Equilibrium of Forces and Friction from Chapter 2 & 3 (PDF)

 Chapter – 2 Equilibrium of Forces If the resultant of a number of forces, acting on a particle is zero, the particle will be in equilibrium. The force, which bring the set of forces in equilibrium is called on equilibrie...

 Chapter – 2 Equilibrium of Forces If the resultant of a number of forces, acting on a particle is zero, the particle will be in equilibrium. The force, which bring the set of forces in equilibrium is called on equilibrient. The equilibriant is equal to the resultant force in magnitude, but opposite in direction. Analytical conditions of equilibrium of a co-planner system of concurrent forces. We know that the resultant of a system of co-planner concurrent forces is given by : R  x2   y 2 , where  y0  x = Algebraic sum of resolved parts of the forces along a horizontal direction.  y = Algebraic sum of resolved parts of the forces along a vertical direction. R 2   x    y  2 2 If the forces are in equilibrium, R=0 0   x    y  2 2 Sum of the squared of two quantities is zero when each quantity is separately equal to zero.   x  0 and  y  0 Hence necessary and sufficient conditions of a system of co-plan concurrent forces are : i) The algebraic sum of the resolved parts of the forces is some assigned direction is equal to zero. ii) The algebraic sum of the resolved parts of the forces is a direction nat right angles to the assigned direction is equal to zero. ( 1 ) Graphical conditions of equilibrium of a system of co-planner concurrent forces : Let a number of forces acting at a point be in equilibrium. Then, it can be said that the resultant of these forces is nil. Hence the length of the closing line of the polygon drawn to represent the given forces taken in orders will be nil. In other words, the length of closing line of the vector diagram drawn with the given forces in orders, is nil. This means the end point of the vector diagram must coin side with the starting point of the diagram. Hence the vector diagram must be closed figure. So, graphical condition of equilibrium of a system of co-planner concurrent forces may be stated as follows : If a system of co-planner concurrent forces be in equilibrium, the vector diagram drawn with the given forces taken in orders, must be closed figure. Lami’s Theorem : If there concurrent forces are acting on a body, kept in an equilibrium, then each force is force is proportional to the sine of the angle between the other two forces and the constant of proportionality is the same. Consider forces, P,Q and R acting at a point ‘O’ mathematically, hom’s theorem is given by the following equilibrium : P Q R   K Sin Sin Sin Since the forces are on equilibrium, the triangle of forces should close, con responding to the forces, P,Q and R acting at a point ‘O’. The angle of triangle are A     B     C     ( 2 ) From the rule for the triangle we get. P Q R   Sin(   ) Sin(   ) Sin(   ) Sin(   )  Sin Sin(   )  Sin Sin(   )  Sin Or we can write the equation (1) according to Lami’s Theorem i.e. P Q R   Sin Sin Sin Example : An electric light fixture weighing 15 N hangs from a paint C, by two strings AC and BC. The string AC is inclined at 600 to the horizontal and BC at 450 to the vertical as shown in figure, using lami’s theorem determine the forces in the strings AC and BC. Solution : Given weight at C = 15 N TAC = Force in the string AC. TBC = Force in the string BC. ( 3 ) According to the system of forces and the above figure, we find that angle between TAC and ISN is 1500 and angle between TBC and 15 N is 1350. ACB  180 0  (400  60 0 )  750 Applying Lami’s Theorem : 15 TAC TBC 0  0  Sin75 Sin135 Sin1500 15Sin.1350  TAC   10.98 N (Ans) Sin750 15Sin.1500 and TBC = TBC   7.76 N (Ans) Sin750 Body Diagram : Free body diagram is a sketch of the isolated body, which shows the external forces on the body and the reaction extended on it by the removed elements. The general procedure for constructing a free body diagram is as follows : 1. A sketch of the body is drawn, by removing the supporting surfaces. 2. Indicate on this sketch all the applied on active forces which send to set the body in motion, such as those caused by weight of the body on applied forces etc. 3. Also indicate on this sketch all the reactive forces, such as those caused by the constrains or supports that tend to prevent motion. (The sense of unknown reaction should be assumed. The correct sense will be determined by the solution of the problem. A positive result indicates that the assumed sense is correct A negative results indicates that the correct sense is opposite to the assumed sense). 4. All relevant dimensions and angles, reference axes are shown on the sketch. Example : ( 4 )  CHAPTER – 3 FRICTION When a body moves or trends to move over another body, an opposing force develops at the contact surface. This force opposes the movement is called frictional force or friction. Frictional force is the resistance offered when a body moves over another body assist the motion. There is a limit beyond which the magnitude of this force cannot increase when applied force more than this limit there will be motion. When applied force less than this limit value, the body remains at rest and such frictional force is called static friction. When body moves (applied force more than limiting friction) the frictional resistance is known as Dynamic friction. Dynamic friction is found less than limiting friction. Sliding friction Dynamic friction Rolling friction Coefficient of friction : Experimentally found that the magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces and this ratio is called coefficient of friction : Co-efficient of friction = F/N = µ F  Limiting friction (Frictional force) N  Normal reaction Laws of friction : 1) Force of friction is directly proportional to the normal reaction and always opposite in the direction of motion. 2) Force of friction depends upon the roughness/ smoothness of the surface. ( 1 ) 3) Force of friction is independent of the areas of the contact. 4) Force of friction is independent of the sliding velocity. Angle of friction : Consider the block ‘A’ resting on horizontal plane ‘B’. Let, P  horizontal force applied. F  Frictional force N  Normal reaction. Let R be the resultant reaction between normal reaction and force of friction acts at angle  to the normal Reaction.   Angle of friction. Tan  = F/N As P increases, F increases and hence  also increases.  can reach the maximum value α when F reaches limiting value Tan α = F/N = µ  Α  Angle of friction. (Angle of limiting friction) Angle of Repose : It is the maximum inclined plane with the horizontal for which a body lying on the inclined plane will be on the point of sliding down. Angle of Repose is equal to the angle of friction. Consider a block of weight ‘W’ resting on a rough inclined plane (angle) Rn Normal reaction f Frictional force F= W Sin  , Rn = W Cos  E Tan  = Rn ( 2 )  E But we knew that tan  = =  Rn Tan  = Tan   = Ex – 1 A body resting on a horizontal plane can be moved slowly along the plane by a horizontal force of 10 KN. A force of a 9 KN inclined at 300 to the horizontal direction will suffice to move the block along the same direction. Determine the co-efficient of friction and weight of the body. Solution : Let  = Co-efficient of friction W = Weight of the body. From fig-1 F1 =  Rn1 = 10 ----------------------(1) Rn1 = W ----------------------(2) So  W = 10 --------------------------- (3) 9 Cos 300 = F2 =  Rn2 --------------- (4) Rn2 + 9 Sin 30 = W -------------------- (5) Rn2 = W - 4.5 -------------------------- (6) Putting the value of Rn2 in equation(4)  [W-4.5] = 9 Cos 300 = 7.794  W -  4.5 = 7.794 ------------------- (7) Putting the value of  W from (3) 10 -  4.5 = 7.794  4.5 = 10 - 7.794  = 0.49 ( 3 ) Putting the value of  in equation (3).49 W = 10 W = 20.408 KN. A body resting on rough horizontal plane, required a pull of 18 KN at 300 to the plane just to move it. It was also found that a push of 22 KN inclined at 300 to the plane just moved the body. Determine the weight of the body and co-efficient of friction. Solution : Let W Weight of the body  Co-efficient of friction From fig - 3 F1 =  Rn1 = 18 Cos 300 = 15.59 KN -------------(1) RN + 18 Sin 300 = W RN = W - 9 ------------------------------------------ (2) Putting the value of equation (2) in equation (1)  (W-9) = 15.59 --------------------------(3) From fig. 4. F2 =  Rn2 = 22 Cos 30 = 19.05 -------------------(4) Rn2 = W + 22 Sin300 = W + 11 ---------------(5) PUtting the value of equation (5) in equation (4)  (W+11) = 19.05 -----------------------------(6) From equation (3) and equaiton (6) 15.59 19.05  ,  W 9 W  11 15.59 W 9 = 19.05 W  11 ( 4 ) W = 99.12 KN Putting the value of W in equation (3)  (99.12  9)  15.59 15.59   0.173 90.12 Ex – 3 What is the value of P in the system shown in fig- 5 to cause the motion of 500 N block to the right side ? Assume the pulley is smooth and the co-efficient of friction between other contact surface is 0.20. Solution : Consider the FBD of fig. 5.1 N1 = 750 Cos 600 N1 = 375 750 Sin 60 +  N1 = T 750 Sin 60 + 0.2 x 375 = T = 724.52 N ---------------------(2) Consider the FBD of fig. 5.2 N2 + P Sin 300 = 500 N2 = 500 - 0.5 P -----------------------------------(3) P cos 30 = T +  N2 = 724.52 + 0.2 (500 -.5P) P Cos 30 +.1 P = 724.52 + 100 P (Cos 30 + 0.1) = 824.52 P = 853.52 N ( 5 ) Ladder friction : It is a device for climbing up or down. Since this system is in equilibrium, the algebraic sum of the horizontal and vertical component of the forces must be zero and the algebraic sum of the moment must also be zero.  Fx  0 ,  Fy  0 ,  M  0 Ex – 5 A ladder of length 4m, weighing 200N is placed against a vertical wall as shown in fig. 7. The co-efficient of friction between the wall and the ladder is 0.2 and that between the floor and the ladder is 0.3. In addition to self weight, the ladder has to support a man weighing 600N at a distance of 3 m. from A calculate the minimum horizontal force to be calculate the minimum horizontal force to be applied at A to prevent slipping. ( 6 ) Solution : The FBD of the ladder is shown in fig. 7.1 Taking moment at A Rw x 4 Sin 60 0 +  w Rw 4 Cos 60 = 600 x 3 cos 60 + 200 x 2 cos 60 866 Rw + 0.2 x 0.5 Rw = 275 0.966 Rw = 275 Rw = 284.68 -----------------------(1) 600 + 200 = Rf +  wRw Rf = 800 - 0.2 x 284.68 = 743.06 ----------------------------------- (2) P +  f Rf = Rw P = 284.68 - 0.3 x 743.06 = 61.76 Ex – 6 The ladder shown in fig. 8 is 6m. long and is supported by a horizontal floor and vertical wall. The co-efficient of friction between the floor and the ladder is 0.25 and between the wall and the ladder is 0.4 The self weight of the ladder is 200 N. The ladder also supports a vertical load of 900 N at C which is at a distance of 1 m from B. Determine the least of value of  at which the ladder may be placed without slipping. Solution :  w = Co-efficient of friction between wall and ladder = 0.4  f = Co-efficient of friction between floor and ladder = 0.25 The FBD of the ladder is shown in fig. 8.1 Rw =  f Rf = 0.25 Rf ---------------- (1) Rf +  w Rw = 900 + 200 = 1100 Rf + 0.4 x.25 Rf = 1100 ( 7 )  Rf = 100 N ---------------------- (2) So Rw = 250 N --------------------- (3) Taking moment about A Rw x 6 Sin  +  w Rw x 6 Cos  = 200 x 3 Cos  + 900 x 5 Cos  250 x 6 Sin  + 0.4 x 250 x 6 Cos  = 600 Cos  + 4500 Cos  1500 Sin  = 4500 Cos  tan  = 3  = 71.5630 Ex – 7 A uniform ladder of 4m length rests against a vertical wall with which it makes an angle of 450. The co-efficient of friction between the ladder and wall is 0.4 and that between ladder and the floor is 0.5. If a man, whose weight is half of that of the ladder ascends it, how height will it be when the ladder slips ? Solution : The FBD is shown in fig. 9.1  f = 0.5,  w = 0.4  f Rf = Rw.5 Rf = Rw -------------------(1) Rf +  wRw = w + 0.5 w = 1.5w ( 8 ) Rf +.5 x.4 Rf = 1.5 w 1.2 Rf = 1.5 w Rf = 1.25 w ----------------------(2) Rw =.5 x 1.25 w = 0.625w -------------------(3) Taking moment about A Rw x 4 sin 45 +  w Rw 4 Cos 450 = w x 2 Cos 45 +.5 wx Cos 450 4 Rw +  w4Rw = 2W +.5 xw 4 x.625 w +.4 x 4 0.625 w = 2 w +.5 x w 2.5 + 1 = 2 +.5x.5 x = 1.5 x = 3 m. Wedge friction : Wedges are generally triangular or trapezoidal in cross section. It is generally used for tightening keys or shaft. It is also used for lifting heavy weights. The weight of the wedge is very small compared to the weight lifted. Let w = weight of the block P = Force requried to lift the load  = co-efficient of friction on all contact surface.  = Wedge angle. ( 9 ) Considering the FBD of fig. 10.1 under the action of 3 forces the system is in equilibrium. 1) R 1  Resultant friction of normal reaction and force of friction between block and wall. 2) W  Weight of the block. 3) R2  Resultant reaction of normal reaction and force of friction between contact surface of block and wedge. Applying Lami’s Theorem W Sin[90  (   )] R1 = Sin[180  (   )] R2 = Sin(90   )] W R1 R or   Cos (  2 ) Sin(   ) Cos Again considering the FBD of Fig. 10.2 under the action of 3 forces the system is in equilibrium. 1) R2  Reaction given by the block. 2) P  Force applied on wedge. ( 10 ) 3) R3  Resultant reaction of normal reaction and force of friction between wedge and floor. Applying lami’s theorem P R R3  2  Sin(  2 ) Cos Cos (   ) ( 11 )  Chapter – 4 Centre d gravity – It is the point where whole of the mass of a body is supposed to act. In other words it is the point through which resultant of the parallel forces of attraction formed by the weight of the body, passes. It is usually denotes by CG or simplify G. Centroid : It is the point where the whole area of a body is assumed to be concentrated. Plane figures (known as laminas) have area only but no mass. The centre of gravity and centroid of such figures are the same point. For plane areas, centroid is represented by two coordinates by selecting coordinate axes in the plane of the area itself. ( 1 ) The two co-ordinate axes are usually selected in the extreme left and bottom of the area. Mathematically (continuous form) x  A x.dA A  y. dA Similarly y  A A  A xdA and  A ydA are known as first moment of area about x & y axes respectively.. Mathematically – (Discrete form) n  AiXi  i 1 x   Ai  n i 1  AiYi  n i 1 y   Ai  n i 1 ( 2 ) Ex - 1 Determine the y - co-ordinate of the centroid of a uniform triangular lamina. From the concept of similar triangles. p h y  b h h y P   b  h  h y  The area of the shaded portion dA =  h b.dy    A y. dA  y  A y h y  y  0 h   b.dy  1  b.h 2 h  y.b 1  y / h b.dy 0 = bh 2  y b   y1  dy  h = bh 2 ( 3 )  2 y2  = h ydy .dy   h  2  y2 1 y3  h =     h  2 h 3 0 2  h 2 1 h 3   2 h 2 2 h3  = h  2  h  3    h  2  h  3h      2 = h h 3 3h  2h = h 3 3 Ex- 2 2  h 2 1 h 3   2 h 2 2 h3          h  2 h 3   h 2 h 3h  x  xdA  dA y  ydA  dA ( 4 ) A - Centre of gravity of some common figures : Sl. No. Plane Figure C.G. location b 1 x 2 yd 2 2 x  b2 y  h ( from base) 3 3 x  b2 h  b  2a  y   3 ba  4 x r 4 r y  3 2 sin  5 y r 3  ( 5 ) B - Centre of gravity of built up symmetrical sections. Built up sections are formed by combining plane sections. The centre of gravity of such built up sections are calculated using method of moments. The entire area of the builtup section is divided into elementary plane areas i.e. a1, a2, a3 etc. Two reference axes are selected and let the distance of c.g. of these areas be x1, x2, x3…… etc. respectively. Let G be the centre of gravity of the entire area A and let the distances be x and y from y axis and x axis respectively. Thus A = a1+a2+a3+....... Now sum of moment of all the individual areas about y-y- axis. = a1x1+a2x2 + a3x3 +..... and the moment of entire area about y y axis = Ax Now equalising the two moment Ax  a1 x1  a2 x2  a3 x3 .......... a1 x1  a2 x2  a3 x3 ..... x  A a1 x1  a 2 x 2  a 3 x 3 ..... a1  a 2  a 3 Similarly a1 y1  a2 y2  a3 y3 .... y a1  a2  a3 ( 6 ) Example : Locate the centre of gravity of a T-Section having dimension 100mm x 150mm x 20mm Two reference axes are selected i.e. y-y and x-x at the extreme left and bottom of the given section. The entire area of the T- Section is divided into two portions. i.e. a1= Area of the rectangle ABCD = 100 x 20 = 2000 mm2 y1 = Distance of c.g. of the rectangle ABCD from x-x axes = 140 mm a2 = Area of the rectangle EDHG = 20 x 130 = 2600 mm2 y2 = distance of c.g. of the rectangle EDHG from x-x axes = 65 mm a1 y1 a2 y2 y  a1  a2 ( 7 )  2000  140  2600  65 280000  169000 =  2000  2600 4600 = 97.61 mm As the given T-section is symmetrical about y-y axis, x  50mm Centre of gravity of section with cutout Many a times some portions of a section are removed and thus the centre of gravity of the remaining portion is shifted to new locations. The c.g. of such a section with cutout can be found out by considering the entire section first and then deducting the cutout section. If the area of the main section and cut out are a1 and a2 respectively and c.g. distances are x1, y1 and x2, y2 from the reference lines, then the C.G. of the out section from the reference lines. a1 x1  a 2 x 2 x a1  a 2 & a y a y y 1 1 2 2 a1  a2 Ex- An isosceles triangle of side 200 mm has been cut from a square ABCD of side 200 mm as per the figure given below. Find out the c.g. of the shaded area. ( 8 ) Here, a1 = 200 x 200 a2 = 1/2 x 200 x 173.2 = 40,000 sqmm = 17,320 x1= 100 mm x2 = 100 y1 =100 mm y2 = 173.2 / 3 = 57.73 mm. a1 x1  a 2 x 2 x  a1  a 2 40,000 100  17320 100 = 40,000  17,320 =100 mm 40,000 100  17,320  57.73 y 40000  17,320 4000000  999883.60 = 22,680 =132.28 mm Ex Locate the centrod of the cut out section (Shaded area) as shown in the figure. ( 9 ) Here 1 a1   100  50 2 = 2500 sqmm h y1  3 50 = 3 = 16.67 mm r 2 a2  2   252 = 2 =981.75 mm2 4r y2  3 4  25  3 = 10.61 mm a1 y1  a2 y 2  y1  a1  a2 2500  16.67  981.75  10.61 = 2500  981.75 41675  10,416.37 = 1518.25 = 20.6 mm Moment of inertia (Second moment) of an area : Moment of interia of any plane area A is the second moment of all the small areas dA comprising the area A about any axis in the plane of ara A. ( 10 ) Referring to the figure : Iyy = moment of interia about the axis y-y =  x 2 dA =  x 2 dA Similarly : Ixx = Moment of interia about the axis x-x 2 =  y dA =  y 2 dA A Parallel axis theorem If the moment of inertia of a plane area about an axis through the centroid is known, the moment of inertia about any other axis parallel to the given centroidal axis can be found out by parallel axis theorem. It states that if 1cg is the moment of inertia about the centroidal axis, then moment of inertia about any other parallel axis say AB at a distance of x from the centroidal axis is given by. I AB  I xx  Ax 2 Perpendicular axis theorem It states that second moment of an area about an axis perpendicular to the plane of the area through a point is equal to the sum of the second moment of areas about two mutually perpendicular axes through that point. ( 11 )  The second moment of area about an axis perpendicular to the plane of the area is known as polar moment of inertia. I zz = Ixx + Iyy Moment of interia of a rectangular section : Let us consider as elementary strip of thickness dy at a distance of y from the centroidal axis x-x.  Area of the strip of this elementary area = b.dy & MI about x x - axis = A.y2 = b. dy. y2 Now M.I. of the entire area. = Ixx d d 2 2 y3 2  by dy  b. = d 3  d 2 2 bd 3 = 12 db 3 Similarly it can be shown that I yy  12 ( 12 ) Let us consider a circle of radius r and an elementary ring of thickness dy a distance of y from the centre ‘0’. Now area of this elemental ring = dA = 2y.dy M.I. of this ring about ‘0’ = 2y.dy. y 2 r 2 and MI of the entire area i.e. Izz   2y.dy. y 0 r 3 = 2    y dy 0 r4 = 2  4 r 4 = 2 I zz  Ixx  Iyy  2 r 4 1 =  2 2 r 4 = 4 d 4 = 64 Moment of inertia of built up section : The second moment of area of a build up section which consists of a number of simple sections can be found out as the sum of the second moments of area of the simple sections. ( 13 ) x  75mm (from the reference axis 1-1) a1 y1  a2 y 2 y (from the reference axis 2-2) a1  a2 150  10  145  140  10  70 = 150  10  140  10 217500  98000 = 1500  1400 = 108.8 mm Moment of inertia of the T section can be found out by adding the M.I. of the two simple rectangular sections. M.I. of the 1st portion about the axis x-x 150 103 =  150 10  (41.2  5) 2 12 = 12,500 + 19,65,660 = 19,78,160 M.I. of the 2nd portion about the axis x-x 10 1403  10  40  (108.8  70) 2 12 = 22,86,666.7 + 21,07,616 = 43,94,282.7 mm4  M.I. of the T section about xx - axis I xx  1978160  4394282.7 = 63, 72, 442.7 mm4 ( 14 )  SIMPLE MACHINES (ENGG. MECHANICS) Chapter 5 Before going to details of Simple Machines, we should know the following terms and their definitions. 5.1. Important Terms : 1. Simple Machine : It is a contrivance in which an external force is applied at some point in order to overcome a bigger resistance at some other point. Example A C B W P Taking moment about C, we get W x AC = P x BC BC > AC W BC  P AC W>P or P Py - Wx or 2Wx > Py Wx 1 or  Py 2 1 Efficiency > 2 or Efficiency > 50% Condition of Reversibility The machine will be revresible when its efficiency is > 50% 11. Self locking Machine If a Machine is not capable of doing any work in the reverse direction after the effort in withdrawn and such a machine is called a non-reversible or self locking machine. For this  < 50% 12. Low of a lifting Machine Where P= effort applied ,W= load lifted ,M and C are constants. C- the effort required to move the machine under no load. i.e., to overcome friction. 13. Maximum Mechanical Advantage of a lifting Machine. P +C we know, M.A  W P = mW P ) Also we knwo, P = MW + C C 2 O  Then putting it W 1 M.A   MW  C M  C W C neglecting W 1 (M.A)max  M 14. Maximum Efficiency of a lifting Machine we know that W P W   VR P  VR Putting for max efficiency W 1 max   (MW  C)VR (M  C )VR W C 1 neglecting , max  W M  VR Example In a lifting machine, an effort of 31 N raised a load of 1 KN. If efficiency of the machine is 0.75, what is its VR ? If on this machine, an effort of 61 N raisd a load of 2 KN, what is now the efficiency ? what will be the effort required to raise a load of 5 KN ? Solution Data given P1 = 31 N, W 1 = 1 KN = 1000 N.  = 0.75, P2 = 61 N, W 2 = 2 KN = 2000 N.  = ?, VR1 = ?, P3 = ?, W 3 = 5 KN = 5000 N. VR MA 1000 1   VR 31 VR 1000 or 0.75  31 VR 1000 or VR   43 (Ans) 31 0.75 2 MA 2000 2    0.763 or 76.3% (Ans) VR 61 43 P3 = ? P = mw +c 31 = m x 1000 +c...... (i) 61 = m x 2000 +c......(ii) substracting 9i) from (ii) 61-31 = 2000 m - 1000m  30  1000m 3 m   0.03 1000 We have 31 = 1000 M + C 31 = 1000 x.03 +C  31 =30 + C C = 1 Then P = 0.03 W +1 Now for 5000N, P= 0.03 x 5000 +1  P = 151 N (Ans)  5.2. Study of Simple Machine Types 1. Simple wheel and axle. 2. Differential wheel and axle. 3. Weston’s differential pulley block. 4. General pulley block. 5. Worm and worm wheel. 6. Worm geared pulley block. 7. Single purchase crab winch. 8. Double purchase crab winch. 9. Pulleys : (a) First system of Pulleys (b) Second system of Pulleys (c) Third system of Pulleys 10. Simple Screw Jack 11. Differential Screw Jack 12. Worm geared Screw Jack. 1. Simple wheel and Axle In Fig 2.1 is shown a simple wheen and axle, in which the wheel A and axle B are keyed to the same shaft. A string is wound round the axle B, which carries the load W to be lifted. A Second string is wound round the wheel A in the opposite direction to that of the string on B. A B d D W P W B- Axle A- Wheel P Fig 2.1 Simple wheel and axle One end of the string is fixed to the wheel, while the other is free and the effort is applied to this end. Since th two strings are wound in opposite directions, therefore, a downward motion of P will raise the load W. Let D = Diameter of effort wheel d = Diameter of the load axle w = Load lifted, and p = Effort applied to the load. Since the wheel as well as the axle are keyed to the same shaft, therefore when the wheel makes one revolution the axle will also make one revolution. In one revolution of effort wheel A, displacement of effort similarly the load displancement  D D D VR   d d W MA  P M.A   VR  If t1 = thickness of the rope used over the wheel. and t2 = thickness of the rope used over the axle. DT then VR  1 dT 2 Example In a wheel and axle machine, the diameter of the wheel is 100cm and that of the axle is 10 cm. The thickness of the load on the wheel is 3 mm and that on the drum is 6 mm. In this machine a load of 500 N is lifted as an effort of 100 N. Determine the efficiency and state whether the machine is reversible at this load. Data given : D = 100 cm, W = 500N d = 10 cm, P=100N T1 = 3mm T2= 6mm To find, n = ? W 500 MA   5 P 100 Solution D  t1 M.A  VR  d  t2 (100  0.3)cm  VR   9.46 (10  0.6)cm MA 5 and   100  52.85% (Ans) VR 9.46 Since efficiency is greater than 50%, the machine is reversible. Double Purchase crab Double or Purchase winch Diagram S1 X A X Bearing C S2 X X B D E X P S3 W Fig.2.2 Single Purchase crab  Single purchase crab consists of a Pinion (A), which is engaged with a Spur gear B. Gear A and the effort handle are fixed to the same shaft(G1). Also, adrum C and the gear b are fixed to the same shaft (S2). Load is lifted by a string wound round the drum. Calculation Let TA = number of teeth of Pinion(A) TB = number of teeth of the gear (B) L = Length of the effort handle D = diameter of the load drum. When the effort handle is rotated once, the distance moved by effort =  2 L when the handle is rotated once, the Pinion (A) also rotates once, i.e., NA=1rpm, where NA = revolution of Pinion A. Let NB=revolution of gear B Then we know for gear Drive N T B A  N T A B T T T  N  A  1 A  A A T T T B B B Revolution of Pinion C is the same as that of gear (B) as both are connected to same shaft. T A So NC = T B Again we know, N T D C  N T C D T  N  N  C D C T D P uttin g N C v alue T T N  A  C D T T B D Super gear (D) and a drum (E) are mounted on a third shaft S3. T T A C So when the effort handle makes one revolution, the gear D and drum E make revolution. T T B D T T So, the load is lifted through a distance  D  A C T T B D Where D= diameter of the load drum E 2 Hence Velocity Ratio  T T A C D  T T B D 2 T T B D VR   D T T A C Example In a double purchase crab, the pinions have 15 and 20 teeth while the spur wheels have 45 and 42 teeth. The effort handle is 40 cm. While the effective diameter of the drum is 15 cm, if the efficiency of the winch is 40%, what load will be lifted by an effort of 250N applied at the end of the handle ? Data Given TA = 15, TB = 45 TC =20, TD = 40 L=40 cm D = 15 cm  =40%, P=250 N To find W =? Solution we know, 2 TB TD VR   0 TA TC 2  40 45  40 Or VR    32 15 15  20 M.A W / 250   VR 15 W Or 0.4  250  32  W  3200N  32KN (Ans)   SINGLE PURCHASE CRAB WINCH TA L A B TB D W Fig 2:3 Single Purchase Crab Winch. In a Single Purchase Crab Winch, a rope is fixed to the drum and wound. A spur gear (B) is mounted on the load drum. Another pinion A is geared with (B) and connect to effor wheel. NB T  A NA TB If, NA = 1 TA NB = TB Distance moved by load in NB revolution TA  D  TB Y 2 2  TA VR     X TA D TB D  TB Example In a Single Purchase Crab, the number of Teeth on Pinion is 25 and on the Spur wheel 250, Radio of the drum and handle are 1500 mm and 300 mm respectively. Find the efficiency of the machine and the effect of friction, if an effor of 20 N can lift a load of 300 N. Data given TA = 25 TB = 250 D = 2 x 150 = 300 mm =30 cm  = 300 mm = 30 cm To Find,  = ? and Friction = ? Solution 2 TB VR   D T A 230 250    20 30 25 MA 300 20    75  75% (Ans) VR 20 W Effort lost in Friction = P VR 300 = 20  20  5N (Ans)  Worm and Worm Wheel Worm (A) Handle D Load Drum (C) Shaft (S) Worm Wheel (B) Description Worm is threaded spindle (A) Worm wheel is a spur gear (B) The threads of the worm are engaged within the teeth of the worm wheel. A load drum (C) is mounted on the same shaft(S) as that of the worm wheel. A string hangs vertically from the load drum string hangs vertically from the load drum (C) and the load (W) to be lifted is attached to the string at its free and as shown above. The effort (P) is applied at the end of a handle fitted at one end of the worm. When the worm is rotated by application of effort at the end of the handle, the worm wheel and the drum (C) rotate winding the string round the surface of the drum. In this way the load is lifted the Worm(A) may have single start threads as ‘multi start threads’. If the worm has single start threads, in one rotation of the worm, one teeth. If the worm has duble start threads is one rotation of the worm, two teeth of the worm wheel will move, and so on. Culculation of VR Let  = Length of the effort handle n = no, of starts of the worm n = 1, single start thread n = 2, double start thread T = number of teeth of the worm wheel. D = Diameter of the load drum (C) Considering on rotation of the effort handle, the distance through which effort (P) moves  2L We Know NA TB  NB TA N B  TA N T A B If N  1 rp m A TA TA  N  1  B TB TB H e re fo r sin g le s tart T 1 A 1 1 N   B TB T  In one rotation of the load drum the load in lifted to a leight D 1 In NB rotation the load lifted will be NB x D = D x T D  n In Multi start thread the load lifted will be = T 2L 2L T VR     D  Tn D n 2LT  VR  nD Example In a worm and worm wheel, the worm in triple threaded and the drum which is rigidly fixed to the wheel having common axis of rotation, has adiameter of 50 cm. Determineter the number of teeth on wheel of 40 turns of the worm move the load up by 60 cm. If the handle attached to the worm has an effective least of 30 cm and the load lifted weighs 25 KN. Calculate the M.A. and take efficiency as 40 %. Solution Given n = 3, D= 50cm, L=30 cm, W= 25 KN, n = 40 % In 40 turns of the worm, distance described by effort = 40 x 2  x 30 = 7539.83 cm 7539.82 Hence VR   125.66 60 2LT 2  30 T But VR    Dn 50 3  T  314.15 As teeth can not be fraction, T = 314. MA   VR MA  0.4 125.66  MA  50.264 (Ans) SIMPLE SCREW JACK w Disc Threaded Vertical Spindle P Body A Simple Screw Jack consists a vertical threaded spindle. The treaded portion of the spindle passes throught a nut cut in the body of the Screw Jack. The Load W is placed on a disc fitted at the top of the vertical spindle. The disc is fitted with a handle. Effort P is applied at the end of this handle. When the screw is truned by P, the screw either moves up or comes down throught a nut, depending upon the direction of rotation of the handle. A Screw Jack is used to raise and hold up heavy machinary like motor car, truck, etc. Calculation Let  = Length of the leaver P = Pitch of the Screw 2 Velocity Ratio = P Relation between Mean diameter Pitch and Heli x angle of a Screw thread Dd Dm = 2 Where D- outside diameter d - Inside diameter Threads of a Screw may be considered to be a metal strip wrapped round a cylinder in the form of helix. The development of this helix in one pitch length is shown here. B 1 P B So tan  P P A )  Dm A1 A Dm Effort Required to Lift a load by means of a Screw Jack When a load is lifted by means of a Screw Jack the case becomes equivalent to lifting a load (W) up an inclined plane of inclination  by applying a horizontal force. The equivalent inclined plane. P1 P W ) Dm Let P1 = horizontal roce required to move a body of weight W up the inclined plane. Then P1 = W tan(    ) Where  = Angle of Friction. In case of Screw Jack, P1, acts at a distance of mean radins from the axis of the screw. DM P    P1 2 DM  W tan(  )  2 DM tan   tan   W 2 (1  tan   tan ) P  DM DM  W 2 ( P ) 1  DM DM P  MDM  W ( 2 DM   P ) D MDM  P or P  W  M  2 DM   P Effort Required to Lower a load by Screw Jack The Equivalent plane is shown below : DM D P1 HereP    P1  W tan(    )  M P 2 2 DM DM  P W or P  W   ) 2 HDM  P Efficiencty when load is lifted Dm o u tp u t w o rk W P   In p u t w o rk P  2  DM A g a in P    P1 2  DYNAMICS Mechanics Statics Dynamics In contrast to statics, that deals with rigid bodies at rest, in dynamics we consider the bodies that are in motion. For convenience dynamics is divided into two branches called Kinematics and kinetics. Dynamics Kinematics Kinetics Kinematics : kinematics is concerned with the space time relationship of given motion of a body without reference to the force that cause the motion. Example : When a wheel rolls along a straight level track with uniform speed, the determination of the shape of the path described by a point on its rim and of the position along with path that the chosen point occupy at any given instant are problems of kinematics. Kinetics : Kinetics is concerned with the motion of a body or system of bodies under the action of forces causing the motion. Example : 1) When a constant horizontal force is applied to a body that rests on a smooth horizontal plane, the prediction of the way in which the body moves is a problem of kinetics. 1 2) Determination of the constant torque that must be applied to the shaft of a given rotor in order to bring it up to a desired speed of rotation in a given interval of time is a problem of kinetics. Particle : The whole science of dynamics is based on the natural laws governing motion of a particle or particles. A particle is defined as a material point without dimensions but containing a definite quantity of mater. But in true sense of term there can be no such thing as a particle, since a definite amount of matter must occupy some space. However, when the size of a body is extremely small compared with its range of motion, it may in certain cases, be considered as a particle. Example : 1) Star & planets, although may thousands of kilometer in diameter are as small compared with their range of motion that they may be considered as particles in space. 2) The dimensions of a rifle bullet are so small compared with those of its trajectory that ordinarily it may be considered as a particle. Whenever a particle moves through space, it describes a trace that is called the path. Which may be either a space curve / tortuous path or a plane curve / plane path. In the simplest case when the plane path is a straight line the particle is said to have rectilinear motion or else, it is a curvilinear motion. Displacement, velocity and acceleration function These are the three quantities that are necessary to completely describe the motion of a particle in dynamics. 2 Displacement : x O A X Assuming the motion of the particle along X-Axis, we can define the displacement of a particle by its x-coordinate, measured from the fixed reference point O. This displacement is considered as positive when the particle is to the right of the origin O and as negative when to the left. As the particle moves, the displacement varies with time, and the motion of the particle is completely defined if we know the displacement x at any instant of time t. Analytically, this relation can be expressed by the general displacement – time equation, x= f(t), where f(t) stands for any function of time. This equation will take different forms depending upon how the particle moves along the x-axis. The displacement is measured by the unit of length. Velocity : In rectilinear motion, considering x the case of uniform rectilinear motion, as A represented graphically by the straight  x2 line BC, we see that for equal intervals of  x1 C time  t, the particle receives equal increments of displacement  x. This B velocity v of uniform motion is given by the equation v =  x/  t. this velocity is considered as positive if the displacement t x is increasing with time and negative if it O t t is decreasing with time. In the more general case of non-uniform rectilinear motion of a particle as represented graphically by the curve OA, in equal intervals of time  t, the particle receives unequal increments of displacement.  x1 &  x2, Thus, if  x denotes the increment of displacement received during the interval of time  t, the average velocity during this time is given by the equation vav=  x /  t. As the time interval  t is taken 3 smaller and smaller, the motion during the interval becomes more & more nearly uniform so that this ratio  x /  t approaches more and more closely to the velocity at any particular instant of time interval. Taking the limit approach we obtain the instantaneous lim x dx velocity of the particle v   t 0   x.  t dt Thus we see that the velocity time relationship of a moving particle can always be obtained by differentiating the displacement time equation. If the derivative is (+)ve i.e. the displacement increases with time, the velocity is positive and has a positive direction of the x-axis other wise it is negative. The velocity is measured by the unit of length divided by the unit of time. Acceleration : If the rectilinear motion of particle is non uniform its velocity is changing with time and we have acceleration. But in case of uniform motion, the velocity remains constant and there is zero acceleration. When the particle receives equal increments of velocity “v in equal intervals of time  t, we have motion with constant acceleration v as given by the equation a . Acceleration is consider positive when the velocity t obtains positive increments in successive intervals of time and negative if the velocity is decreasing. However the acceleration of a particle may be positive when its velocity is negative of vice versa. It the more general case when in equal intervals of time  t, the increments of velocity  v1 and  v2 are unequal, we have a motion of the particle with variable acceleration. To obtain the acceleration of the particle at any instant t in such a case, we take the limit approach using the equation lim v d 2 x a  t 0   x. t dt2 Thus, in any case of rectilinear motion of a particle, the acceleration time relationship can be expressed analytically by taking second derivative with respective time of the displacement time equation. The acceleration is measured by the unit of length divided by square of unit of time. 4  X Problem 1: B A slender bar AB of length l which remains always in the same vertical plane has its ends A & B constrained to remain in contact x l with a horizontal floor and a vertical wall respectively as shown in the figure. The bar A starts from the vertical position and the end A O v0t is moving along the floor with constant velocity v0 So that its displacement OA = v0t. Write the displacement – time, velocity-time and acceleration-time equations for the vertical motion of the end B of the bar. Solution : Let us choose x-axis along the vertical line of motion of end B, i.e. along OB with origin at O. From the geometrical relationship of the figure, the displacement x of the end b is given by the equation. x  l 2  (v0t ) 2 ------------- (1) which is the desired displacement time equation. Differentiating w.r.t. time once, we get v02t x  () ---------- (2) l 2  v02t 2 the velocity time equation. Differentiating w.r.t. time twice, we get v02 l 2 x  ( ) 3 ---------- (3) l 2  v02 t 2  2 the acceleration time equation. These expressions are valid in the interval (0 < t < l/v0) 5 Problem 2 : A particle starting from rest moves rectilinearly with constant acceleration a and acquires a velocity v in time t, traveling a total distance s. Develop formulae showing the relationship that must exist among any three of these quantities. Solution : Let us draw the velocity time v diagram. Here A i) The acceleration a is represented by the slope of the straight line OA. ii) The velocity v by the ordinate BA. a = pe iii) The time t by the abscissa OB. v iv) The distance traveled s by the area slo S OAB. t From geometry of the figure we may O B t write v = at ------ (1), s = 1/2 vt ------- (2) Putting (1) in (2), s = 1/2 at2 -------- (3) v 1 v Eliminating t from equations (1) & (2), t  & s  v. or v  2as ---------- (4) a 2 a But these formulae do not account for any initial displacement or initial velocity. Problem 3 : A particle moving with initial velocity u moves with constant acceleration a and acquires a velocity v after time t during which it travels a total distance s. Develop formulae showing relationship between these v four quantities. Also give the expression for displacement, if it has an initial displacement B s0 before it starts moving. a e = slop v Solution : A C Acceleration a = slope of AB S u Initial velocity u = OA along velocity t O D axis. t 6  Velocity v = ordinate BD Time t = ordinate OD Distance s = area of OABCD From the geometry of the figure, we may write, v = CD + CB = OA + CB = u + at --------- (1) s = ABC + OACD = ut + 1/2 at2 ----------- (2) 2 vu  v u  1 vu  uv  u 2 (v  u ) 2 Eliminating t, t  , s  u    a    =  a  a  2  a  a 2a 2uv  2u 2  v 2  u 2  2uv v 2  u 2 = = 2a 2a or v 2  u 2  2as ---------- (3) When there is initial displacement, s= s0+ ut + 1/2 at2. Principles of dynamics There are several axioms called the principle of dynamics which are concerned with the relationship between the kind of motion of a particle and the force producing it. These axioms are in fact, broad generalization of Kepler’s observation on the motion of heavenly bodies and of carefully conducted experiment with the motion of earthly bodies. The first reliable experiments were made by Galileo in this connection who discovered the first two laws of motion of a particle. However, the complete set of principles and their final formulation were made by Newton after the name of which they are commonly called the Newton’s laws of motion. Newton’s Laws of Motion : First Law : This law also some times called the law of inertia is stated as “Every body continues in a state of rest or of uniform motion in a straight line unless and until it is acted upon by a force to change that state. For practical problems, we consider the surface of the earth as immovable and refer the motion of the particle with respect to the earth i.e. surface of the earth is taken as inertial frame of reference. But for the motion of heavenly bodies (planets & satellites), a system of coordinates defined by stars is taken as the immovable system of reference and their motion with respective these stars considered. From the first law it follows that any change in the motion of a particle is the effect of a force, that form the concept of force. However, the relation 7 between this change in velocity and the force that produces it given by the second law of dynamics. Second Law : The observations made by Galileo on the basis of his experiments on falling bodies and bodies moving along inclined planes, was verified by experiment on pendulums of various materials by Newton and generalized as second law of dynamics which states that the acceleration of a given particle / body is proportional to the force applied to it and acts in the direction of force. In this law, as formulated above, there is no mention of the motion of the particle before it was acted upon by the force and hence the acceleration of a particle produced by a given force is independent of the motion of the particle. Thus a given force acting on particle produces the same acceleration regardless of whether the particle is at rest or in motion and also regardless of the direction of motion. Again, there is no reference as to how many forces are acting on the particle. So if a system of concurrent forces is acting on the particle, then each force is expected to produce exactly the same acceleration as it would have if acting alone. Thus, the resultant acceleration of a particle may be obtained as the vector sum of the accelerations produced separately by each of the forces acting upon it. Moreover, since the acceleration produced by each force is in the direction of the force and proportional to it, the resultant acceleration act in the direction of, and proportional to, the resultant force. By using Newton’s Second law, the general equation of motion of particle can be established as follows : Let the gravity force i.e. weight of the particle W, acting alone produces an acceleration equal to g. It instead of weight W, a force F acts on the same particle, then from the 2nd Law it follows that the acceleration a produced by this force will act the direction of the force and will be in the same ratio to the gravitational acceleration g as the force F to gravity force W. 8  a F W F W   or  or F  a g W g a g This is the general equation of motion of a particle for which its acceleration at any instant can be obtained if the magnitude of force F is known. Also it is evident that for a given magnitude of force f, the acceleration produced is inversely proportional to the factor W/g, which is called mass of the particle denoted by m. This factor measure the degree of sluggishness (inherent resistance against motion) with which the particle yields (responds) to the action of applied force and is a measure of the inertia of the particle. Thus the second law of dynamics forms the basis of concept of mass. Thus using the notation m = W/g, the general equation of motion of a particle becomes F=ma. Third Law : By using the first two laws formulated above the motion of a single particle subjected to the action of given forces can be investigated. However in more complicated cases where it was required to deal with a system of particles or rigid bodies the mutual actions & reactions among them must be taken into account, which is given by Newton’s third law. It states that “To every action there is always an equal and opposite reaction or in other words, the mutual action of any two bodies are always equal and oppositely directed”. This implies that if one body presses another, it is in turn is pressed by the other with an equal force in the opposite direction and if one body attracts another from a distance, this other attracts it with an equal and opposite force. Example : Sun attracts the earth with a certain force and therefore the earth attracts the sun with exactly the same force. This holds good not only for the forces of gravitation but also for the other kind of forces such as magnetic or electrical forces. Thus a magnet attracts a piece of iron no more than the iron attracts the magnet. 9  The equation of motion can be used so solve two kinds of problems. 1) Motion of particle is given, to find the force necessary to produce such a motion. 2) The force is given and it is required to find the motion of the particle. Let us discuss the first kind of problem : Problem 4 : A balloon of gross weight W falls vertically downward with constant acceleration a. What amount of ballast Q must be thrown out in order to give the balloon an equal upward acceleration a. Neglect air resistance. Solution : Case 1 : When the balloon is falling The active forces on the balloon are : i) The total weight ‘W’ including the ballast, that acts vertically down ward and. ii) The buoyant force P, acting upward representing the weight of the volume of air displaced. The equation of motion in this case may be written as  W a  W  P ---------------- (1) g Case II : When the balloon in rising : The active forces on the balloon are : i) The balance weight (W-Q) excluding the ballast acting vertically downward. ii) The same buoyant force P acting upward as in the previous case. The equation of motion in this case may similarly be written as : W a  P  (W  Q ) ------------- (2) g Solving equations (1) & (2), we obtain 2Wa Q ------------------------- (3) ga 10 Particular cases : i) When a=0, if the balloon was in equilibrium, Q= 0 & W=P  No ballast would have to be thrown out to rise with the same zero acceleration. ii) When a=g, if the balloon was falling freely, Q=W & P=0  It is impossible to throw out sufficient ballast to cause the balloon to rise. Problem 5 : Two equal weights W and a single weight Q are attached to the ends of the flexible but in extensible cord over hanging a pulley as shown in the figure. If the system moves with constant acceleration a as indicated by arrows, find the magnitude of the weight Q. Neglect air resistance and the inertia of the pulley. Solution : Here we have a system of two particles for which there are two equations of motion. For the weight on the left : The active forces are the tension in the cord ‘S’ acting upward and the weight of the body ‘W’ down ward. The equation of the motion in this case may be written as W a  S  W ------------------ (1) g For the weight on the right : The active forces are the same tension in the cord ‘S’ acting upwards but a down ward force of (W+Q). The equation of motion in this case may similarly be written as : W  Q   a  (W  Q)  S ------------- (2)  g  Solving the equations (1) & (2), we have 2Wa Q ----------------------- (3) ga 11 Particular Case : (Interpretation of the solution of the problem) To produce an acceleration a=g of the system would require a weight Q =  , which implies that it is impossible for the weight on the RHS to attain practically a free fall condition. Motion of a particle acted upon by a constant force : Now, let us discuss the second kind of dynamics problem, in which the acting force is given and the resulting motion of the particle is required. To start with, let us consider the simplest case of a particle acted upon by a constant force, the direction of which remains unchanged. In this case, the particle moves rectilinearly in the direction of force with a constant acceleration. Let the line of motion be along x-axis and the magnitude of force be denoted by X. Also the particle starts from A with initial displacement x0, when t = 0. The equation of motion can be written as : W W X d 2x X a  x or x   a  2 ----------------(1) g g W/g dt To find the velocity x and displacement x as function of time, integrating this differential equation,  dx  dx we get, d    d  x   adt or  x  at  C1 , where C1 is the constant of integration.  dt  dt Boundary condition : At t = 0 , x  x0  C1  x0  x  x0  at -----------(2) 12 This is the general velocity time equation for the rectilinear motion of a particle under the action of a constant force X. dx Writing equation (2) in the form -  x  at or dx  x0 dt  atdt and integrating again, dt 1 2 we obtain, x  x 0t  at  C 2 ------------------(3) 2 Where C2 again a constant of integration. Boundary Condition : At t = 0, x  x0  C2  x0 1 2  x  x0  x0 t  at 2 This is the general displacement time equation for the rectilinear motion of a particle under the action of a constant force X. Thus the initial conditions influence the motion of a particle quite as much as does the acting force. Infect the initial condition represents the heredity of motion, while the acting force represents its environment. Particular case : Case – I – Freely falling body : Acting force X= W and a = g Then from equations (2) & (3), x  x0  gt ------------------------ (2)’ x  x0  x 0t  12 gt 2 ----------------(3)’ Case II – Body starts to fall from rest and from the origin : In this case x0 = 0 and x0  0 and the equations (2) & (3) reduces to x  gt -------------------- (2)” x  12 gt 2 ----------------------(3)” 13 Problem 6 : A particle projected vertically upward is at a height h after t1 seconds and again after t2 seconds. Find the sight h and also the initial velocity v0 with which the particle was projected. Solution : Neglecting air resistance, the active force in the particle is its own gravity force W which is always directed vertically downward. Taking x-axis along the vertical line of motion, the origin at the starting point and considering the upward displacement as positive, from displacement time equation we have : 1 2 When t = t1, h  v0t1  gt1 ------------------------(1) 2 & When t  t2 , h  v0t 2  12 gt 22 ------------------- (2) Multiply equation (1) with t2 & equation (2) with t1 & subtracting 1 1 ht 2  t1   gt1t2 (t 2  t1 ) or h  gt1t 2 --------------- (3) 2 2 1 2 1 Equating equations (1) & (2), v0t1  gt1  v0t 2  gt22 2 2 1 1 or v 0 t1  t 2   g t1  t2 t1  t 2  or v 0  g t1  t 2  ----------- (4) 2 2 Problem 7 : From the top of a tower of height h = 40m a ball is dropped at the same instant that another is projected vertically upward from the ground with initial velocity v0 = 20 m/s. However from the top do they pass and with what relative velocity ? Solution : Case 1 : For the ball dropped from the top : The top of the tower is taken as the origin and downward displacement is considered as positive. The initial displacement and initial velocity are zero. Acceleration due to its own gravitational force is downward. Thus the displacement time equation will be 14  1 2 x1  gt ------------------------- (1) 2 Case 2 : For the ball projected from the ground : The ground is taken as the origin and the upward displacement is considered as positive. Here initial displacement = 0 and initial velocity = v0 upward while the acceleration due to gravity is down ward. Thus the displacement time equation will be 1 x2  v0t  gt2 ------------------------- (2) 2 When the balls pass each other, we must have x1  x 2  h ----------------------------- (3) Substituting the values of x1 and x2 from equation (1) & (2) in (3), we obtain, 1 2 1 gt  v0t  gt 2  h or h  v0t ----------------------(4) 2 2 v0 40 Putting numerical values, we have t    2 sec h 20 Putting this value of t in equation (1) we find x1 t 2  1  9.81 4  19.62m. 2 Relative Velocity : The velocity time equation for case - I, dx1 1 x1   g.2t  gt  dt 2 At time t = 2 Sec, Velocity x1  9.81 2  19.62m / s  The velocity time equation for case - II, dx 2 x 2   v 0  gt  dt At time t = 2 Sec, Velocity x2  20 19.62  0.38m / s.   Their relative velocity vR  x1  x2  19.62   0.38  20m / s Hence the balls pass 19.62 m. below the top of the tower with a relative velocity of 20 m/s. i.e. same as the initial velocity. 15 Problem 8 : A small block of weight W is placed on an inclined plane as shown in the figure. What time interval t will be required for the block to traverse the distance AB, if it is released from rest at A and the coefficient of kinetic friction on the plane is µ=0.3. What is the velocity at B. Solution : From the free body diagram of the block, the resultant force along the plane X  W (sin    cos  ) acting down ward. X Resulting acceleration a   g (sin    cos ) W/g For sliding of the block, a must be positive  (sin   cos )  0 or tan    Here sin   3 5 and cos   4 5 & putting other numerical values 3 4 a  9.81  0.3    3.53m / s 2 5 5 Again the initial displacement = 0 and initial velocity =0 1 2 1 2 The displacement time equation becomes x  0  at  at 2 2 1 2 Putting numerical valuces 50   3.53  t or t 2  28.32 or t = 5.32 s 2 Substituting the value of t in the velocity time equation, Velocity at B, xb  0  at  at  3.53  5.32  18.78m / s D’ Alembert’s Principle : The differential equation of rectilinear motion of a particle X  mx can be written in the form X  mx  0 , where x denotes the resultant in the direction of X-axis of all applied forces and m, the mass of the particle. This equation of motion of particle is of the same form as an equation of static equilibrium and may be considered as an equation of dynamic equilibrium. For writing this equation, we need only to consider in addition to the real forces acting on the 16 particle, an imaginary force  mx. This force equal to the product of the mass of the particle and its acceleration and directed oppositely to the acceleration, is called inertia force of the particle. D’ Alembert was first to point out this fact that equation of motion could be written as equilibrium equations simply by introducing inertia forces in addition to the real forces acting on a system. This idea is known as the D’ Alembert’s principles and has definite advantage in the solution of engineering problems of dynamics. Due to application of this principle, when dealing with a system having one degree of freedom, we need to write only one equation of dynamic equilibrium in stead of writing as many equation of motion as there are particles. It particularly becomes very useful when used in conjunction with the method of virtual work. Problem 9 : Two unequal weights W1 and W2 are attached to the ends of a flexible but inextensible cord overhanging a pulley as shown in the figure (W1 > W2). Neglecting air resistance and inertia of the pulley, find the magnitude of acceleration of the weights. r Solution : Here we have a system of particle connected between themselves and so constrained that each particle can have only a rectilinear motion. Assuming motion of the system in the direction shown by the arrow on the pulley, there will be upward acceleration x of the weight W 2 and downward acceleration x of the weight W 1. Denoting the masses by m1 and m2 respectively, the corresponding inertia forces act as shown in the figure. Assuming the tension in the string (String Reaction) to be S on both sides of the pulley, neglecting friction on the pulley, we can have a system of forces in equilibrium for each particle by adding the inertia forces to the real forces. S  W2  m2 x  0 ------------------ (1) W1  S  m1x  0 ------------------ (2) 17 Eliminating S from these equations, W2  m2 x  W1  m1x -------------- (3) Again this problem can be conceived in another way i.e. since each particle is in equilibrium we can say that the entire system of forces is in equilibrium. So instead of writing separate equations of equilibrium for each particle, we can write one equation equilibrium for the entire system by equating to zero the algebraic sum of moment of all the forces (including the inertia forces) with respect of the axis of the pulley. In this case, we need not consider the internal forces or reaction S of the system and can directly write. W2  m2 xr  W1  m1xr or W2  m2 x  W1  m1x  W  W2   m  m2  from which x   1  g   1  g W  1  W2  m  1  m2  Problem 10 : Let us solve the previous problem No. 5 by D’ Alembert’s Principle. Solution : Here also we have a system of two particles so connected that it has single degree of freedom. r Taking moment about the centre of the pulley O,  W   a  r  W  Q   W  Q  a r we have W    g   g  W a  W  Q   W  Q  a or W  g g W Wa Qa 2Wa  a or aQ  or  Q1   g g g g  g 2Wa  g a Wa or  Q  or Q  2 g  g  g  a  Problem 11 : Two bodies of weight W 1 & W 2 are connected by a thread and move along a rough horizontal plane under the action of a force F applied to the first body. If the coefficient of friction between the sliding surface of bodies and the plane is 0.3, determine the acceleration of the bodies and tension in the thread. 18  W1 W2 a a g a g F S S W1 W2 R1 R2 Given m1 = 80 Kg., m2 = 20 Kg.,  = 0.3, F= 400 N. Solution : Here also we have a system of two particles so connected that, it has a single degree of freedom. So we can apply D’ Alembert’s principle to the system and write only one equation of equilibrium for the entire system taking only the active forces but without considering the internal forces (reactions). W1 W Along horaizontal direction, F  a  2 a  W1   W2 g g  W  W2    1 a   W1  W2   g   W1  W2  or,   a  F   W1  W2   g  F   W1  W2  F  g m1  m2  400  0.3  9.81(80  20) or, a  or, a   1.06m / s 2  W1  W2  m1  m2  (80  20)    g  Considering the equilibrium of second body W2 a  S  a   W 2  W 2      m 2 ( a   g )  20 1. 06  0. 3  9. 81   80 N g g  ** ** ** 19  6.2 Work, Energy and Power Work: When a forceis acting on a body and the body undergoes a displacement then some wok is said to be done. Thus the work done by a force on a moving body is defined as the product of the force and the distance moved in the direction of the force. S S F F θ Body moving with direction Body not moving with of force direction of force When force F acts on the body and body displaces in the direction of force as shown in fig. The work done by the force F = Force x Distance = F x S When the force acts on the body at an angle θ to the horizontal and the body moves in horizontal direction by distance s, then work done by the force F = Component of the force in the direction of displacement x distance= Fcosθ x S. From definition of work it is obvious that unit of work is obtained by multiplying unit of force by unit of length. Hence, if unit of force is newton and unit of distance is meter then unit of work is N.m. So one N-m of work is denoted by one Joule(J). Hence one joule may be defined as the amount of work done by one newton force when the particle moves one meter in the direction of the force. Energy: Energy is defined as the capacity to do work. There are many forms of energy like heat energy, mechanical energy, electrical energy, and chemical energy. In mechanics mostly discussed about mechanical energy. The mechanical energy may be classified in to two forms i.e Potential energy and Kinetic energy. Potential Energyis the capacity to do work due to position of the body. A body of weight ‘W’ held at a height ‘h’ possesses an energy Wh. Kinetic Energy is the capacity to do work due to motion of the body. Consider a car moving with a velocity m/s. If the engine is stopped, it still moves forward doing work against frictional resistance and stops at a certain distance s. From the kinematic of the motion, From D’Alembert’s principle, Then, Work done = This work is done by the energy stored initially in the body. Kinetic Energy= Where, is the mass of the body and is the velocity of the body. Unit of energy is same as that of work, since it is nothing but capacity to do work. It is measured in joule (N-m) or kilo Joule (kN-m). Power: It is defined as the time rate of doing work. Unit of power is watt (w) and is defined as one joule of work done in one second. In practice kilowatt is the commonly used unit which is equal to 1000 watts. Work Energy equation: Consider the body shown in figure below subjected to a system of forces and moving with acceleration in x-direction. Let its initial velocity at A be and final velocity when it moves distance AB=s be. Then the resultant of system of the forces must be in x- direction. Let F1 W F4 x A B F2 F3 From Newton’s second law, Multiplying both side of the equation by elementary distance ds, then Integrating both sides for the motion from A to B, Work done = Final kinetic energy- Initial kinetic energy (It is called work energy equation) This work energy principle may be stated as the work done by system of forces acting on a body during a displacement is equal to the change in kinetic energy of the body during the same displacement. Conservation of Energy: Conservative force: A force is said to be conservative if the work done by the force on a system that moves between two configurations is independent of the path of the system takes. Non-Conservative forces:A force is said to be non-conservative if the work done by a force on a system depends on the path of the system follows. Principle of conservation of energy, “ when a rigid body or a system of rigid bodies, moves under the action of the conservative forces, the sum of the kinetic energy (T) and the potential energy (V) of the system remains constant. or or Consider a conservative system; say a particle moving from position 1 to position 2. As the particle moves from position 1 to position 2 it undergoes a change in potential energy, Where: are the potential energies in positions 1 and 2 respectively. As the particle displace from position 1 to 2, let be the work of the forces that act on the particle. The work does not depend on the path or the speed with which the particle moves on the path. Consequently, Then, the principle of work energy ), yields Where; is change in kinetic energy from position 1 to position 2. Thus relationship may be written as Where is called the mechanical energy of the particle. 6.3 Define Momentum & Impulse, Explain Conservation of energy and Linear momentum, Explain collision of elastic bodies and different co-efficient of restitution. Solved Examples: Example-1: A pump lifts 40m3 of water to a height of 50m and delivers and delivers it with a velocity of 5m/sec. What is the amount of energy spent during the process? If the job is done in half an hour , what is the input power of the pump which has an overall efficiency of 70%? Solution:Output energy of the pump is spent in lifting 40m3 of water to a height of 50m and deliver it with the given kinetic energy of delivery. Work done in lifting 40m3 water to a height of 50m = Wh Where W=weight of 40m3 of water=40 x 9810 N (Note: 1m3 of water weighs 9810 Newton) Work done= Wh = 40x9810x50 =19620000Nm Kinetic energy at delivery = = =500000Nm Total Energy spent= 19620000+ 500000= 20120000Nm= 20120kNm= 20.12 x106 Nm. This energy is spent by the pump in half an hour i.e. in 30x60=1800sec. Output power of pump = Output energy spent per second = =11177.8watts=11.1778kW. Input power= = Example 2:For the same kinetic energy of a body, what should be the change in its velocity if its mass is increased four times? Answer: Let 'm1' be the mass of a body moving with a velocity 'v1'. When mass is increased four times let the velocity be v2. v2 = ? Substitute the value of m2 in the above equation Given The velocity of the body is halved when its mass is increased four times. Example 3: Calculate the time taken by a water pump of power 500 W to lift 2000 kg of water to a tank, which is at a height of 15 m from the ground? Solution: Given g = 10 m/s2 Power of the water pump = 500 W Since the water is lifted through a height of 15 m, work done is equal to the potential energy. Mass of water (m) = 2000 kg Height (h) = 15 m t=? t = 40 x 15 = 600 s Time required to lift water = 600 s. Example 4: A car weighing 1000 kg and travelling at 30 m/s stops at a distance of 50 m decelerating uniformly. What is the force exerted on it by the brakes? What is the work done by the brakes? solution: Mass of the car (m) = 1000 kg Initial velocity (u) = 30 m/s Distance traveled (S)= 50 m Since the car stops, final velocity (v) = 0 Work done by the brakes = kinetic energy of the car W =F xS The force exerted by the brakes is 9000 N.  6.3 Momentum and Impulse It is clear from the discussion of the previous chapters that for solving kinetic problems, involving force and acceleration, D’Alembert’s principle is useful and that for problems involving force, velocity and displacement the work energy method is useful. The Impulse- Momentum method is useful for solving the problems involving force, time and velocity. Momentum:Momentum is the motion contained in a moving body. It is also defined as the product of an object's mass and velocity. It is a vector. The SI units of momentum are kg * m/s. Impulse:Impulse is defined as a force multiplied by the amount of time it acts over. In calculus terms, the impulse can be calculated as the integral of force with respect to time. Alternately, impulse can be calculated as the difference in momentum between two given instances. The SI units of impulse are N*s or kg*m/s. Impulse-Momentum Equation: As stated earlier impulse can be calculated as the difference in momentum between two given instances. If is the resultant force acting on a body of mass in the direction of motion of the body, then nd according to Newton’s 2 law of motion, But,acceleration, , is velocity of body. Then, i.e. If initial velocity is and after time interval the velocity becomes , then The term is called the impulse. If the resultant force is in newton and time in second, the unit of impulse is N*sec. If the resultant force is constant during time interval then, impulse is equal to. The term [ ] is called change in momentum. So, Impulse= Change in momentum= Final momentum- Initial momentum. Impulse momentum equation can be stated as: The component of the impulse along any direction is equal to change in the component of momentum in that direction. Note: Since the velocity is a vector, impulse is also a vector. The impulse momentum equation holds good when the direction of , and are the same. The impulse momentum equation can be applied in any convenient direction and the kinetic problem involving