Vidyalanakar F.Y. Diploma : Sem. II [AE/CE/CH/CR/CS/CV/EE/EP/FE/ME/MH/MI/PG/PT/PS] Engineering Mechanics

Vidyalanakar F.Y. Diploma : Sem. II [AE/CE/CH/CR/CS/CV/EE/EP/FE/ME/MH/MI/PG/PT/PS] Engineering Mechanics Time : 3 Hrs.] Prelim Question Paper Solution [Marks : 100 Q.1 Attempt any TEN of the following : [0] Q.1(a) Define funicular polygon. [] (A) Funicular Polygon : The polygon is so constructed by drawing the lines in the respective spaces of space diagram; parallel to the rays of polar diagram by maintaining the order is called as funicular polygon. Q.1(b) State any two advantages and any two disadvantages of friction. [] (A) Advantages of friction (1) Man cause can easily walk on horizontal surface due to friction. () There is safe movement of vehicles by applying breaks. (3) One can hammer the nail into wall due to friction. Q.1(c) State Newton s Laws of motion? (First, Second & Third) [] (A) Newton first law : Everybody continues to be in its state of rest or of uniform motion in a straight line, unless it is acted upon by some external agency. Second Law : The rate of change of momentum is directly proportional to applied forces. Third Law : To every action there is equal and opposite reaction. Q.1(d) What is efficiency of a machine? [] (A) Efficiency () : The efficiency of a machine is the ratio of output to the input of a machine and is generally expressed as a percentage. Output % = 100 input Q.1(e) What is Bow's notation? Explain with a sketch. [] (A) Bow s notation Bow s notation is used designate a force as per this notation, each force is designated or named by two spaces one on each side of the line of action of a force. This space are generally named by capital letter s as A, B, C serially. Explanation A force say F acting on rigid body divided space above or below it into two parts, say A and B hence the force F is named as AB. Q.1(f) Define Resolution of force [] (A) Resolution of force : The way of representing a single force into number of forces without changing the effect of force on body is called resolution of force. Q.1(g) State Parallelogram Law of forced with neat sketch. [] (A) Parallelogram Law of forces : If two forces acting at and away from the point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram passing through the point of intersection of two forces, represents the resultant in magnitude and direction. 1

: F.Y. Diploma Mechanics B C O Q P R Q A D Q sin Q.1(h) State principle of transmissibility of force. [] (A) Principle of transmissibility of force : If force acts at a point on a rigid body, it is assumed to act at any other point on line of action of force within the body. Q.1(i) Define Lami s theorem. [] (A) Lami s theorem states that, if three forces acting at a point on a body keep it at rest, then each force is proportional to the sine of the angle between the other two forces. As per Lami s theorem, F1 F F 3 sin sin sin Q.1(j) State any two types of beams with diagram of each. [] (A) Following are the different types of beams (1) Simply supported beam () Cantilever beam (3) Over hanging beam (4) Fixed Beam (5) Continuous beam Q.1(k) Explain meaning of self locking machine. State the condition for it. [] (A) SelfLocking Machine A machine which is not capable of doing work in reverse direction even on removal of effort, then the machine is called as selflocking or Nonreversible machine. Condition for SelfLocking Machine Efficiency < 50 % < 50 % Q cos

Prelim Question Paper Q.1(l) Define cone of friction. [] (A) Cone of friction It is an imaginary cone generated by revolving resultant reaction about the normal reaction, when body is in limiting equilibrium condition. OR When body placed on horizontal surface is in limiting equilibrium by a horizontal force P, then the resultant reaction makes an angle with normal reaction. If we gradually change the angle of force P through 360, then imaginary cone is generated as resultant reaction revolves around normal reactions. This imaginary cone is called cone of friction. Q. Attempt any FOUR of the following : [16] Q.(a) Resolve the force 19 MN along and 3 on either side of it. (A) Resolve the force 19 MN along and 3 on either side of it. Let 1 = and = 3. Referring figure, Fsin F 1 = sin 1 F 1 = 1.44 MN Fsin1 F = sin 1 F = 8.79 MN. 19sin3 = sin 3 19sin = sin 3 Q.(b) In a simple axle and wheel, the diameter of wheel is 180 mm and that of axle 30 mm. If the efficiency of the machine is 80%, find the effort required to lift a load of 100 N. (A) Given: D = diameter of wheel = 180 mm, d = diameter of axle = 30 mm = efficiency of the machine = 80% W = load = 100 N To find : V.R:? Formula: V.R = D d Solution: V.R. = D d = 180 30 = 6 Effort (p) required to lift a load of 100N: = M.A 100 V.R. W = P V.R. 100 = W.A 100 V.R. MA = W P 80 = 100 100 P 6 100100 P = 80 6 = 0.83 N 3

: F.Y. Diploma Mechanics Q.(c) The diameters of bigger and smaller pulleys of Weston's differential pulley block are 50 mm and 100 mm respectively. Determine the effort required to lift a load of 3 kn with 80% efficiency. (A) Given data : D = 50 mm W = 3 kn = 3000 N d = 100 mm % = 80% (i) V.R. = D = D d (ii) To find effort D M.A. % = 100 V.R. W/P % = 100 V.R. 80 = 3000 / P 3.33 100 80 = 3000 3.33 P 100 3000 100 P = 3.33 80 50 = 3.33 50 100 P = 11.6.1 N Q.(d) The resultant of two forces is 8KN and its direction is inclined at 60 to one of force whose magnitude is 4KN. Find magnitude and direction of other force. (A) F 1 = 4 KN F F = 8 KN = 60 F =? F = 8KN =? We know that, Fsin F 1 = sin 8sin 4 = sin 60 4 sin(60+) = 8 sin 4(sin60 cos + cos60sin) = 8 sin 3464 cos + sin = 8 sin 3464 cos = 6 sin sin cos =3.464 6 tan = 0.58 = 30 = 60 F 1 = 4KN also, F = Fsin sin ( = 8 0.866 sin90 F = 6.93 KN = 8sin60 sin 6030 4

Prelim Question Paper Q.(e) A screw jack has an effort wheel diameter of 0 cm and pitch is 5 mm. Find velocity ratio. If a load of 1000 N is lifted by an effort of 150 N. Find the efficiency of the machine. (A) (1) VR of simple screw jack is given by D VR = p x00 VR = 5 VR = 15.66 () MA = W 1000 6.66 P 150 M.A. 6.66 (3) % = x100 x 100 = 5.31% V.R. 15.66 Q.(f) Write the different types of force system. (A) Classification of force system : Based on Line of Action, Force system may be classified as following : (i) Collinear forces System : The forces acting in same line of action is called collinear forces. A collinear system is necessarily coplanar Weight W and tension T are in same line of action. (ii) Concurrent forces : The system in which all the forces act at same point is called as concurrent forces. A concurrent force system may be either coplanar or non-coplanar provided that there are more than two forces. All the forces F 1, F, F 3, F 4, F 5, F 6 are meeting at point O. (iii)nonconcurrent forces : The system in which the forces act at different points is called nonconcurrent forces. A nonconcurrent system may be either coplanar or non-coplanar In the diagram F 1, F, F 3, F 4 are acting at different point. Parallel forces The system of in which line of action are parallel to each other are called as parallel forces. A parallel force system may be either coplanar or noncoplanar. (i) Like parallel forces : Like forces acting in same direction are called as like parallel forces. (ii) Unlike parallel forces : Parallel forces acting in opposite direction are called as unlike parallel forces. F 3 F 4 A F 4 D F 5 T W F 1 O F 3 P Q R S F 1 F F 3 F 4 B F F 6 C F F 1 5

: F.Y. Diploma Mechanics Q.3. Attempt any FOUR of the following : [16] Q.3(a) Four forces 0N, 15N, 30N, & 5N are acting at 0, 60, 90 & 150 from xaxis taken in order. Find resultant by analytical method. (A) Fx = 0 + 15cos 60 5 cos 30 30N `F X = 5.85 N 15 sin60 F y F y = 15 sin60 + 30 + 5 sin30 = 55.49 N Resultant force is given by, R = F F x = 5.85 55.49 y R = 55.79 N Direction: 1 FX = tan 1 55 49 = tan F y 5.85 = 83.98 with horizontal. Q.3(b) Determine the magnitude of resultant and position of it wrt point A for the force system shown in Figure. Solve it graphically. (A) Fig. 5N 5 cos30 5 sin30 15 N 15 cos60 30 60 0N Resultant (R) = (ae) scale = 100 = 00 kn x = 5.5 1 = 5.5 m Position (x) = 5.5 m 6

Prelim Question Paper Q.3(c) Calculate the moment about point B for the force system as shown in Figure. (A) Taking moment @ point B M B = (15 x 0) + (10 x 3) (0 x ) + (30 x 3) + (40 x ) = 0 + 30 40 + 90 + 80 = +160 N- m ( ) = 160 N-m (Clockwise moment) Q.3(d) Four forces 6N, 4N, N and 1N are acting along the sides of a square AB, BC, CD and AD respectively. Find the resultant in magnitude and direction only. (A) Fx = 6 = 4N Fy = 1 4 = 5N A B 6N R = Fx Fy = 4 5 = 6.4N 1 Fy = tan = tan 1 5/4 = 51.34 Fx Q.3(e) Calculate the magnitude and direction of resultant for concurrent force system as shown in Figure. Use analytical method. (A) (1) Resolving all forces Fx = +(50 cos 30) (70 cos 45) + (100 cos 180) + (60 cos 70) = + 43.30 49.50 100 + 0.5 = - 85.68 Fy = +(50 sin 30) + (70 sin 45) + (100 sin 180) (60 sin 70) = + 5 + 49.50 + 0 56.38 = + 18.1 N () Magnitude of Resultant R = Fx Fy R = ( 85.68) (18.1) R = 87.58N (3) Direction and position of resultant As Fx is ve and Fy is +ve, resultant lies in nd quadrant. Fy θ = tan 1 1 tan 18.1 Fx 85.68 θ = 11.94º N 1N D C 4N 7

: F.Y. Diploma Mechanics Q.3(f) ABCD is a rectangle such that AB = 3m and BC = m. Along sides AB, CB CD and AD, the forces of 100 kn, 00 kn, 50 kn and 150 kn are acting respectively. Find the magnitude, direction and position of the resultant of the forces from C. Use analytical method only. (A) Drawing figure from given data. Resolving the given system of force horizontally. We get, ± ; Fx = 100 50 = 150 KN Resolving the given system of forces vertically we get + Fx = 150 00 = 50 KN Magnitude of resultant (R): R = Fx Fy = 150 50 R = 158.11 KN Magnitude of resultant is 158.11 KN Q.4 Attempt any FOUR of the following : [16] Q.4(a) Two men carry a weight 670 N by means of ropes fixed to the weight. One rope is inclined at 40 and other at 50 with the vertical. Find the tension in each rope analytically. (A) Drawing the F.B.D in fig as follows: Let T 1 and T be the tension in ropes. Applying Lami s theorem, we get. T 1 sin190 = T sin140 = 67 sin 90 T 1 = 670 sin130 sin 90 = 513.49 670 T = sin140 sin 90 = 430.66 N Q.4(b) Forces of 3, 6, 9, and 1 kn respectively acts on a regular pentagon as shown in Figure. Find the resultant in magnitude and direction. Use analytical method only. (A) 8

Prelim Question Paper (A) (1) Exterior angle = 360/No. of angular points = 360/5 = 7º Interior angle = 180º 7º = 108º Angle BAC = Angle CAD = Angle DAE = 108º/3 = 36º () Magnitude of Resultant Resolving all forces Fx = (1 cos 0) + (9 cos 36) + (6 cos 7) (3 cos 7) = +1 + 7.8 + 1.85 0.93 = +0. N Fy = (1 sin 0) + (9 sin 36) + (6 sin 7) + (3 sin 7) = 0 + 5.9 + 5.71 +.85 = +13.85 N (3) Magnitude of resultant R = Fx Fy R = (0.) (13.85) R = 4.49N (4) Direction and position of resultant As Fx = +ve and Fy = +ve, R lies in 1 st quadrant θ = tan 1 Fy tan 1 13.85 Fx 0. θ = 34.44º Q.4(c) Check whether a wire having capacity of 600 N can lift a load of 800N if it is attached as shown in Figure. (A) Apply Lami s theorem : T 1 sin105 = T sin130 = 800 sin15 T 1 0.966 = T 0.766 = 976.6 T1 taking, 0.966 = 976.6 T 1 = 0.966 976.6 = 943.34N > 800N T taking 0.760 = 976.6 T = 0.766 976.6 = 748.1N < 800N W W = 800 N 50 130 T 1 50 15 0 W = 800 N 105 15 T Since wire has a capacity of 800N. But tension in one part is 943.34N, it cannot lift a load of 800N 9

: F.Y. Diploma Mechanics Q.4(d) Explain the different types of beam with neat sketches. (A) (i) Simply supported beam A beam which is freely supported on the walls or columns at its both the ends is called as a simply supported beam. (ii) Cantilever beam A beam fixed at one end and free at other is called as a cantilever beam. (iii)overhanging beam If the end portion of the beam extends beyond the supports it is called as an overhanging beam. A beam may be overhanging on one side or both sides. Overhung on right side Overhung on both sides Overhung on Let side (iv) Fixed beam A beam whose both the ends are rigidly fixed in walls is called a fixed beam, constrained beam, built-in beam or an encastre beam. (v) Continuous beam A beam which is supported on more than two supports is called a continuous beam. Two span continuous beam Three span continuous beam Q.4(e) A sphere weights 100N. It is supported by two planes at 35 and 50 to the horizontal respectively. Calculate the support reactions. (A) Using Lami s theorem, W RA RB sin85 sin145 sin130 10

Prelim Question Paper 100 RA RB sin85 sin145 sin130 (1) () (3) Using term (1) and () 100 RA sin85 sin145 RA = 100 x sin145 sin85 RA = 690.9 N Using term (1) and (3) 100 RB sin85 sin130 sin130 RB = 100x sin85 RB = 9.77 N Q.4(f) Find the beam reactions for the beam loaded and supported as shown in Figure. (A) Applying condition of equilibrium F y = 0 R A (1 4) + R B 3 = 0 R A + R B = 7 kn M@A = 0 4 1 4 + (3 4) 8R B = 0 8 + 1 = 8R B R B =.5 kn From equation (i) R A + R B = 7 R A = 7.5 R A = 4.5 kn R A = 4.5 kn R B =.5 kn (i) 11

: F.Y. Diploma Mechanics Q.5 Attempt any FOUR of the following : [16] Q.5(a) A block weighing 50 N is lying on a horizontal table for which the coefficient of friction is 0.40. Determine limiting force of friction, normal reaction, resultant reaction and angle of friction (A) (i) F y = 0 Normal reaction R 50 = 0 R = 50 N (ii) F x = 0 Force of friction P F = 0 P R = 0 P 0.4 50 = 0 P = 100 N F = 100 N (iii) Angle of friction = tan tan 1 () = = tan 1 (0.4) = 1.80 (iv) Resultant reaction s R F (50) (100) s = 69.58 N Q.5(b) A block weighing 40 kn resting on a rough horizontal plane can be moved by a force of 0 kn applied at an angle of 40 with horizontal. Find coefficient of friction. (A) Drawing Fig from given data Fig: Force are acting on a block Resolving the force horizontally, we get Fx = 0 cos 40 F. (By sign convention ±) Fx = 0 cos 40 R ( F = R) 0 = 0 cos 40 40 (From equilibrium Fx = 0) 0 = R 7.145 R = 7.145 N Putting the value of R in equation (i), we get 0cos 40 = = 0.56 7.145 1

Prelim Question Paper Q.5(c) A body of weight 300 N, resting on an inclined plane inclined at an angle of 30 with the horizontal, just started to move down the plane. Calculate : (i) coefficient of friction (ii) angle of friction (iii)angle of repose (A) Given : W = 300 N Find : (i) (ii) (iii) (i) F x = 0 F 300sin30 F = 150 N R = 150 N = 150 R (ii) F y = 0 R 300csos30 = 0 R = 59.807 N Put R in equation (i) 150 = 59.807 = 0.577 (iii)tan = = tan 1 () = tan 1 (0.577) = 9.98 say 30 ( F = R) (iv) Angle of friction = Angle of repose = = 30 (1) Q.5(d) A Ladder of weight 400N and length 10m is supported on smooth well with its lower end 4m from the wall. The coefficient of friction between the flower and the ladder is 0.3. Show the forces acting on the ladder and find frictional force at floor. (A) sin = 4/10 = sin 1 (4/10) F W = R W W = 0 = 3.58 R W A Fx = 0 R W E F = 0 5m R W = F F smooth wall Fy = 0 W = 0 F W + R F W = 0 G 0 + R F 400 = 0 5m R F = 400 N R F MA = 0 F F L cos R F 4 + 400 + R W 0 = 0 D F F 10 cos 3.58 400 4 + 800 = 0 B m m C Ff = F R F = 0.3 R F F F 10 cos 3.58 = 1600 800 = 800 800 F F = = 87.5 N W = 400N 10cos3.58 L cos Ans. : (i) R W = F F = 87.9N (ii) R F = 400 N (iii)f W = 0 13

: F.Y. Diploma Mechanics Q.5(e) For a certain machine an effort of 100 N and 150 N can lift a load of 1 kn and kn respectively. Find the law of machine. Also calculate maximum efficiency if VR is 0. (A) Effort (P) = 100 N and W = 1 kn = 1000 N Effort (P) = 150 N and W = kn = 000 N (i) Law of machine P = mw + C 100 = m 1000 + C (1) 150 = m 000 + C () Multiplying equation (1) by 1000 m + C = 100 000 m + C = 150 Subtracting equation () from equation (1) 000 m + C = 00 (1) 000 m + C = 150 () C = 50 N put in equation () 150 = m 000 + 50 150 50 = 000 m 100 = 000 m m = 0.05 (ii) Law of machine P = (0.05W + 50) N (iii) Max.M.A. = 1 m = 1 0.05 Max M.A. = 0 Max.M.A. 0 (iv) Max. = 100 100 V.R. 0 Max. = 100 % Q.5(f) A heavy stone of mass 450 kg is on a hill slope of 40 incline. If the between the ground and the stone is 0.65, is the stone stable? (A) By Given : W = mg = 450 x 9.81 = 4414.5 N Weight of stone = 4414.5 N Resolving the forces perpendicular to plane, We get Fy = R 4414.5 cos 40 0 = R 3381.70 (Fy = 0 For the stone in limiting equilibrium) R = 3381.70 N Considering the stone is just on the print of moving down the plane. We know, F = R = 0.65 3381.70 = 193.105N ( R = 3381.70 N) F = Frictional Force = 198.105 N. 14

Prelim Question Paper Downward component of weight of stone = 4414.5 sin 40 = 837.58N 837.58 N > 198.105 N, stone will not be stable Q.6 Attempt any FOUR of the following : [16] Q.6(a) Locate the centroid of angle section 90 mm 100 mm 10 mm. (90 mm side is vertical.) (A) (i) a 1 = 80 10 = 800 mm a = 100 10 = 1000 mm (ii) x 1 = 10 = 5 mm x = 100 = 50 mm y 1 = 10 + 80 = 10 + 40 = 50 mm y = 10 = 5 mm ax x 1 = 1 1 a x a1 a x = 30 mm ay y = 1 1 a y a1 a y = 5 mm = 800 5 1000 50 800 1000 = 800 50 1000 5 800 1000 Q.6(b) Find the centroid of an inverted T-section with flange 00 mm 10 mm and a web of 300 mm 10 mm. (A) (1) Figure is symmetric @ y-y axis and hence, x = Maximum horizontal dimension/ = 00/ = 100mm () Area calculation A 1 = 00 x 10 = 000mm A = 300 x 10 = 3000mm A = A 1 + A = 5000mm ) Location of y y 1 = 10/ = 5mm y = 10 + (300/) = 160mm 15

: F.Y. Diploma Mechanics AyA 1 1 y y = A (000x5) (3000x160) y = 5000 y = 98mm Hence, centroid (G) for given section lies at G ( x,y ) = (100mm from OB and 98mm from OA) Q.6(c) Find the centroid of the I-section with following details. (i) Top flange = 00 mmx 10mm (ii) Bottom flange = 100 mm 0 mm (iii)web thickness = 15 mm (iv) over all depth = 50 mm (A) x = x 1 = x = 00 = 100 mm 00 mm a 1 = 00 10 = 000 mm 10 mm 1 a = 15 0 = 3300 mm a 3 = 100 0 = 000 mm A = a 1 + a + a 3 = 000 + 3300 + 000 = 7300 mm 15 mm 50 mm y 1 = 0 + 0 + 10/ = 45 mm 3 0 mm 0 y = 0 = 130mm, 100 mm y 3 = 0/ = 10 mm ay y = 1 1 a y ay 3 3 00035330013000010 = A 7300 y = 91900 7300 = 15.89 mm. ax,y = (100 mm, 15.89 mm) Q.6(d) Locate the position of centre of gravity for a composite body in which a cone having 400 mm diameter and 800 mm height is placed on a cube of 400 mm side coaxially. 1 1 6 3 (A) (i) V1 r h 00 800 33.5110 mm 3 3 3 6 3 V 400 6410 mm VV1 V = 97.51 10 6 mm 3 V = 97.51 10 6 mm 3 (ii) Distance of C.G. from base 800 y1 400 600 mm 4 400 y 00mm vy 1 1 vy Let y v y = 6 6 33.51 10 600 64 10 00 97.51 10 (33.51600 6400) y 6 97.5110 y 337.46 mm 6 6 400 (iii) x 00mm x 00 mm 16

Prelim Question Paper Q.6(e) A sphere of 300 mm diameter is placed on a cube of 500 mm side coaxially. Locate the position of centre of gravity of the assembly. (A) (i) V 1 = Volume of cube V 1 = 500 3 = 15 10 6 mm 3 4 3 4 V = R 150 3 3 3 V = 14.136 10 6 mm 3 V = 139.136 10 6 mm 3 (ii) Since the body is symmetrical about vertical line 500 x 50 mm x 50 mm (iii)distance of c.g. from x 1 axis 500 y1 50 mm y 500 150 650 mm 6 6 vy 1 1 vy 15 10 50 14.136 10 650 y v 6 1 v 139.13610 y 90.639 mm Q.6(f) A wall of height 6m has one side vertical and other inclined. The top thickness is 1 m and bottom thickness is 4 m. Find its centroid. (A) From given data, Divide the section of retaining wall into rectangle (1) and triangle () and taking the complete section of retaining wall in first quadrant A 1 = 1 6 = 6m A = 1 3 6 = 9m X 1 = 1 = 0.5m 1 X = 1 = 3 = m wrt to OY Y = 1 6 = m. wrt to OX. 3 AX 1 1 AX X = = 60.59 = 1.4m A A 6 9 1 AY Y = 1 1 A Y 639 A1 A 6 9 G1.4m,.4m = Gx,y = =.4m 17