force F and the resultant moment M should be zero. This gives us six scalar equations of equilibrium viz.       x F  M  0 0 x y F  M  0 0 y zF  M  0 0 z SUPPORTS FOR SPACE STRUCTURE a)Ball and Socket Support This support allows rotation in all the three direction, unlike a hinge which allows rotation about one axis only. A ball and socket support does not allow any linear movement in any direction. This results in a reaction force having components along x, y and z axis. Fig. shows a ball and socket support offering three components of reaction force. b)Fixed End Support A fixed end will not allow rotation in any direction, nor will allow any linear movement. This results in a reaction moment and a reaction force having components along x, y and z axis. Refer Fig. 10

  11. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com Equilibrium of Concurrent Space Force System For a concurrent system, only the resultant force needs to be zero. This is the necessary and sufficient condition of equilibrium. This gives us three scalar equations of equilibrium, viz.    x F  0 y F  0 zF  0 We can also use the other three scalar equations of equilibrium viz.    M  0 x M  0 y M  0 z EQUILIBRIUM OF NON-CONCURRENT SPACE FORCE SYSTEM For equilibrium of both parallel and general space force systems, all the six scalar equations of equilibrium are applicable viz.       x F  M  0 0 x y F  M  0 0 y zF  M  0 0 z 11

  12. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 3 1.A boom AB supports a load of 1000 N as shown. Neglect weight of the boom. Determine tension in each cable and the reaction at A. 2.A vertical load of 1100 N is supported by the three rods shown in figure. Find the force in each rod. Points C, O and D are in XZ plane while point B lies 5 m above this plane. 3.A vertical load of 1OOO N is supported by three bars as shown. Find the force in each bar. Point C, O and D are in the x-z plane while B is 1.5 m above this plane. 4.A vertical tower DC shown is subjected to a horizontal force P = 5O kN at its top and is anchored by two similar guy wires BC and AC. Calculate a.Tension in the guy wires. b.Thrust in the tower pole. Co-ordinates of the points are as below, O (0, 0, 0), B (0, 0, -4), D (3, 0, 0), A (0, 0, 4), C (3, 20, 0) 12

  13. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 5.A uniform thin plate ABCD, square in shape is suspended by strings GA (2a m length), GD (2a m length), GE  3amlength and GH (a m length) so that it  remains in horizontal plane. If the size of the plate is 2a x 2a and E is the midpoint of BC, find the tension in each of the string if the plate weighs 400 N. 6.A square steel Plate 2400 mm  2400 mm has a mass of 1800 kg with mass centre at G. Calculate the tension in each of the three cables with which the plate is lifted while remaining horizontal. Length DG = 2400 mm. 7.A crate is supported by three cables as shown. Determine the weight of the crate, if the tension in the cable AB is 750 N. 8.Determine the tension in cable BC and BD and reactions at the ball socket joint A for the mast as shown in figure. 13

  14. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 9.A vertical mast OD is having base ‘0’ with ball and socket. Three cables DA, DB and DC keep the mast in equilibrium. If tension in the cable DA is 100 kN, find tensions in the cables DB and DC and force in the mast. Refer figure. 10.A load of 500 N is held in equilibrium by means of two strings CA and CB and by a force p as shown in figure. Determine tensions in strings and magnitude of P. 11.A uniform equilateral triangular plate weighing 2.4 kN is held in horizontal plane by means of three cables at A, B and C. An additional load of 1.2 kN acts at midpoint of edge AC of the plate. Determine the tensions in the three cables. 12.A plate ACED 1O mm thick weighs 76O0 kg/ms. It is held in horizontal plane by three wires at A, B and C. find tensions in the wires. 14

  15. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 13.Figure shows steel Pipe ABCD of length 1.8 m having 900 bend at B. The Pipe weighs 30 N/m and supported by three vertical wires attached to the Pipe at A, C and D. Find tension in wires. EXERCISE 4 THEORY QUESTIONS Q.1 How is a space force represented in magnitude and direction. Q.2 Write the equations of equilibrium to be satisfied for non co-planar concurrent force system. (SPCE Nov 12) Q.3 State conditions of equilibrium for non-concurrent forces in space. UNIVERSITY QUESTIONS 1.A pole is held in place by three cables. If the force of each cable acting on the pole is as shown in fig. determine the resultant. (6 Marks) 2.A force of 10 kN acts at a point P(2,3,5) m and has its line of action passing through Q(10, -3, 4)m. Calculate moment of this force about a point S(1,-10,3) m. (4 Marks) 3.Explain conditions for equilibrium for forces in spare. (6 Marks) 15

  16. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 4.Force F = (3i – 4j + 12k) N acts at point A (1,-2,3) Find -- (4 Marks) (i) Moment of force about origin (ii) Moment of force about point B (2,1,2)m. 5.A rectangular parallelepiped carries Three forces shown in fig. Reduce the force system to a resultant force applied at the origin and a moment around origin. (6 Marks) 6.The lines of action of three forces concurrent at origin ‘O’ pass respectively through points A(-1,2,4), B (3,0,-3) and C(2,-2,4)m. The magnitude of forces are 40N, 10N and 30N respectively. Determine the magnitude and direction of their resultant. (6 Marks) 7.A force of 1200N acts along PQ, P(4,5,-2) and Q (-3,1,6)m. Calculate its moment about a point A(3,2,0)m (4 Marks)    8.A force passes through a point A Compute its moment F 80 i 50 j 60 k about point B (8,1,4). (6 Marks) 16

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