LOM-03-SUBJECTIVE SOLVED

SOLVED EXAMPLES (SUBJECTIVE) 1. A block of mass 2kg is pushed against a rough vertical wall with a force of 40N, coefficient of static friction being 0.5. Another horizontal force of 15N, is applied on the block in a direction parallel to the wall. Will the block move ? If yes, in which direction and with what minimum acceleration ? If no, find the frictional force exerted by the wall on the block Solution: The force may cause the tendency of motion or motion in the body is its own weight and the applied horizontal force of 15N. The resultant of the forces F = In a direction  25N tan1 15   370 20 with the vertical.   The friction, by its very virtue of opposing the tendency will act in a direction opposite to the resultant. Now, the acceleration (minimum) = F  μ N m (as, µN is the maximum frictional force) 25 - 0.5  40 =  5 N 2 2 5 So, Minimum acceleration is 2 = 2.5 m/s2 2. A 4 kg block is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings become taut as shown in the figure. (a) How many revolution per second must the system make in order that the 2m tension in the upper cord shall be 60 N? (b) What is then the tension in the lower cord? ( g = 10 ms-2) Solution: Let  be the angular velocity of rotation of rod. Along vertical T1 cos   mg  T2 cos  Net force towards centre (Centripetal force) T1 sin   T2 sin   m2r ...(i) ...(ii) T2 cos Given : cos   1  4 1.25 5 sin   3 5  r  1.25sin   0.75 m substituting the values of cos , sin , r and T1  60N, m = 4kg in (i) and (ii), we get; ω = 3.74 rad/s T2  10N (from equation (i)) 3. A block of mass M is suspended through a light string A horizontal force F  3 mg is applied at the middle point of string. Find the angle of the string with the vertical in equilibrium and tension in two points of string Solution: Let middle point of string be P, force acting on point P is shown below. Let tension in the string is given for two parts as shown. F.B.D. of block F.B.D. of point P of string F T1 Mg Block is under equilibrium Consider vertical T1  Mg  0 Direction (Y)  T1  Mg For horizontal equilibrium (Along X-axis) F  T2 sin   0 T2 F ...(i)  T2 sin   F ...(ii) For vertical equilibrium (Along y-axis) T2 cos   T1 dividing (ii) by (iii) ...(iii) tan   F T1  F  Mg    60º. 4. A particle slides down a smooth inclined plane of elevation θ , fixed in an elevator going up with an acceleration a0. Initial velocity of block with respect to lift is zero and velocity of lift is also zero. The base of the incline has a length L. (a) Find the time taken by the particle to reach the bottom. (b) Velocity of block when it reaches near floor of lift in ground frame. Solution: (a) Let us work in the elevator frame a pseudo force ma0 in the down- ward direction is to be applied on the particle of mass m together with the real forces. Thus, the forces on m are (i) N normal force (ii) mg downward (by the earth) L (iii) ma0 downward (pseudo). Let a be the acceleration of the particle with respect to the incline. Taking components of the forces parallel to the incline and applying Newton’s law, mg sinθ + ma0 N sinθ = ma or a = (g + a0)sinθ This is the acceleration with respect to the elevator. In this frame, the distance trav- elled by the particle is L/cos θ . Hence, mg L cosθ   1 (g  a 2 0 2L ) sin θ.t2 1/ 2 or t  (g  a ) sin θcos . (b) In lift frame when reaching the floor Vbl  → Vbl is velocity of block with Vl  a 0 t Vl respect to lift → → → → Vbg  Vbl  Vl V  . Vbl Vl is velocity of lift with respect to ground. 5. All the surfaces shown in figure are assumed to be frictionless. The block of mass m slides on the prism which in turn slides backward on the hori- M zontal surface. Find the acceleration of the smaller block with respect to  the prism. Solution: Let the acceleration of the prism be a0 in the backward direction. Consider the motion of the smaller block from the frame of the prism. The forces on the block are (i) N normal force, (ii) mg downward (gravity), mg (iii) ma0 forward (psuedo). The block slides down the plane. Components of the forces parallel to the incline give ma0 cos θ + mg sinθ = ma or a = a0 cos θ + g sinθ ...(i) Components of the force perpendicular to the incline give N + ma0 sinθ = mg cos θ ...(ii) N ma0 Now consider the motion of the prism from the lab frame. No pseudo force is needed as the frame used is inertial. The forces are N' (i) Mg downward, (ii) N normal to the incline (by the block) (iii) N’ upward (by the horizontal surface). Horizontal components give, N N sinθ = Ma0 or N = Ma0/sinθ ...(iii) Putting in (ii) Ma 0  ma sin θ 0 sin θ  mgcosθ or a0  mg sin θcosθ M  m sin 2 θ mg sin θcos2θ From (i) a  M  m sin 2 θ  g sinθ (M  m)g sin θ = M  msin 2 θ 6. The coefficient of static friction between the two blocks shown in figure is µ s and the table is smooth. What maximum horizontal force F can be applied to F the block of mass M so that the blocks move together ? Solution : Two cases are possible either there is relative motion between blocks or not. Let us assume that there is no relative motion, means both blocks move together. The only horizontal force on the upper block of mass m is that due to the friction by the lower block of mass M. Hence this force on m should be towards right. The force of friction on M by m should be towards left by Newton’s third law. Consider the motion of m. The forces on m are F.B.D. of m (a) mg downward by the earth (gravity) N (b) N upward by the block M (normal force) and (c) f (friction) towards right by the block M. In the vertical direction, there is no acceleration. This gives f N = mg ...(i) In the horizontal direction, let the acceleration be a, then mg f = ma For M, F - f = Ma ...(ii) F.B.D. of M Ng Solving (i) and (ii) F f F a = M  m . since f is static f   NmM =  mg ...(iii) N from (i) and (ii) we get F  (M  m)g For F  (M  m)g there is relative motion between blocks and friction will become kinetic. A 7. Consider the situation shown in figure. The horizontal surface below the big- ger block is smooth. The coefficient of friction between the blocks is µ. Find the minimum and the maximum force F that can be applied in order to keep the smaller blocks at rest with respect to the bigger block. Solution: If no force is applied, the block A will slip on C towards right and the block B will move downward. Suppose the minimum force needed to prevent slipping is F. Taking A + B + C as the system, the only external horizontal force on the system is F. Hence, the acceleration of the system is F a = M  2m Now take the block A as system. The forces on A are, (i) tention T by the string towards left, (ii) friction f by the block C towards left, (iii) weight mg downward and (iv) normal force N upward. For vertical equilibrium N = mg. ...(i) F.B.D. of A N mg T As the minimum force needed to prevent slipping is applied, the friction is limiting. Thus, f = µN = µ mg. As the block moves towards right with an acceleration a, T - f = ma or T - µmg = ma ...(ii) Now take the block B as the system. The forces are (i) tension T upward, (ii) weight mg downward, (iii) normal force N ’ towards right, and (iv) friction f ’ upward. As the block moves towards right with an acceleration a, N’ = ma. As the friction is limiting, f ’ = µN ’ = µma For vertical equilibrium T + f ’ = mg or T + µ ma = mg ...(iii) Eliminating T from (ii) and (iii) F.B.D. of B f' N' mg amin  1 μ g 1  μ When a large force is applied the blocks A slips on C towards left and the block B slips on C in the upward direction. The friction on A is towards right and that on B is downwards. Solving as above, the acceleration in this case is amax  1 µ g 1µ Thus, a lies betweeen 1µ g and 1 µ g 1 µ 1µ From (i) the force F should be between 1µ (M  2m)g and 1µ (M  2m)g . 1µ 1µ 8. In the shown figure masses of the pulleys and strings as well as friction be- tween the string and pulley is negligible. Find the acceleration of the masses m1 Solution : and m2 Let the lengths of the strings passes over A is l1 and of that passes over B is l2 from the ground level and yA and yB be distance of pulleys.  (y - y ) + (y - y ) = l A 1 A B 1  2yA - y1 - yB = l1 yB  yA is constant  y1 + yB = 2yA - l1 = constant Differentiating twice this equation w. r. t. time we get d2y d2y 1  B  0  a  a ...(i) Similarly dt2 dt 2 1 B yB + yB - y2 = l2  2yB - y2 = constant d2y d2y  2 B  2  0  2aB  a2 ...(ii) dt 2 dt2 Since mass of the pulley is negligible hence net force on it is zero.  T1 = 2T Let all the masses moves up, F.B.D. of the masses 2T T m1 m1g Equation of motion 2 a1  1 dt2 m2 m2g d2y a 2  2 dt 2 2T - m1g = m1a1 ...(iii) T - m2g = m2a2 ...(iv) Solving equation (i) and (ii), we get (2m2 - m1)g = m1a1 - 2m2a2 ...(v) d2 y  dt 2  d2yB dt 2 Using Eq. (i), (ii) and (v) we get  (2m2 - m1)g = m1(-aB) - 2m2(2aB)  a  (m1  2m2 ) g  a  (2m2  m1) g B (m  4m ) 1 (m  4m ) 1 2 1 2 a  2(m1  2m2 ) g (m1  4m2 ) 9. Two masses m1 and m2 are connected by means of a light string, that passes over a light pulley as shown in the figure. If m1 = 2kg and m2 = 5kg and a vertical force F is applied on the pulley then find the acceleration of the masses and that of the pulley when (a) F = 35 N (b) F = 70 N (a) F = 140N Solution: Since string is massless and friction is absent hence tension in the string is same, every where. (a) Let acceleration of the pulley be ap. For F.B D of the pulley ap to be non zero. F  2T ...(1) also T  m1g F  T  2g ...(2) From (1) and (2), we get F  2  (2g)  F  40 N Therefore when F = 35 N ap = 0 and hence a1 = a2 = 0. T T (b) as mass of the pulley is negligible F - 2T = 0 F  T = F/2  T = 35N to lift m2 T  m2g  T  50 N Therefore block m2 will not move F.B.D of m1  T - m1g = m1a1  15 = 2a1  a  15 m / s2 1 2 Constraint equation yp + yp - y1 = constant  2yp - y1 = c d2 y  2 p  dt2 d2y dt 2 1  0 F  a  a1  15 m / s2 . p 2 4 (c) When F = 140 N T = 70N F.B.D. of m  T - m1g = m1a1 ...(i)  70 N - 20N = 2  a1  a1 = 25 m/s 2 T - m2g = m2a2 ...(ii)  70 N - 50N = 5a2  a2 = 4 m/s 2 Constraint equation yp - y2 + yp - y1 = c  2yp - y2 - y1 = c d2 y  2 p  dt 2 2 1  dt 2 d2 y dt 2 2  0  a  a1  a 2  29 m / s2 . p 2 2 10. In the shown figure the wedge A is fixed to the ground. The prism B of mass M and the block C of mass m is placed as shown. Find the accelera- tion of the block C w.r.t. B when the system is set free. Neglect any friciton. x Solution: Let a be the acceleration of B towards, then F.B.D. of C relative to B. N masin   ma macos  mg  mg - N - masin  = 0, as C is always in contact with B.  N = mg - ma sin  ...(i) and ma cos  = ma’, where a’ = acceleration of C relative to B  a’ = a cos  ...(ii) F.B.D of B N Nsin  Ncos  Mg sin  + N sin  = Ma Putting the value of N from (i), we get  Mg sin  + mg sin  - ma sin2  = Ma ...(iii)  a  (M  m)g sin  M  msin 2  from equation (ii), a'  (M  m)g sin  cos . M  m sin 2  N’ Mgsin Mgcos Mg 11. Two blocks A and B of masses m1 and m2 respectively are placed on each other and their combinations rests on a fixed horizontal surface. A massless string passing over a smooth pulley as shown in the figure is used to connect A and B. Assuming the coefficient of sliding friciton between all surfaces to be µ, show that both A and B will move with a uniform speed, if A is dragged with a force F = µ(3m1 + m2)g to the left. Solution: For vertical equilibrium F.B. D. of m1 N1 - m1g = 0  N1 = m1g ...(1) as m1 slides on m2 hence f f1 = µN1 = µm1g ...(2) 1 For horizontal equilibrium F - T - f1 = 0 m1g  F - T - µm1g = 0 ...(3) F.B.D m2 For vertical equilibrium of m1 N2 - N1 - m2g = 0  N2 = N1 + m2g = m1g + m2g ...(4) f1 Since f2 = µN2 = µ(m1 + m2)g ...(5) f For horizontal equilibrium f1 + f2 - T = 0 m2g  µm1g + µ(m1 + m2)g - T = 0 ...(6) Subtracting (3) from (6), we get 3µm1g + µm2g - F = 0  F = µ(3m1 + m2)g. 12. Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a light and inextensible string and a fricitonless pulley as shown in the figure. The wedge is inclined at 450 to the horizontal on both sides. The coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3. If the system of A and B is released from rest, then find, (i) the acceleration of A (ii) tension in the string (iii) the magnitude and direction of the frictional force acting on A. Solution: (i) In absence of friction the block B will move down the plane and the block A will move up the plane. Frictional force opposes this motion. F.B.D. of the blocks. (A) (B) N1 T T mgsin45 0 f1 mg cos450 2mg cos450 2mgsin450  T - mg sin 450 - f = ma ...(1) and 2mg sin 450 - f - T = 2ma ...(2) mg sin450 - (f + f ) = 3ma 2 for ‘a’ to be non-zero mg sin 450 must be greater than the maximum value of (f + f2)  (f + f ) = (µ m + 2µ m)g cos450 = 4 mg cos 450 1 2 max 1 2 3  mg sin450 < (f + f2)max Hence blocks will remain stationary (ii) F.B.D. of the block B T 2 f(2) max  1 2mg cos 450  3 2mg 2 mg f2 2mg cos450 2mgsin450 W|| = 2 mg sin 450 =  2mg sin450 > f , therefore block B has tendency to slide down the plane. for block B to be at rest T + f = 2mg cos450 T  mg  2  2      3   T = mg . (iii) mg cos450 =  T(tension) is greater than mgcos450 Hence block A has tendency to move up the plane, therefore frictional force on the block A will be down the plane. F.B.D of A N T For A to be at rest mg sin 450 + f = T  f = T - mg sin450 mgsin450 f mg cos450  2  f  2 mg  3 . 13. A 5 kg block is projected upwards with an initial speed of 10 m/s from the bottom of a plane inclined at 30º with horizontal. The coefficient of kinetic friction between the block and the plane is 0.2. (a) How far does the block move up the plane? (b) How long does it move up the plane? (c) After what time from its projection does the block again comes back to the bottom? With what speed does it arrive ? Solution: While the block is moving up, the frictional force acts downward. As the block is slowing down, the velocity acceleration must be in opposite directions. Velocity in this case is upwards so acceleration is in downwards direction and hence negative. N The magnitude of acceleration  mg sin 30º mg cos 30º m  a  gsin 300  μg cos 300  6.6m / s2 For the motion of block from the bottom to up the plane. u  10 m / s , using v 2  u 2  2as , we get 02  102  2(6.6) (s)  v  u  at s  7.58 m 0 = 10 – 6.6  t  t = 1.5 seconds Hence the block moves up the plane for 1.5 sec covering 7.58 m. For the motion of block down the plane: N mg sin 300  μmg cos 300 the magnitude of acceleration m a = gsin 300  μg cos 300  a  3.2 m / s 2 As acceleration is in downward direction, a = - 3.2 m/s2 s = – 7.58 m (down the plane) and u = 0 m/s s  ut  1/ 2at 2 – 7.58 = (0) + ½ (–3.2) t2  t  2.18 sec. So the total time taken to come back: t up  t down  1.5  2.18 v  u  at = 3.68 seconds v  0  (3.2)(2.18)  6.8 m/s v  6.8 m / s . So the block arrives at the bottom with a speed of 6.8 m/s. 14. A small block of mass m is rotated in a horizontal circle with the help of a string of length l connected to m. The other end of the string is fixed to a point O vertically above the centre of the circle so that the string is always inclined with the vertical at an angle  . This arrangement is known as a conical pendulum. Calculate the time period of motion. Solution: From the force diagram of the block: r  l sin  Along the vertical: Net force towards centre : T cos   mg T sin   ma ...(i) from (i) and (ii), we have, Tsin θ  mω2r ...(ii) ω 2  g tanθ  g tanθ  g  Note: r  T  2π ω  2π 𝑙sinθ 𝑙 cosθ T cos  (i) If h is the height of point O above the centre of the circle, then period = 2 (ii) For a conical pendiulum, 2l cos   g  ω  (because cos   1 ). 15. If the system shown in the figure is rotated in a horizontal circle  with angular velocity , find (a) the minimum value of  to start relative motion between the two blocks. (b) tension in the string connecting m1 & m2 when slipping just starts between the blocks. R The coefficient of friction between the two masses is 0.1 and there is no friction between m2 & ground. The dimensions of the masses can be neglected. Take R = 10m, m1 = 10kg, m2 = 5 kg. (g = 10 ms-2) Solution: Owing to the larger value of centrifugal force on m1, it will have a tendency to move outward relative to m2  frictional force on m1 acts inward while that on m2 acts outward, Drawing the F.B.D.s of two masses For equilibrium m  2R = T + µm g 1 1 and T = µm g + m  2R ...(1) 1 2  m  2R = 2µm g + m  2R   (m - m )  2R = 2m µg 1 2 1  min = T T T Tf R m12R f m 2R Putting values of m1, m2 µ, g and R, we get min = 0.63 rad/s Substituting the values in (1) we get, T = 30N. min = Putting values of m1, m2µ, g and R, we get min = 0.63 rad/s Substituting the values in (1) we get, T = 30 N. 16. A smooth semicircular wire track of radius R is fitted in a vertical plane. One end of a mass less spring of natural length 3 R/4 is attached to the lowest point O of the wire track. A small ring of mass m, which can slide on the track is attached to the other end of the spring. The ring is held stationary at point P, such that the spring C makes an angle of 60º with the vertical. The spring constant K  mg . Consider the R O instant when the ring is released and (i) draw the free body diagram of the ring (ii) determine the tangential acceleration and normal reaction. Solution: CO = CP OCP  COP  60º  OCP is an equilateral triangle. OP = R is extended length of the spring. As initial length of the spring = 3 R/4,  Extension, 1  R  3R  R 4 4 mg R mg C N Elastic force in the spring, F = K (1)  R  4  4 . 60º F P For equilibrium along CP, N + F cos 60º = mg cos 60º N = (mg – F) cos60º N  mg  mg  1  3mg 60º O 60º mg N    4  2 8 Along the downward perpendicular to CP kx Tangential force, f = F sin 60º + mg sin 60º Mg  (F  mg) sin 60º   mg  mg      4  2 8  Tangential acc, a  f m  5 3 g. 8