Engineering Mechanics

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 : 00 Q. Attempt any TEN of the following : [20] Q.(a) Difference between mass and weight? [2] Difference between mass and weight Mass Weight. Mass is the quantity of matter contained in a body. Weight of a body is the force with which the body is attracted by the earth towards its centre. 2. It is scalar quantity. it is a vector quantity. 3. S.I. unit of mass is ky S.I. hit of weight is N. Q.(b) State Newton s Laws of motion? (First, Second & Third) [2] Nowtons 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.(c) Define Resolution of force. [2] 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.(d) Define effort and effort lost in friction. [2] Effort(P) : The force applied to lift the heavy loads is known as effort. Effort lost in friction (P f ): It is the effort by considering the wear and tear effect while use of machine. OR It is the effort obtained by subtracting ideal effort from an effort. Q.(e) What is polar diagram? [2] Polar Diagram : In case of non-concurrent or parallel force system the point of application of resultant can be found out by constructing polar diagram. Polar diagram is obtained from the vector diagram. To construct a polar diagram, any point O known as pole is chosen near the vector diagram and the points on the vector diagram are joined to it. The lines joined in this way are known as rays.

Vidyalankar : F.Y. Diploma Engineering Mechanics Q.(f) State types of friction. [2] (i) Static friction : The friction experienced by a body when it is in equilibrium. (ii) Dynamic friction : The friction experienced by a body when it is in motion. (iii) Rolling : The friction experienced by a bodies when one rolls over the another body. (iv) Sliding : The friction experienced by a bodies when one slides over the another body. Q.(g) State principle of transmissibility of force. [2] 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.(h) What is Bow's notation? Explain with a sketch. [2] 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.(i) Explain meaning of self locking machine. State the condition for it. [2] 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.(j) State velocity ratio for screw jack with meaning of term involved. [2] Velocity Ratio of Simple Screw jack is given by VR = 2L / P When handle of length L is provided OR VR = 2R / P When effort wheel is provided Where, L = length of handle P = pitch of screw R = radius of an effort wheel. Q.(k) What is coefficient of friction? [2] Coefficient of friction : Coefficient of friction is defined as the ratio of limiting friction to the normal reaction at the surfaces of contact. = F/R Q.(l) What is efficiency of a machine? [2] 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 % = 00 input 2

Prelim Question Paper Solution Q.2 Attempt any FOUR of the following : [6] Q.2(a) Resolve the force 9 MN along 22 and 32 on either side of it. Resolve the force 9 MN along 22 and 32 on either side of it. Let = 22 and 2 = 32. Referring figure, Fsin2 F = sin F = 2.44 MN Fsin F 2 = sin 2 2 F 2 = 8.79 MN. 9sin32 = sin 2232 9sin22 = sin 2232 Q.2(b) In a simple axle and wheel, the diameter of wheel is 80 mm and that of axle 30 mm. If the efficiency of the machine is 80%, find the effort required to lift a load of 00 N. Given: D = diameter of wheel = 80 mm, d = diameter of axle = 30 mm = efficiency of the machine = 80% W = load = 00 N To find : V.R:? Formula: V.R = D d Solution: V.R. = D d = 80 30 = 6 Effort (p) required to lift a load of 00N: M.A = 00 V.R. = W 00 PV.R. W.A = 00 V.R. 00 80 = 00 P 6 00 00 P = 80 6 = 20.83 N MA = W P Q.2(c) Write the different types of force system. Classification of force system : Based on Line of Action, Force system may be classified as following : 3

Vidyalankar : F.Y. Diploma Engineering Mechanics (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. T (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. F 4 F 5 W F O F 2 All the forces F, F 2, F 3, F 4, F 5, F 6 are meeting at point O. F 3 F 6 (iii) Nonconcurrent forces : The system in which the forces act at different points is called nonconcurrent forces. A non-concurrent system may be either coplanar or non-coplanar A F 4 B F In the diagram F, F 2, F 3, F 4 are acting at different point. F 3 D C 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 non-coplanar. P Q R S F 2 (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 F 2 F 3 F 4 Q.2(d) A screw jack of pitch 8 mm a lever of 250 mm length if the efficiency of machine is 30%, find the effort required to lift a load of 500 N. Pitch = 8 mm, L = 250 mm, = 30 %. W = 500N find effort (P) =? V.R = 2L P = 2250 8 M.A. = W/P = 500 P M.A = 00 V.R P = 25.46N = 96.35 500 30 = 00 P 96.35 Q.2(e) For a general pulley block number of cogs on effort wheel is 24, that of on load wheel is 6 No. of teeth on the pinion is 4 and that of on spur is 36. If the maximum effort, which can be applied is 60 N, calculate the maximum load that can be lifted, if efficiency of machine is 80%. Given : Geared pulley block machine No of cogs on effort wheel (N ) = 24 No of cogs on load wheel (N4) = 6 No of cogs on pinion (N2) = 4 4

Prelim Question Paper Solution No of cogs on spur (N3) = 36 Max effort (P) = 60 N, () = 80% Find : Max load lifted by machine (W) Solution : (i) for given machine VR is given by V. R. = N X N 3 / N 2 X N 4 = 24 36 / 4 6 = 36 VR = 36 (ii) Efficiency () = MA / VR X 00 80 = MA / 36 00 MA = 28.8 But, MA = W/P 28.8 = W/60 W = 728 N Q.2(f) A crank ABC with system of forces acting on it is shown in figure. Find force P to maintain equilibrium. Given : 70 N force acting at 30 inclination as shown Find : P, if equilibrium is maintained Taking moment @ point B and considering equilibrium condition M B = 0 = Fy 250 P 300 = 70 sin 30 250 = 300P P = 29.7 N Q.3 Attempt any FOUR of the following : [6] Q.3(a) Determine the magnitude of resultant and position of it wrt point A for the force system shown in Figure. Solve it graphically. 5

Vidyalankar : F.Y. Diploma Engineering Mechanics Resultant (R) = (ae) scale = 2 00 = 200 kn x = 5.5 = 5.5 m Position (x) = 5.5 m Q.3(b) Calculate the moment about point B for the force system as shown in Figure. Taking moment @ point B M B = (5 x 0) + (0 x 3) (20 x 2) + (30 x 3) + (40 x 2) = 0 + 30 40 + 90 + 80 = +60 N- m ( ) = 60 N-m (Clockwise moment) Q.3(c) Calculate the magnitude and direction of resultant for concurrent force system as shown in Figure. Use analytical method. () Resolving all forces Fx = +(50 cos 30) (70 cos 45) + (00 cos 80) + (60 cos 70) = + 43.30 49.50 00 + 20.52 = - 85.68 Fy = +(50 sin 30) + (70 sin 45) + (00 sin 80) (60 sin 70) = + 25 + 49.50 + 0 56.38 6

Prelim Question Paper Solution = + 8.2 N (2) Magnitude of Resultant R = Fx Fy 2 2 2 2 R = ( 85.68) (8.2) R = 87.58N (3) Direction and position of resultant As Fx is ve and Fy is +ve, resultant lies in 2 nd quadrant. θ = tan θ =.94º Fy 8.2 tan Fx 85.68 Q.3(d) Four forces 20N, 5N, 30N, & 25N are acting at 0, 60, 90 & 50 from xaxis taken in order. Find resultant by analytical method. 30 5 sin60 25 5 N 25N 5 cos60 25 30 60 20N Fx = 20 + 5cos 60 25 cos 30 `F X = 5.85 N F y = 5 sin60 + 30 + 25 sin30 F y = 55.49 N Resultant force is given by, R = F 2 2 F x y 2 2 = 5.85 55.49 R = 55.79 N Direction: = F X 55 49 tan = tan F y 5.85 = 83.98 with horizontal. Q.3(e) Find the angle between two forces of magnitude 20 N each, such that their resultant is 60 N. Given : P = Q = 20 N R = 60 N To find : Solution : Using Law of parallelogram of forces 7

Vidyalankar : F.Y. Diploma Engineering Mechanics R 2 = P 2 + Q 2 + 2PQ cos (60) 2 = (20) 2 + (20) 2 + 2X20X20cos 3600 = 4400 + 4400 + 28800 cos 3600 = 28800 + 28800 cos 3600 28800 = 28800 cos 25200 = 28800 cos = cos 25200 28800 = cos ( 0.875) = 5.04 Q.3(f) What are the components of 60 N force acting horizontal, in two directions on either side at an angle of 30 each? F = F2 = Fsin 60sin30 sin sin 30 30 = 34.64N Fsin 60sin30 sin sin 30 30 = 34.64N Q.4 Attempt any FOUR of the following : [6] Q.4(a) A sphere of weight 400 N rests in a groove of smooth inclined surfaces which are making 60 and 30 inclination to the horizontal. Find the reactions at the contact surfaces. Apply Lami s theorem : R A R B sin30 = sin60 = 400 sin(6030 ) = 40 sin90 R A = 0.5 400 = 200N R B = 0.866 400 = 346.4N R A A 60 O 60 30 B 400 H 30 R B Q.4(b) Check whether a wire having capacity of 600 N can lift a load of 800N if it is attached as shown in Figure. 5 W W = 800 N 8

Prelim Question Paper Solution Apply Lami s theorem : T sin05 = T 2 sin30 = 800 sin25 T 0.966 = T2 0.766 = 976.62 T taking, 0.966 = 976.62 T = 0.966 976.62 = 943.34N > 800N W = 800 N T2 taking 0.760 = 976.62 T 2 = 0.766 976.62 = 748.N < 800N Since wire has a capacity of 800N. But tension in one part is 943.34N, it cannot lift a load of 800N 50 30 T 50 25 0 05 T 2 Q.4(c) A beam AB of 9m span is simply supported at ends. The beam carries point load of 2KN upwards at 2m from A and uniformly distributed load of 000 N/m downwards on a length of 6m from B. Determine support reactions analytically. 2KN 000 N/M= KN/M A 2m C m D 6m R A RB Converting u.d. l to its equivalent load; 2KN 6KN (000 6) = 6000 N A C D 3m 3m B 2m m 6m R A R B taking moments about A, M A = 0 (R A 0) (2 2) + (6 6) (R B 9) = 0 4 + 36 9R B = 0 9R B = 32 R B = 3.55KN also, F y = 0 R A + R B + 2 6 = 0 R A + R B = 4 R A = 4 3.55 R A = 0.45 KN. Q.4(d) Find the tensions in the string as shown in figure. T 2 T 75 9 50

Vidyalankar : F.Y. Diploma Engineering Mechanics Using Lami's theorem, 25 T T 2 sin75 sin35 sin50 () (2) (3) Using term () and (2) Using term () and (3) 25 sin 75 T sin35 T = sin35 X25 sin 75 25 sin 75 T 2 = T2 sin50 sin50 X25 sin75 T = 8.30N T 2 = 3.940N Q.4(e) Two men carry a weight 200 N by means of ropes fixed to the weight. One rope is inclined at 45 and other 30 with the vertical. Find tension in each side of rope. Using Lami's theorem, W T T 2 sin 75 sin50 sin35 200 T T2 sin 75 sin50 sin35 () (2) (3) Using term () and (2) 0

Prelim Question Paper Solution T 200 sin 75 sin50 sin50 T = 200 sin 75 T = 03.527 N Using term () and (3) 200 T 2 sin 75 sin35 sin35 T 2 = 200 sin 75 T 2 = 46.40 N Q.4(f) Find the support reactions of simply supported beam shown in figure. 0 N 20 N 5 3 m 4 m 3 m (i) Equivalent point load and it's position Equivalent point load = Intensity of udl = 5 3 = 5 N Position from RA= 7m + Span of udl / 2 = 7 + (3/2) = 8.5 m (ii) Applying equilibrium conditions Fy = 0 ( +ve, ve) and M = 0 ( +ve, ve) Fy = 0 RA + 0 20 5 + RB = 0 RA + RB = 25 N () M A = 0 Taking moment of all forces @ point A (RA 0) (0 3) + (20 7) + (5 8.5) (RB 0) = 0 RB = 23.75 N Putting value of RB in equation () RA + 23.75 = 25 RA =.25 N Q.5 Attempt any FOUR of the following : [6] Q.5(a) A Ladder of weight 400N and length 0m is supported on smooth well with its lower end 4m from the wall. The coefficient of friction between the flower and the ladder is

Vidyalankar : F.Y. Diploma Engineering Mechanics 0.3. Show the forces acting on the ladder and find frictional force at floor. sin = 4/0 = sin (4/0) F W = R W W = 0 = 23.58 Fx = 0 R W A R W E F = 0 R W = F F 5m Fy = 0 F W + R F W = 0 smooth wall W = 0 0 + R F 400 = 0 G R F = 400 N 5m MA = 0 R F F F L cos R F 4 + 400 2 + R W 0 = 0 D F F 0 cos 23.58 400 4 + 800 = 0 F F 0 cos 23.58 = 600 800 = 800 B 2m 2m C Ff = F R F = 0.3 R F 800 F F = 0cos23.58 = 87.25 N W = 400N Ans. : (i) R W = F F = 87.29N (ii) R F = 400 N (iii) F W = 0 L cos Q.5(b) For a certain machine an effort of 00 N and 50 N can lift a load of kn and 2kN respectively. Find the law of machine. Also calculate maximum efficiency if VR is 20. Effort (P) = 00 N and W = kn = 000 N Effort (P) = 50 N and W = 2 kn = 2000 N (i) Law of machine P = mw + C 00 = m 000 + C () 50 = m 2000 + C (2) Multiplying equation () by 2 000 m + C = 00 2 2000 m + C = 50 Subtracting equation (2) from equation () 2000 m + 2C = 200 () 2000 m + C = 50 (2) C = 50 N put in equation (2) 50 = m 2000 + 50 50 50 = 2000 m 00 = 2000 m m = 0.05 (ii) Law of machine P = (0.05W + 50) N (iii) Max.M.A. = m = 0.05 Max M.A. = 20 2

Prelim Question Paper Solution Max.M.A. 20 (iv) Max. = 00 00 V.R. 20 Max. = 00 % Q.5(c) A block of weight 500 N is placed on a inclined plane at an angle of 20 with the horizontal. If coefficient of friction is 0.4, find the force P applied Parallel to the plane, just to move the body up the plane. Fy = 0 R 500 cos 20 = 0 R = 469.85 N Fx = 0 P 500 sin 20 R = 0 P = 500 sin 20 + R = 7.0 + 0.4 469.85 P = 236.789N 500 sin 20 F = R Q.5(d) The velocity ratio of a certain machine is 72. The law of machine is P = W 30 N. Find the maximum mechanical advantage and maximum efficiency. 48 State also whether the machine is reversible or not. VR = 72 P = W 30 N 48 m = and c = 30N 48 (i) Maximum M.A. =? (ii) Maximum =? (iii) To decide whether the machine is reversible or not. 20 R 20 500N P motion 500 cos 20 (i) Maximum M.A. = = m /48 = 48 (ii) Maximum = 00 mv.r. = 00 = 66.67% > 50% / 4872 (iii) Since the maximum efficiency is more than 50%, the machine is reversible. Q.5(e) A block of 80 N is placed on a horizontal plane where the coefficient of friction is 0.25. Find the force at 30 up the horizontal to just move the block For limiting equilibrium ( +ve, ve) 3

Vidyalankar : F.Y. Diploma Engineering Mechanics ( +ve, ve) Fy = 0 + R W = 0 R = W = 2000 N R = 2000 N Fx = 0 + P F = 0 + P = F P = R Since F = R P = 0.4 2000 P = 800 N Q.5(f) Find the horizontal force required to drag a body of weight 00 N along a horizontal plane. If the plane is raised gradually upto 5, the body will begin to slide. We know, = tan = tan 5 = 0.27 For limiting equilibrium ( +ve, ve) Fy = 0 + R W = 0 R = W = 00 N R = 00 N ( +ve, ve) Fx = 0 + P F = 0 + P = F P = R Since F = R P = 0.27 00 P = 27 N Q.6 Attempt any FOUR of the following : [6] Q.6(a) Locate the centroid of angle section 90 mm 00 mm 0 mm. (90 mm side is vertical.) (i) a = 80 0 = 800 mm 2 a 2 = 00 0 = 000 mm 2 (ii) x = 0 2 = 5 mm x 2 = 00 2 = 50 mm y = 0 + 80 2 = 0 + 40 = 50 mm y 2 = 0 2 = 5 mm ax ax 2 2 x = a a 2 x = 30 mm = 800 5 000 50 800 000 4

Prelim Question Paper Solution y = ay ay 2 2 a a 2 y = 25 mm = 800 50 000 5 800 000 Q.6(b) Find the centroid of the I-section with following details. (i) Top flange = 200 mmx 0mm (ii) Bottom flange = 00 mm 20 mm (iii) Web thickness = 5 mm (iv) over all depth = 250 mm x = x = x 2 = 200 = 00 mm 2 a = 200 0 = 2000 mm 2 0 mm a 2 = 5 220 = 3300 mm 2 a 3 = 00 20 = 2000 mm 2 A = a + a 2 + a 3 = 2000 + 3300 + 2000 = 7300 mm 2 y = 20 + 220 + 0/2 = 245 mm y 2 = 20 220 2 = 30mm, y 3 = 20/2 = 0 mm ayay ay 2 2 3 3 20002353300302000 0 y = = A 7300 y = 9900 7300 = 25.89 mm. ax, y = (00 mm, 25.89 mm) 200 mm Q.6(c) A wall of height 6m has one side vertical and other inclined. The top thickness is m and bottom thickness is 4 m. Find its centroid. From given data, Divide the section of retaining wall into rectangle () and triangle (2) and taking the complete section of retaining wall in first quadrant A = 6 = 6m 2 A 2 = 36= 9m 2 2 2 3 00 mm 5 mm 20 mm 250 mm X = 2 = 0.5m X 2 = = 2 3 = 2 m wrt to OY Y 2 = 6 = 2m. wrt to OX. 3 X = AX AX 2 2 A A 2 AY AY 2 2 Y = A A 60.592 = 6 9 6392 = 2 6 9 Gx,y = G.4m,2.4m = 2.4m =.4m Q.6(d) A square of 400 mm side from which a circle of 400 mm diameter is cut-off from the centre. Find centroid of the remaining area. 5

Vidyalankar : F.Y. Diploma Engineering Mechanics (i) Area calculation A = 400 400 = 60000 mm 2 A 2 = ( / 4) (400) 2 = 25663.706 mm 2 A = A = A 2 = 34336.293 mm 2 (ii) Location of x x = 400 / 2 = 200 mm x 2 = 400 / 2 = 200 mm AxAx x 2 2 A x = 200mm (iii) Location of y y = 400 / 2 = 200 mm y 2 = 400 / 2 = 200 mm Ay Ay y 2 2 A y = 200mm Hence, centroid (G) for given section lies at G x, y = (200 mm from ob and 200 mm from OA) Q.6(e) Locate the position of centroid of an ice-cream cone as shown in figure. 6

Prelim Question Paper Solution Note : Considering Centroid (i) Figure is symmetric @ y y axis and hence, x = Maximum horizontal dimension / 2 = 200 / 2 = 00 mm (ii) Area Calculation A = bh 200 600 = 60000mm 2 2 2 r 2 00 A 2 = = 5707.96mm 2 2 2 A = A + A 2 = 75707.96mm 2 (iii) y calculation 2 2 y = h 600 = 400 mm 3 3 2 4r 4 00 y 2 = h + 600 = 642.44mm 3 3 Ay Ay 2 2 y = A y = 450.30 mm Hence, centroid (G) for given ice cream cone lies at G x, y = (00 mm from OB and 450.30 mm from OA) OR Note : Considering Center of Gravity of ice-cream cone. (i) Figure is symmetric @ y-y axis and hence, x = Maximum horizontal dimension / 2 = 200 / 2 = 00 mm (ii) Volume Calculation V = (/3) r 2 = (/3)(00) 2 600 = 6.2838 0 6 mm 3 V 2 = (2/3) r 3 = (2/3)(00) 3 = 2.094395 0 6 mm 3 2 V = V + V 2 = 8.377575 0 6 mm 3 (iii) y calculation h 600 y = h 600 4 4 = 450mm 3r2 300 y2 = h 600 = 637.5mm 8 8 7

Vidyalankar : F.Y. Diploma Engineering Mechanics Vy Vy 2 2 y = V y = 496.875mm Hence, Centre of Gravity (G) for given ice cream cone lies G x, y = (00 mm from OB and 496.875 mm from OA) Q.6(f) The frustum of a cone has top diameter 40 cm and bottom diameter 60 cm with height 8 cm. Calculate Y only. Let, Full cone as figure and cut cone as figure 2 (i) Figure is symmetric @ y y axis and hence, x = Maximum horizontal dimension / 2 = 60 / 2 = 30 cm h = 8 cm, h 2 = Height of cut cone In triangle, ABE and CDE h h 2 60 40 60 h = h 2 40 h =.5h 2 h + h 2 = h h + h 2 =.5h 2 h =.5h 2 h 2 h = 0.5h 2 8 = 0.5h 2 h 2 = 36cm h = 8 + 36 = 54cm (ii) Volume Calculation V = (/3) V 2 = (/3) 2 r h = (/3)(30) 2 x 54 = 50.86 0 3 cm 3 2 2 r h+ = (/3)(20) 2 36 = 5.07 0 3 cm 3 V = V V 2 = 35.82 0 cm 3 8

(iii) y calculation h 54 y = 4 4 = 3.5cm h2 36 y 2 = h 8 = 27 cm 4 4 Vy Vy 2 2 y = V y = 7.85 cm Hence, centre of gravity (G) for given frustum of cone lies at G x, y = (30 cm from AQ and 7.85 cm from AP) Prelim Question Paper Solution 9