<p> 78AM-1 Sr. No. 6 EXAMINATION OF MARINE ENGINEER OFFICER</p><p>APPLIED MECHANICS</p><p>CLASS -I (Time allowed - 3 hours)</p><p>INDIA (2001) Morning Paper Total Marks 100</p><p>N.B. - (1) Attempt SIX questions only . (2) All questions carry equal marks. (3) Neatness in handwriting and clarity in expression carries weightage</p><p>1. At a point P on a vertical mast three forces F1, F2 and F3, act: F1=50i F2 = -30 i - 15 j F3 = -25 i - 10 j + 5k. Determine the resultant force at P.</p><p>2. A semicircle is rotated about its diameter to generate a sphere. Calculate the volume of a sphere of radius R. 3. A particle is projected to move along a parabola y2=4x. At a certain instant, when passing through a point P(4, 4) its speed is 5 m/s and the rate of increase of its speed is 3 m/s2 along the path. Express the velocity and acceleration of the particle in terms of rectangular coordinates.</p><p>4. A quick-return shaping mechanism consists of a crank CA rotating clockwise, as shown in Fig. Ex. 7.6, at 50 revolutions per minute. At an instant when the crank makes 30° with the -t-axis, determine the velocity of the ram F moving in the horizontal direction. Determine also the stroke length of the ram and the velocity of the ram, which carries the cutting tool, during the cutting stroke of the ram when the oscillating link OE and the crank CA are vertical.</p><p>5. Two pendulum of length 1.44 and 1m length start swinging together. After how many vibration they will again start swinging together.</p><p>1/7  r  6. The velocity distribution in pipe flow of radius R is given by the expression v = V1  where V  R  is the maximum velocity at the center and v is the velocity at radius r. Find the energy correction factor. 7. A differential mercury manometer is used to measure pressure difference due to flow of oil of relative density 0.85 in a pipe line shown in figure. Find the difference in piezometric heads and the pressure intensities between points A and B. The manometer columns are filled with oil except for the </p><p> mercury.</p><p>8. A brass liner is fitted over and firmly attached to a solid steel shaft 80 mm diameter, Find the outside diameter of the brass liner so that when a torque is applied to the composite shaft it will be equally shared by the two materials. If the torque is 16 KNm, calculate the maximum shearing stress in each material. G steel = 80 GN/m2. G brass = 40 NG/m2. </p><p>9. A venturi meter is tested with its axis horizontal and the flow measured by means of a tank. The pipe diameter is 76 mm, the throat diameter is 38 mm and the pressure difference is measured by U-tube containing mercury, the connections being full of water. If the difference in levels in the U - tube remains steady at 266mm of mercury while 2200 kg of water are collected in 4min, what is the coefficient of discharge? (Sp. gr. of mercury is 13.6). </p><p>10. A horizontal jet of fresh water 25mm in diameter with a velocity of 15 m/s strikes a flat plate, which is inclined at 450 to the direction of the jet. Assuming no splash back of the water, calculate the normal force exerted on the plate when the plate is:- (a) stationary (b) moving, at 10 m/s in the direction of the jet. </p><p>------X------78AM-1 Sr. No. 6 EXAMINATION OF MARINE ENGINEER OFFICER</p><p>APPLIED MECHANICS</p><p>CLASS -I (Time allowed - 3 hours)</p><p>INDIA (2001) Morning Paper Total Marks 100</p><p>N.B. - (1) Attempt SIX questions only . (2) All questions carry equal marks. (3) Neatness in handwriting and clarity in expression carries weightage Answers Answer for Question No. 1</p><p>Solution The resultant of the concurrent forces is given by R = F1 + F2 + F3, = 50 i - 30 i - 15 j - 25 i - 10 j + 5 k = -5 i - 25 j + 5 k This single force when applied at P results in the same dynamic action' as that exerted by the given system of concurrent forces.</p><p>Answer for Question No. 2</p><p>Solution: The theorems of Pappus-Guldinus can be used to great advantage in this case. The centroid of the semicircle is 4R/3 above the diametrical axis. The area of semicircle is R2/2. According to the Pappus-Guldinus, the volume of the body of revolution generated should be same as that which be obtained if the entire area R2/2 were concentrated at a radius 4R/3. The volume of the sphere is therefore, 2(4R/3) x (R2/2) = (4/3) R2</p><p>Answer for Question No. 3</p><p>Solution Since the data relate to the path of the particle, the path coordinates may be used to advantage. The unit vectors are related as follows:</p><p> et= cos i + sin i (i)</p><p> en = sin i - cos i (ii) dy where tan  = — at P dx From the equation of the path, ' y2 =-4x differentiation with respect to x yields 2y(dy/dx) = 4 and (dy/dx)=2/y=1/x1/2 which, at P, is 1/41/2 or 0.5 and  = tan-1 0.5 = 26.57° = 0.464 rad Equations (i) and (ii) at point P become et = 0.894 i + 0.447 j en = 0.447 i-0.894 j The velocity of the particle is given by V = V et = 5(0.894 i + 0.447 j) = 4.47 i + 2.235 j m/s The tangential component of acceleration is 2 at = 3(0.894 i + 0.447 j) = (2.68 i + 1.34 j) m/s The normal component of acceleration is 2 an = (V /)en The radius of curvature r is given by = _(l+(dy/dx) 2 ) 3/2 d2lyldx2 =(1+0.52)3/2 / 0.0625 =22•36 m because d2y/dx2 = (-l/2x-3/2) = 0.0625 The normal component of acceleration is 2 an = (V /)en =(52/22.36) x (0.447 I - 0.894 j) =0.5I - j The acceleration is, therefore, given by a = an+ at =2.68i+1.34j +0.5 i - j =(3.18i+0.34j) m/s2 </p><p>Answer for Question No. 4 Solution The link diagram is first drawn to a suitable scale as shown in Fig. Ex. 7.6 (Solution). The angular velocity of the crank CA is =2 x 50/60 =5.24 rad/s the linear velocity of point A is</p><p>VA = 5.24 x 0.1 = 0.524 m/s</p><p>Consider P as a point on link OE coincident with the point A on the crank CA. The points A and C have, therefore, a relative motion of sliding in the slot s1s2 shown in the line diagram. At the instant shown,  =30° OA =OP=41cm For the velocity diagram, draw ca = 0.524 m/s perpendicular to CA to represent the velocity of A. Since P can slide with respect to A along the slot, draw pa parallel to the slot as shown in Fig. Ex. 7.6 (Solution). Knowing that P is on the rotating link OPE, the absolute velocity of P must be perpendicular to OPE; this is shown by drawing op perpendicular to OPE and thus locating P. From the link diagram,</p><p>VE = OE ______Vp OP oe is, therefore, mode 55/41 times op. oe = 0.355 X 55/41 = 0.48 m/s Since F can only move horizontally, of is drawn a horizontal line. The velocity of F with respect to E on the link EF must be perpendicular to EF; ef is drawn thuslocating the point f. By measurements from the velocity diagram drawn to scale,</p><p>VF =of= 0.55 m/s towards right The stroke length of the ram is</p><p>F1F2 =E1E2 which, by measurement in the link diagram is 31.5 cm. The velocity of the ram in the cutting stroke when the crank is in the vertical position as shown by CA^ is now determined. The velocity diagram at that instant would be as shown in Fig. Ex. 7.6. A quick reflection will show that the velocity of P equals the velocity of A and also that the velocity of F equals the velocity of E. Thus, cp = oa; pa = 0</p><p>VF = VE; ef = 0 and oe = of = 55/45 x 0.524 = 0.64 m/s The cutting velocity of the shaper machine at that instant is, therefore,</p><p>VF = 0.64 m/s This is very nearly the maximum velocity of the tool.</p><p>Answer for Question No. 5 Sol: The pressure at P and Q are equal. Writing the equilibrium equation.</p><p> pA + 1(x + 0.1) = pB + 1(1.0 + x) + 2 x 0.1 </p><p> pA – pB = 0.12 + 0.9 1 As long as the distance x is filled with the same liquid in both limbs, the magnitude is of little importance. Introducing the given values, p  p     13.6  A B  2  hA – hB = = 0.1  0.9 = 0.1  0.9 = 2.5 m. 1  1   0.85  2 2 pA – pB = 2.5 x 0.85 x 9810 = 20.8 x 10 N/m = 20.8 kN/m2.</p><p>Answer for Question No. 6 Sol: Consider unit length of tank. Horizontal component of water pressure 3.7 = 9.81 x 3.7 x 1 x 67.2 kN 2 3.7 and it acts at m from the base. 3 Vertical components of water pressure = 9.81 x 27 x 1 x 1 acting 0.5 m from O l2 1 4 1 +9.81 x acting from O. 4 3 Taking moments about the hinge C, 3.7 1 4 H x 3.7 = 67.2 x + 26.5 + 7.7 x . 3 2 3 H = 27 kN.</p><p>Answer for Question No. 8 (a) 105.3 mm</p><p>Answer for Question No. 9 (a) 0.9651</p><p>Answer for Question No. 9 (a) 78.1 N (b) 8.6776 N</p>