Engineering Mechanics

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2 Engineering Mechanics Solutions for Vol I _ Classroom ractice Questions Chapter- orce and Moment Systems. ns: (b) Let the angle between the forces be R Q. ns: (b) R =8 (8 ) R = 6 here, R is the resultant of the two forces. Q R x ssume = ( > ) x = R = 4 cos 6 = 4 4 cos 6 = 5 4 cos () R = x x cos 8 =.. cos8 4 8 = 5 4 cos () 6 = 5 4 cos 8 = 5 4 cos 6 8 =, 6 = 5() +4() cos = 6.89 here angle between two forces. If Q is doubled i.e., Q then resultant (R) is perpendicular to. Qsin tan 9 Qcos + Q cos = = Q cos (i) lso, R Q Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata R = Q [using eq.(i)] Qcos. ns: (b) Since moment of about point is zero. passes through point, m Y y x 6m x

3 : : ME GE _ Vol I _ Solutions M M 8 m 9 m M M 8 x y x = 6. () M 6 9 x 6 6 y = -9 7 y = 6 y = 45 = x y = y 6 45 = 75 84d = = 9 x.5 x 84d = 9.5 d = 9.6 m 5. ns: (c) Moment about O M = sin 6 x.dx. x.5 dx 6 = = 5 4. ns: (a) 6 /m dw dx x = ns: (a) 5 5 w dw 6 wdx 6 x w = 9 xdx = 9 6m = 9 x / 6 6 = 6 (6) / w = 84 he moment due to average force should be equal to the variable force R d = dw x C D.9m.m.75m R = y R = (upward force ositive downward force negative) R = 45 or equilibrium M = (since R = resultant) Let R is acting at a distance of d 45d = d =.55m (from ) Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

4 : : Engineering Mechanics. ns: (c). ns: (d) Y Chapter- Equilibrium of orce System X 45 o 75 o C 6 o 45 o D 45 o sin o o cos C Resolve the forces along the inclined surface x = cos45 sin = sin = =. cos 45. ns: (a) 6 o o C 6 cos6 = C cos = C sin6 + C sin = 6 C C 6 = 5 ; C = ig: ree body diagram at CD C ig: ree body diagram at C or Equilibrium of oint C sin(6 75) sin(6 45) C =.7 rom Sine rule at C. CD C sin(75 45) sin(6 75).7sin5 = sin5 = 4.7 sin() sin5 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

5 : 4 : ME GE _ Vol I _ Solutions 4. ns: (d) R x or body, y = + = = a mg 5 tan = = sin = mg. L y = for entire system R sin4.45 = (59.8) 89.5 R x = cos4.45 = R y = R + ( ) = R = () or equilibrium M = 5. ns: (c) L = ( ) a L = a a L +a = a (L+ a) = a m ++ = mg 4 = mg m = 4/g a = L a substitute in equation () a R = L a 6. ns: (b) L a = = = R = (L a) a L a L a a L a L a L a (L a) L a Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

6 : 5 : Engineering Mechanics 7. ns: (c) = 6 R C = R R H cos = R H = 5 y = R V +sin = C R V = 5 m 5 m D R D = R = 6 9. ns: 4.5m y = 6 R C + R D 6 = R C = R D = R m.5m M = 6 5 = R R = = R C = R D =6 m m 8. ns: (a)..d R H M = R V C y = = = 6 M = + 4= an = 4 8 = 6.4 sin 4 ( ) ( ) 6 ( ) = = 79.5 ow, x =, Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

7 : 6 : ME GE _ Vol I _ Solutions Chapter- riction ree body diagram for block (). ns: (c) he D of the above block shown D of the block Y = + = = = 98 = =. (98 ) X = = = =. (98 ) = 48. ns: (c) Given an = 4 sin = /5 cos = 4/5 4 5 ree body diagram for block () rom D of block () x = = cos 4 = = () 5 y = + sin = = sin = 5.6 ut = =.(5.6) = () rom () & ().8 = = 5 = 5. = 5.6 = 5.6 (5.) = 4.8 = =.4.8=.4 rom D of block () y = = = + = = 4.8 = =. 4.8 = 7.4 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

8 : 7 : Engineering Mechanics x = = = + = = ns: (d)..d of both the books are shown below.. ns: (b) ree ody Diagram cm cm f m g f f m g 5 cm = = = = M = cm = ( )+ ( )( )+( a )( ) = + + = = 5 y = = = 5 x = = + = = (5 5) 5 where, f is the friction between the two books. f is the friction between the lower book and ground. ow, maximum possible acceleration of upper book. f max m g a max g m m =. 9.8 =.94 m/s or slip to occur, acceleration (a ) of lower book. i.e, a a max f f.94 m [ f = f max =.94 and f = (m + m ) g =. 9.8].77 min =.77 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

9 : 8 : ME GE _ Vol I _ Solutions 5. ns: (d) = 6 an = sin = cos = 5 D for bar () 5 4 ut, = =.4 6 = 4 x = = + = = m 4 m 8 m H 6 m 6. ns: (a) Given, =, =. =, =.5 D for block. Y X D for block () Given = 8, = 4 ow, M = 4 ( ) + 8( ) 6 ( ) = = = ut, = =.4 = 8 rom D of block () y = = = + = +4 y = = cos = cos ut, = =.5 cos = 5 cos x = + sin = = sin = 5 cos sin () D for block Y X Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

10 : 9 : Engineering Mechanics y = cos = = cos = =. cos ut, = 4 cos x = + sin = ut, = = x = 5.5 cos45 + sin45 = cos 45 = = = sin = sin 4cos ut from equation () = 5 cos sin 5cos sin = sin 4cos 9 cos = 5 sin 9 tan = 5 8. ns: (a) D of block = = x = = = ( = ) C r =.8 o 7. ns: (d) D for the block 45 o Y = 5 y = sin45 sin45 = 5 = X = y = + = + = + = ( = ) (+ ) = = = Couple = ( + ) r = r ( + ) r ( = f) Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

11 : : ME GE _ Vol I _ Solutions 9. ns: 64 -m D of shoe bar : D 8 48 H 6 6 cm C V C V D of Drum rake : V C C 6 6 cm M = V C 48 + C 8 = C = V C =. V C 48V C +.V C = 8 5V C = 8 V C = 6 C =. V C =. 6 = M =. C =. = 64 -m. ns: (a) = 6 cos = = 6 = 6 4 = 4 = + = 8 = 8 = = 6 = (hen moves upwards) or min calculation, > = = min e = e 6 e e e 4 =.. min = 6 e min = or max calculation e Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

12 : : Engineering Mechanics 6 = e = 8.6 e 4 = 8.6 e = max e 6 max = e max = 5. ns: (b) Given =., tan = 4 rom D () y = cos = = cos =.8 =.8 = =..8 =.6 x = sin = = + sin = =.76 rom D () y = cos = cos = = + cos sin = 5 = y X = = =. (.8 +8) =.6 +6 sin cos = y ig: D () sin cos = ig: D () X e = e =.76 e. =.4 x = + + = sin = 5.74 = 44 = 5.87 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

13 : : ME GE _ Vol I _ Solutions. ns: (d) R R 4. ns: (b) R R.8 t equilibrium R = R =,. aking moment about pin, 5. ns: (b) = Y = = 9.8 s = = =.98 he External force applied =.8 < s rictional force = External applied force =.8 ig: D () rom D () y = = = rom D () e. = e = e =.9 rom D () y = = = = =. = x =, + =.9 + = = 6.9 = 6.4 ig: D () ig: D () Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

14 : : Engineering Mechanics Chapter- 4 Kinematics of article Rectilinear and Curvilinear Motion. ns: (d) x t t t dx V 6t t dt dv a t dt t t = V = and a =. ns: (a) V = kx 4x + 6x V at x = if k = = 4() + 6() = 4 a = dv dx dx dx k.x 8x 6 dt dt dt dt a = x (V) 8x(V) + 6(V) = () 4 (8 4)+6(4) = 8 m/s. ns: (d) Given, a = 6 V dv 6 dt V dv 6dt V V 6t C Given, at t = sec, V = 6 6 = 6() + C C = V = 6t V = 9t ds ut V = 9t dt ds 9t dt S = t + C t, t = sec, S = m = () + C C = 6 S = t + 6 t t = sec S = () + 6 S = 87 m 4. ns: (a) Given = 8S dv d s = 8s = a dt dt e know that, V dv a ds V 8s ds V 8 C S Given, at S = 4m, V = m/sec 8 C 4 C = V 8 S Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

15 : 4 : ME GE _ Vol I _ Solutions V = ds dt 4 s 4 s s ds 4 dt s / 4t C t t =, S = 4 / (4) 4() C 6 4 C = 4 = s / 4t C / s 4t 4 t t = sec / 4 s 4() s = 5.88 m 8 8 a = = s ns: (c) Given, a = 4t dv 4t dt dv = (4t ) dt 4t v = t C dx dt 4t t C =.7 m/sec 4t dx t C dt 4 4t t x =. Ct C 4 4 t x = t Ct C Given condition, t t =, x = m = C t t=, x = m 4 = 4() ( ) 9 C = x = t 4 9 t t at t = 4 sec x = (4) = 8.67 m 6. ns: (b) u = m/sec a = 5 m/sec u = 6 m/sec a = m/sec t t & S S Let S be the distance traveled by Let S be the distance traveled by Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

16 : 5 : Engineering Mechanics S = S +84 u t a t u t a t 84 t 5t 6t t 84 4t 4t 84 = t = 6 sec (or) t = 6 sec t = 6 sec 8. ns: (c) h = 6 u = u = 8 m/sec x x 7. ns: (b) ake, y = x 4x + Initial velocity, V = If V x is constant V y, a y at x = 6 m dx V x = V x = 4 dt V y = dy dt dx x dt 4 (V y ) = x (4) 4(4) V y = 8x 6 4î 6ĵ dx dt (V y ) at x = 6 = 8 (6) 6 = m/sec dv d a y = (xvx 4Vx ) dt dt = a y = dx V x = V x. V x dt V x (a y ) x = 6 = 4 = m/sec ( V x = constant) Let at distance of x ball () crossed ball () x + x = 6 x = (t) + x = x = gt gt () 8(t) gt x + x = 6 ( s = ut + gt 8t gt 6 8 t = 6 t = sec x = (9.8). at ( a = g moving upward) = 9.6 m (from the top) x = = 6.8 m (from the bottom) ) Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

17 : 6 : ME GE _ Vol I _ Solutions 9. ns: (b) V = u + at V = (5) V = 49.5 m/sec V = velocity with which stone strike the glass Velocity loss = % of V 49.5 = = 9.8 m/sec Initial velocity for further movement in glass = = 9.4 m/sec Distance traveled for sec of time is given by S = ut at S = 9.4() (9.8)() S = m S u = t = 5sec V = u +at g V x = V cos = = 86.6 m/sec V y = V sin = = 5 m/sec y = Voyt a yt 6 5t ( )t t 5t 6 = t = 6 (or) sec t = 6 sec x = V t a xt x = (86.6 6) + ( 4)6 x = m 448 m. ns: (a) Given, V = m/sec x = m, y = 8. m y. ns: (a) V y V V y V = m/sec V x o V x x =? a x = 4 m/sec, a y = m/sec 6 m V x = Vcos, V y = Vsin x = Vx t at ( a = along x direction ) x = Vcos t = cost x Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

18 : 7 : Engineering Mechanics t = cos y = Vyt gt 8. = 8. = Vsin () t gt sin 9.8 cos cos 8 = tan 4.9 sec 8 = tan 4.9 (+ tan ) 4.9 tan tan +.9 = tan =.8, tan =.8 = 7.4 ; = ns: (d) Range = maximum height V sin V sin g g sin = sin sin sin cos = tan = 4 = tan (4) = 76. ns: (a) V y V V x V y a =a y V x V x = t / V y = + t t = (t )(t+5) = t = sec V x at t = V x = / Radius of curvature, r = r = 4. ns: (a) V y = 68.7 m/sec V a dvy here a = a y = dt a = ( 4t) t= a = m/sec V x 68.7 = V y V V a 6 g V x Given, v = m/sec v x = vcos6 = / v x = 5 m/sec v y = v sin6 = V x v y = 86.6 m/sec = 55.8 m at tsec Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

19 : 8 : ME GE _ Vol I _ Solutions v y = v y gt (use V = u+at) = () v y = 76.8 m/sec v x = v x = 5 m/sec Chapter- 5 Kinematics of Rigid odies ixed xis Rotation and General lane Motion v at t= = = v x v y = 9.6m/sec. v = tan - y 76.8 v = an - x 5. ns: (a) a r a y = 4 m/sec V x = m/sec = 56.9 a = gcos = 9.8cos56.9 = 5.5m/sec r =m 4 a V r = V a 9.6 = m 5.5 tan = 4 5. ns: (d) V=5 m/sec V y V x V y V x a = g v x = v cos = 4. m/sec a = g = a r = V x 4. = 9. m a 9.8 = an - /4 = 6.6 a y = a cos a sin ote: Velocity will always act in the tangential direction V x = Vsin V = a = sin 6.6 V r =. m/sec. a =. m/sec a y = a cos a sin 4 = a cos6.6.sin6.6 a = 5.8 m/sec a = r = a = r 5.8 =.58 rad/sec Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

20 : 9 : Engineering Mechanics. ns: (c) Given = 4 = radians at t = sec =? =? at t = sec d = d dt dt = 4 = t dt 8 / t t c () rom given condition, at t =, = rad 8 / () = c c 8 / = t t t = sec, = t = =.8rad = t = = d d(4 t ) dt dt 8 / ().5rad / sec. ns: (b) r = cm, = rad/sec, a = cm/s a = r = () = 8 cm/sec Since total acceleration a = a = a a a 8 a = 4 cm/sec a = r = 4 4 = = rad/sec t a a 4. ns: (d) Given angular acceleration, = rad/sec ngular displacement in time t and t = rad = t = rad/sec t =? t t t t 4 t t t = 5. ns: (c) Given retardation = t d t dt d = t dt = t + c rom given condition at t =, = 7 rad/sec 7 = +c c = 7 = t + 7 heel stops at =, = t + 7 t = sec Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

21 : : ME GE _ Vol I _ Solutions 6. ns: (c) angular speed, = 5 rev/sec = 5 rad/sec = rad/sec 8. ns: (b) V b =? r o = 6 rad/sec Radius, r =.m If is constant, d = r o m = a = (since a = r) Since a = m V = m/sec a = a a a = a = r v = r 7. ns: a = 4m/s Q = r r =. = m/sec angential acceleration a = r = = 4m/s ormal acceleration, a = r = 4 = m/s he resultant acceleration a m a a a a = rad/s =4 rad/s 4 4m / s a V = = r o ro m/s ro 6 = m 4 = + r o r o = m V = ro = 6 V = m/sec 9. ns: (a) Instantaneous centre will have zero velocity because the instantaneous centre is the point of contact between the object and the floor.. ns: (a) O 6 m I Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

22 : : Engineering Mechanics V a = m/s V a = along vertical V b = along horizontal So instantaneous center of V a and V b will be perpendicular to and respectively I is the instantaneous centre. rom sine rule Q IQ I sin 45 sin 7 sin 65 I sin 65 IQ sin 7 I O l cos cos6 I O l sin sin 6 V a I m m V Q V IQ V Q IQ I I V IQ Q sin 65 sin 7 V a I rad / sec = ns: (d) Refer the figure shown below, by knowing the velocity directions instantaneous centre can be located as shown. y knowing velocity (magnitude) of Q we can get the angular velocity of the link, from this we can get the velocity of using sine rule. 45 I V Q =m/sec Q V 7 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

23 : : ME GE _ Vol I _ Solutions Chapter- 6 Kinetics of article and Rigid odies. ns: (a). ns: (b) u =, v =.88 m/sec, S =.85 m, v u = as.88 = a a = a =.94 m/sec Direction of motion or the left cord, y = a..() g or the right cord y = Q a g Q () rom () & () Q a = +Q a g g Q a = +Q a a g g g Q g Qa = g a Q g a g a g a g Q = Q a g Q +Q Direction of motion a g a or equilibrium, y = = + a g 4448 = = ns: (a) tan = 4 Direction of Inertial force = tan / net g a x = ma y Direction motion 4 x Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

24 : : Engineering Mechanics x = a g cos6.86 = a g 4 g.8 =.g.8 = = = () y = + y = = y (since = ) = = ( y ) =.(4 sin 6.86) = () rom () & () = (444.8.).5 = = = 967 rom static equilibrium condition y = = = = rom dynamic equilibrium condition x = = ma = a g a = g a = g.() Since v u = as (9.6) = ( a).689 a =.4.() rom () & ().4 = (9.8) =. 5. ns: (a) ma 4. ns: (d) u = 9.6 m/s V= cos sin Q ma Y ma s mg.sin mg cos X Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

25 : 4 : ME GE _ Vol I _ Solutions y = (static equilibrium) cos = = cos = mgcos Since = = mgcos.() x = (Dynamic equilibrium) +ma sin = = ma+mgsin = mgsin ma () rom () & () mg cos = mgsin ma a = gsin gcos a = gcos(tan ) Given Q = s s = ut+ at s = (t)+ at = 6. ns: (a) t = s g cos tan s a 5 x a =.() ork done by & equal S = S S = S S = S a = a.() or body = m a + m g.() or body = m g m a.(4) (), () & (4) sub in () m g m a = (m (a ) + m g) m g m a = 4m a + m g m a + 4m a = m g m g m g m g a = m 4m = = 5 (5) = = ns: 4.95 m/s S =.4 ; K =. D of the block = 5 m g m a m a m g = t Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

26 : 5 : Engineering Mechanics.r.t free body diagram of the block: S = S ; K = K y = = = = Limiting friction or static friction ( S ) =.4= 8 Kinetic riction ( K ) =. = 4 he block starts moving only when the force, exceeds static friction, S hus, under static equilibrium x = S = t = 8 8 t 8 sec he block starts moving only when t >8seconds During 8 seconds to seconds of time: ccording to ewton s second law of motion orce = mass acceleration dv dv K m (t 4) dt 9.8 dt 8 ( t 4)dt 9.8 V dv 5t 4t.87V V 8 Velocity (V) = 4.95 m/s 8. ns:.98 m/s.r.t. D of the crate: X = sin = 98sin = 7.4 Y = cos = 98 cos = Y = Y = = Y = 966.9; = = =89.88 X = + X = = = = ma = a.98 m/s 9. ns: m x = sin 45 = 98. sin 45 = y = cos 45 = Y D of the crate X =9.8=98 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

27 : 6 : ME GE _ Vol I _ Solutions Y = Y = = Y = = K = = 4.68 x = (Dynamic Equilibrium D lembert principle) x ma = a = a.468m / s S = ut + at t is unknown we can not use this equation So use V u = as V = m/s ; u = ; a =.468m/s V as V S 57.67m a.468. ns:.5 rad/s m x =sin 45 m 45 7 m 45 =mg = 98. ma y = cos 45 = 9.8 = 9.4 M = I M = 9.4 = 88.9-m I I m 8 d md 6 7 4kg m M I. ns: (d) V L/ y = V +ma = V = m(g a) () L here, a = Since, M = I L = ml mg L = L m 4mL a L a = g () 4 from () & () V = m g g = 4 V = 4 L ma.5 rad / s mg 4 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

28 : 7 : Engineering Mechanics. ns: (d) I = 5kg.m R =.5m = 8 mr Mass moment of inertia, I x = I y = 4 mr I z = M = I 8.5 = 5 =.4 = = (.4) (since for half revolution = ) =.58 rad/sec. ns: 4.6 seconds M = 6 m L = m, =, = rpm = 6 rad =.94 sec 4. ns: (a) hread Reel r mg a = linear acceleration, k = radius of gyration or vertical translation motion mg = ma () or rotational motion r = I r = mk a = mk r mk a () r mk mg a ma a r gr k r Moment, M = I ml 6 = 6 = 4 = 4.5rad/sec = +t.94 = 4.5t t = 4.65 sec Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

29 : 8 : ME GE _ Vol I _ Solutions Chapter- 7 ork-energy rinciple and Impulse Momentum Equation. ns: (a) L=.48m o L Lcos o u u = 69.6 m/sec u & u = Initial velocity of shell and block respectively V & V = inal velocity of block & shell. ns: (b) = 6. = 4.448, u =? he loss of KE of shell converted to do the work in lifting the sand box and shell to a height of L Lcos o i.e., d = mv here d = L Lcos o =.48.48cos =.4 m = V 9.8 V =.8 m/sec here V is the velocity of block & shell y momentum equation L Lcos o S S = KS Strain energy in spring = rea under the force displacement curve. = s = (ks) s = ks ks Gain of KE ks mv v ks ks = = g m w kg w v.s m w g S m u + m u = m v + m v here v = v = V & u =?, u = Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

30 : 9 : Engineering Mechanics. ns: (a) Given, m = kg osition at any time is given as x = t + 5t + t t t =, x =, t t = sec, x = + 5( ) + ( ) = m dx Velocity, V = t 6t dt Initial velocity i.e., t =, is v i = m/s inal velocity i.e., at t = sec, is v f = + () + 6() = 85m/s ork done = change in KE = mv 4. ns: (a) Given force = e -x ork done = dx 5. ns: (b) = 4x x f mv i = 85 = x x = 74 J.5 x.5 x e e dx =.J.. otential Energy at x =.7 = work required to move object from to.7m.7 E = dx.7 = 4 x = x dx.7 x x 4 6. ns: (c) x = x x. 7 = (.7) (.7) =.867 J here w = weight per unit meter dw = a small work done in moving small elemental dx of chain through a d/s x ork done = change in KE b b dw x wdx.x w(l b)b wb wb wl g wl b b v wlv g wlv wl bb g wlb wb wb wlb d = wdx b wlv g wlv g L b Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

31 : : ME GE _ Vol I _ Solutions b Lv bl g v = gb b L v = b gb L 7. ns: (d) = V = 4m/s m = kg, m = kg,(since g = m/sec ) Velocities before impact v = 4 m/sec, v = m/s Velocities after impact u =? u =? Coefficient of restitution e =.6 rom momentum equation m v +m v = m u +m u (4) + (-) = (u ) + (u ) u + u =..() u u relative velocityof Seperation e v v relative velocityof approach.6 = u u 4 ( ) = V = m/s u u = () rom & u =. m/sec u = 6.66 m/sec 8. ns: (b) Given, m = kg, m = 6 kg Velocities before impact u = 4 m/s, u = m/s Velocities after impact v = m/s, v =? rom momentum equation m u + m u = m v + m v (4) + 6( ) = () + 6(v ) 6 = 6v v = m/s v Coefficient of restitution, e = u 9. ns: (c) KE = mv + I here, = KE = e = = 4 ( ) 5 V R I = m R R R mv R = 5 mr v u 5 mr V R Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

32 : : Engineering Mechanics KE = KE =. ns: (a) mv 5 mr 5 4 kg m/s kg C m V 4R V R = mv mr = mv mv mv Method I : y conservation of linear momentum,we get = ( +) V cm (where, V cm = velocity of centre of mass) V cm = m / s pplying angular momentum conservation about an axis passing through the contact point () and perpendicular to the plane of paper, we get I cm [ngular momentum about any axis passing through can be written as, L L m r V ] cm = rad/sec cm Method II : pplying angular momentum conservation about an axis passing through centre of wheel and perpendicular to the plane of paper. = I cm = rad/sec. ns: (a) d (m+m) g m = m mass of bullet m = M mass of block u = V bullet initial velocity u = block initial velocity v = v = v velocity of bullet and block after impact. d = (M+m)a = (M+m)g a = g rom momentum equation m u + m u = m v + m v mv + m() = (m + M)V mv v = m M ow from v u = as mv gs m M (m+m) a Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

33 : : ME GE _ Vol I _ Solutions m M V = gs m. ns: (a) K =.6k/m =.4 =.44.m. Chapter- 8 Virtual ork 5k 5k u =, u = rom momentum equation m u +m u = m v +m v =.4V +.44V..() ks m v m v = v + v () rom & v =.98 m/s, v =. m/s Let R & R be the reactions at support & respectively. Let y displacement be given to the beam at without giving displacement at he corresponding displacement at C & D 4 are y and y 7 7 y virtual work principle, 4 R 5 y 5 y R y R y 7 5 Since y, R = 7 R = m C m D m 5 7 k /7 y 4/7 y y Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

34 : : Engineering Mechanics ow let us give virtual displacement at as y, herefore corresponding displacement at C & D are 5 7 ' y & 7 ' y 4/8 y 6/8 y C y virtual work principle, D y R y virtual work principle, R y y 5 7 y + R = R 5/7 y m C /7 y m ' y D m R R y +R 8 6 y 5 y = 5 R y y 5 y 4 5 R 5 (since y ) 4 R = 9 ow, Let the virgual displacement at as y. ns: y, R 7 R = k 7 R 4m Let the virtual displacement at D as y, then corresponding displacement at different point as shown below (ssume no displacement at ). 5k m C R 5k m D y 4m y 6 he corresponding displacement at C & D are y and y 6 6 ow by virtual work principle, R y 5 y R 5 y 6 6 Since y, 5 5 R = R = k Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata C m m D y 6

35 : 4 : ME GE _ Vol I _ Solutions. ns: (b) t joint CD Chapter- 9 nalysis of russes H E = y = R E +R = M = a + a + R E a = R E = 4 (downward) R = 4 (upward) D 6 o D C CD D y = CD sin6 = CD = sin 6 x = CD = = CD (ositive indicate CD in tension) CD = 54. ns: (d). ns: (d) a a Sectioned R S h h U h Q h H E E C D a 6 k k R R Q aking moments about point R Q h h 6 h= R Q h = h R E R R Q = 4 k x = R +R Q = 6 + H E = R = 9 4 Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

36 : 5 : Engineering Mechanics R = 5 k t joint y = R p = R sin45 o R = = R p sin 45 5 / R = 5 (compression) x = = R cos45 = 5 = 5 k (ension) R RS R R ns: (b) Qsin45 Rsin Q 45 R Qcos45 Rcos orce in member Q considering joint Q cos45 = R cos Q =.4 R Q sin45 + R sin =.4R.77.5R R =.7 ow, considering joint R R cos R R h h U SU 6 k R Rsin QR QR = R cos =.7 cos =.6 (ensile) M u = RS h ( ) + 6 h ( ) R h( ) = RS h + 6 h h = RS h = 4 h RS = 4 k (Compression) y = SU + R 6 = SU = SU = 4 k (ension) 5. ns: (a) y R M R R L, R D at oint : y R L L L L L Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

37 : 6 : ME GE _ Vol I _ Solutions E sin 45 R L x C L E cos 45 E R L 45 C C sin ns: (a) =9K m x x sin 9.6k Q =K m R D at oint C: y EC y V m EC = C CD L C C CD S.5m m.5m 6. ns : k m tan.5. k tan 6.44 rom the Lami s triangle sin m C C sin 9 sin 6.56 k 6.56 C sin 6.44 sin 6.44 k y = V +V 9+ = M R = V = V = k () V = + 9 = 4 k () dopting method of sections section x-x adopted and RHS taken. tan 5..5 y = (.r.t. RHS of the section x-x) V + V y = x V sin 5. = + 4 =.5 k (ension) orce in member QS =.5 k (ension) Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata

38 : 7 : Engineering Mechanics 8. ns: (c) R = h E C h h R = 4 h 9. ns: (c) R H R H R V = M = R V 5 k C m D m E 5 k m E 5 ( ) +5 6 ( ) R H = 5 + = R H C R H = 45 M = R = / h ( ) h ( ) (h)( ) +R h( ) = h h h +hr = hr = h R = R +R = 4 R = = y = ( at the joint C) C sin45 + R = C sin45 + = CD R H = 5 k X = R H + R H = R H = R H R H = 5 k (egative indicate R H is left side) t joint R H x = D = 5 k y = = D C = C = Hyderabad Delhi hopal une hubaneswar Lucknow atna engaluru Chennai Vijayawada Vizag irupati Kukatpally Kolkata