Newton s Law of motion

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1 5-A / 9, WEA, Sat Nagar, Karol Bagh New Delhi M : , P : E : keshawclasses@gmail.com W: Newton s Law of motion Q. 1. Two sphere A and B are placed between two vertical walls as shown in figure. Draw the free body diagrams of both the spheres A B Solution: F.B.D f sphere A : N 2 N AB (exerted by B) N 1 m A g F.B.D. of sphere B : (exerted by A) N BA N 3 m B g Note: Here N AB and N BA are the action-reaction pair (Newton s third law). Q. 2. The system shown in figure is in equilibrium. Find the magnitude of tension in each string: T 1, T 2, T 3 and T 4. (g = 10 m/s -2 ) 60 0 T 4 B T T 2 T 1 A 10kg Solution: F.B.D. of block 10 kg T 0 T 0 = 1o g T 0 = 100 N

2 10g F.B.D of point A F y = O Y T 2 cos 30 0 = T 0 =100 N 30 0 T 2 T 2 = 200 N T 1 x F x = O T 0 T 1 = T 2 sin 30 0 = 200 = 100 N. F.B.D of point of B y F y = O T 4 cos60 0 = T 2 sin T 4 And F x = O T 3 +T 2 sin30 0 = T 4 sin60 0 T 3 x T 3 = 200 N, T 4 = 200 N B 30 0 T 2 Q. 3. A small body A starts sliding down from the top of a wedge (Fig. 1.12) whose base is equal to I = 2.10 m. the coefficient of friction between the body and the wedge surface is k = at what value of the angle α will the time of sliding be the least? What will it be equal to? S o l. The total down ward force on the body = mg sinα kmgcosα Hence, acceleration = g sinα kg cosα A = (sinα k cosα) g A Now, from simple kinematics, l secα = 0 + at 2 t = α i.e. t = 2l b ( + k cos 2 α) g For minimum t, ( - k cos 2 α) g = x (say) should be maximum,

3 = + 2k cosα sinα = 0 Tan2α = - α = 49 0 Ans. Q. 4. A particle slides down a smooth inclined plane of elevation, fixed in an elevator going up with an acceleration α 0 (figure). The base of the incline has a length L. find the time taken by the particle to reach the bottom. a 0 L Solution: Let us work in the elevator frame. A pseudo force mα 0 in the downward direction is to be applied on the particle of mass m together with the real forces. Thus, the force on m are (figure) (i) N normal force (ii) mg downward (by the earth), (iii) mα 0 downward (pseudo). N ma 0 mg Let α 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, M g sin + mα 0 sinα = m α or Α = (g + α 0 ) sin. This is the acceleration with respect to the elevator. In this frame, the distance travelled by the particle is L cos. Hence, = (g + α 0) sin.t 2 Or t = ( 2L 1/2 (g + α 0 ) sin cos 5. A body of mass m is suspended by two strings making angles α and β with the horizontal. Find the tensions in the strings.

4 Solution: take the body of mass m as the system. The force acting on the system are (i) mg downwards (by the earth), (ii) T 1 along the first string (by the first string) and (iii) T 2 along the second string (by the second string). T 1 T 2 β α mg These forces are shown in figure. as the body is in equilibrium, these forces must add to zero. Taking Horizontal components T 1 cosα T 2 cosβ + mg cos = o T 1 cosα = T 2 cosβ. Or, Taking vertical components, T 1 sinα + T 1 sinβ mg = o. Eliminating T 2 from (i) and (ii), T 1 sin α + T 1 sin β = mg Or, T 1 = mg =. Sin α + sin β Fiom (i), T 2 =. 6. A force F is applied horizontally on mass m 1 as shown in figure A B Find the contact force between m 1 and m 2. F m 1 m 2 smooth Solution; considering both blocks as a system to find The common acceleration. Common acceleration F m1 m2 a A = F.(1) (m 1 + m 2 ) To find the contact force between A and B we draw F.B.D. of mass m 2.

5 F.B.D of mass m 2 R F x = ma x N =m 2. a N m 2 a N = m 2 F (m 1 + m 2 ) M 2 g 7. A horizontal force is applied on a uniform rod of length L kept on a frictionless surface. Find the Tension in rod at a distance x from the end where force is applied L F X Solution: considering rod as a system, we find acceleration of rod a = now draw F.B.D of rod having length x as shown in figure. F T Using Newton s second law X F T = ( ) X.a T = F -. t = f(1 - ). 8. Two blocks A and B of same m attached with a light spring are suspended by a string as shown in figure. Find the acceleration of block ;A and B just after the string is cut. A m Solution: when block A and B are in equilibrium position F.B.D of B T 0 T 0 = mg..(i) B m Mg F.B.D of A T T = mg + T 0 (ii) T 2 mg

6 Mg T 0 When string is cut, tension T becomes zero. But spring does not change its shape just after cutting. So spring force acts on mass B, again draw F.B.D. of blocks A and B as shown in figure T 0 = mg F.B.D. of B T 0 mg = m. a B a b = o mg F.B.D. of A Mg + T 0 = m. a A 2 mg = m. a A a A = 2g (downwards) Mg T 0 = mg 9. A liquid of density p is flowing with a speed V through a pipe of cross-sectional area A. The pipe is bent in the shape of a right angle as shown. What force should be exerted on The pipe at the corner to keep it fixed? SOLUTION: V Consider a mass Δm of liquid flowing across the corner in time Δt. We will apply Newton s IInd law of motion to the mass Δm. V Magnitude of initial momentum = p i = (Δm) V force exerted on the liquid = = Magnitude of final momentum = p f = (Δm) V = = (volume per sec) = Change in momentum = Δp = p f - p i = 2A V 2 ρ Δp can be calculated by vector subtraction geometrically. Hence the pipe must be pushed at the corners with Force 2A V 2 ρ at an angle of 45 0 with the horizontal. As pi = pf > = 45 0 Δp (p i 2 + p f 2 ) = 2 (Δm) 2 v 2 = -p i p f

7 10. At the moment t = 0 a stationary particle of miss m experiences a time-dependent force F= at (T-t), Where a is a constant vector, T is the time during which the given force acts. Find: (a) the momentum of the particle when the action of the force discontinued: ype e uation here. (b) the distance covered by the particle while the force acted. SOLUTION: We can see that F = 0 at t = 0 and at t = T m dv = a t t) dt The force acts from t = 0 to t = mv (t) = a( t2 T/2 t 3 /3) m = a(t 2 T/2 t 3 /3) a From Newton s second law, we have: = F F m dx = a )dt dp = Fdt Ava = Δp Δp at -t) 0 T t mx(t) = a ( ) = a[ ] = x(t) = at 4 (b) F = at ( T t) = m 12m 11. A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s 2 by an external force F 0. F 0 (a) What is F 0? 5kg (b) What is the net force on rope? (c) What is the tension at middle point of the rape? )g = 10 m/s 2 ) 2 kg Solution: for calculating the value of F 0, consider two blocks with the rope as a system 3 kg F.B.D. of whole system

8 (a) F 0 2m/s 2 10 g = 100 N F = 10 2 F0 = 120 N (1) (b) According to Newton s second law, net force on rope. F = ma = (2) (2) = 4 N.(2) (C) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown. T T 4g = 4.(2); T =48 N 4g 12. A rod of mass 2m moves vertically downward on the surface of wedge of mass as shown in figure. Find the relation between velocity of rod that of the wedge at any instant. m solution: Using wedge constraint. Component of velocity of rod along perpendicular to Inclined surface is equal to velocity of wedge along that direction u cos = tan 2m v = tan u m u = v tan perpendicular 13. With what acceleration α should the box of figure descend so that the block of mass M exerts α force

9 Mg /4 on the floor of the box? M a Solution: The block is at rest with respect to the box which is accelerated with respect to the ground. Hence The acceleration of the block with respect to the ground is α downward. The forces on the block are (i) Mg downward (by the earth).and (ii) N upward (by the floor). The equation of motion of the block is, therefore, Mg N = Mα. If N = Mg/4, the above equation gives α = 3g/4. The Block and hence the box should descend with an acceleration 3g/ The figure shows mass m moves with velocity u. find the velocity of ring at that moment. Ring is restricted to move on smooth rod. = 60 0 m u Answer: V R = V R = 2u Solution: Velocity along string remains same V R cos = u V R =

10 Ѳ 60 0 V R = 2u 15. An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the Surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an Angle α with the vertical, the maximum possible value of α is given (a) cot α = 3 (b) tan α = 3 (c) sec α = 3 (d) cosec α = 3 Equilibrium of insect gives N = mg cos α μn N μn mg sin α mg cosα mg mg sinα From Eqs. (i) and( ii), we get Cot α = 1/μ = System shown in figure is in equilibrium and at rest. The spring and sring are massless, new the String is cut. The acceleration of mass 2m and m just after the string is cut will be (a) g/2 upwards, g downwards (b) g upwards, g/2 downwards (c) g upwards, 2g downwards (d) 2g upwards, g downwards 2m M

11 Initially under equilibrium of mass m T =mg Now, the string is cut. Therefore, T = mg force is decreased on mass m upwards and downwards on mass 2m. a m = = (downwards) and a 2 = = (upwards) 17. Two particles of mass m each are tied at the ends of a light string of light 2a. the whole system is Kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance afrom the centre P (as shown in the figure). F m p m a a Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. as a Result, the particles move towards each other on the surface. The magnitude of acceleration, when The separation between them becomes 2x, is (a) a (b) x a 2 x 2 a 2 x 2 (c) (d) 2T cos = F T = sec F a a x x Acceleration of particle = = = x a 2 x 2

12 18. A block A of mass m is tied to a fixed point C on a horizontal table through a string passing round a massless smooth pulley B (figure). A force F is applied by the experimenter to the pulley. That if the pulley is displaced by a distance x, the block will be displaced by 2x. find the acceleration of the block and the pulley. C B F A m Solution; suppose the pulley is displaced to B and the black to A (figure). The length of the string is CB + BA and is also equal to CB + BB + BA Hence, CB + A A= BB + B B + BA or, A A 2 BB. c B B A A The displacement of A is, therefore, twice the displacement of B in any given time interval. Differentiating twice, we find that the acceleration of A is twice the acceleration of B. To find the acceleration of the block we will need the tension in the string. That can be Obtained by considering the pulley as the system. The forces acting on the pulley are (i) F towards right by the experimenter, (ii) T towards left by the portion BC of the string and (iii) T towards left by the portion BA of the string. The vertical force, if any, add to zero as there is no vertical motion. As the mass of the pulley is zero, the equation of motion is F 2T = 0 giving T = F/2. Now consider the block as the system. The only Horizontal force acting on the block is the tension T Towards right. The acceleration of the block is, therefore, A = T/m =. The acceleration of the pulley is a/2 =.

13 19. The figure shown one end of a string being pulley down at constant velocity v. find the velocity of mass m as a function of x. Solution: using constrain equation b b 2 + y = length of string = constant Differentiating w. r. t time: 2. 2x ( ) + ( ) = 0 v 2 x ( ) = v ( ) = - m b b y v x m 20. All surfaces are smooth in the adjoining figure. Find F that block remains stationary with respect to wedge. F m m Solution: Acceleration of (block + wedge) is a = Let us solve the problem by using both frames. Ѳ N From inertial frame of reference (Ground) F.B.D. of block w. r. t. ground (Apply real forces) mg y With respect to ground block is moving with an acceleration a.. F y = 0 N cos = mg (i) x and F x = ma N sin = ma (ii)

14 from Eqs. (i) and (ii) a = g tan.. F (M + m) a = (M + m) g tan From non-inertial frame of reference (Wedge): F.B.D. of block w. r. t. wedge (real forces + pseudo force) w. r. t. wedge, block is stationary.. F y = 0 N cos = mg (iii) F x = ma N sin = ma (iv) From Eqs. (Iii) and (iv), we will get the same result i. e. F = (M + m) g tan. 21. The breaking strength of the string connecting wall and block B is 175 N, find the magnitude of weight of Block A for which the system will be stationary. The block B weighs 700 N. (g = 10 m/s 2 ) Solution: FBD of block B B 30 0 N A 175N B 175N Mg FBD of point in figure T sin N T cos 30 0 Equating forces in horizontal direction T cos30 0 = 175 T T = N 3 In vertical direction

15 T sin 30 0 = T so, T = = 175 N FBD of block A T 3 3 So T = W = 175 N 3 W 22. A body hangs from a spring balance supported from the ceiling of an elevator. (a) if the elevator has an upward acceleration of 2.45 m/s2 and balance reads 50 N, what is the true Weight of the body? (b) under what circumstances will the balance read 30 N? (c) what will be the reading in the balance if the cable of the elevator breaks? SOLUTION: (a) Reading in the spring balance is equal to the tension in the spring = 50 N. As the elevator is accelerating in upward direction with m/s 2 t The acceleration of the block = a = m/s 2 = g/4 T mg = ma acc mg 50 mg = m(g/4) Mg = 40N = true weight Note: when the elevator has an upward acceleration, reading is greater than the actual weight. Acceleration of the block = a = m/s 2 = g/4 T (b) reading of balance = T = 30 N As T < mg (actual weight), the block and elevator acc. =g Must have a downward acceleration a. Mg T = ma A = 2.45 m/s2 is in downward direction. It is possible in two ways: 1. the elevator is going up and slowing down or 2. the elevator is going down and its speed is increasing. Note: The reading of balance is less then actual weight if elevator has downward acceleration. (c) if the cable breaks, the acceleration of the block and the elevator = g (downwards) Net force = mass x acceleration Mg T = mg

16 , T = 0 The reading of the balance = 0 N 23. In the figure a ball and a block are joined together with an inextensible string. The ball can slide on a smooth horizontal surface. If v 1 and v 2 are the respective speeds of the ball and the block, then determine the constraint relation between the two. V 1 m 1 v 2 SOLUTION: distance are assumed from the centre of the pulley as shown in figure. Constraint: Length of the string remains constant. x h x 2 = constant. X 1 Differentiating both the side w. r. t. time we get 2x 1 dx 1 + dx 2 = 0 h 2 x 2 2 x h 1 2 dt dt V 1 v 2 Since the ball moves so as to increase x 1 with the time block moves so as to decrease x 2 with time, Therefore, dx 1 = + v 1 and dx 2 = - v 2 also x 1 = cos v 2 dt dt x h 1 Thus, v 1 cos v 2 = 0 or v 2 = v 1 cos Ans Alternatively, the problem can be solved very easily if we look at the problem form a different Viewpoint and identify a different constant, i.e. velocity of any two points along the string is Same obviously, from the figure V 1 cos = v 2 Ans

17 24. A triangular block of mass M rests on a horizontal table. A block of mass m rests on tis inclined side. What horizontal acceleration a must M have relate to table to keep m stationary relative to triangular block, assuming frictionless contacts? a m M SOLUTION: First consider an inertial observer. To this observer, m will appear to move horizontally with an Acceleration a. drawing the F.B.D. of m, we have Y N N cos N sin X mg mg Equations of motion are N sin = ma and N cos mg = 0 m so that a = g tan M Next,consider an observer which rides the triangular block. To this observer, the mass m will be at rest. Drawing the F.B.D. of m with respect to this observe we have N cos N (pseudo force) ma N sin ma mg mg Equations of motion are N sin mg = 0 And N cos mg = 0 So that a = g tan

18 25. A 60 kg painter stands on a 15 kg platform. A rope attached to the platform and passing Over an overhead pulley allows the painter to over an overhead pulley allows the painter to raise himself and the platform. (i) to get started, he pulls the rope down with force of 400 N. find the acceleration of the platform as well as that of the painter. (ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1s? (iii) What force should he apply now to maintain the constant speed of 1 m/s? N T T a (M + mg) SOLUTION: The free body diagram of the point and the platform can be drawn as shown in the figure Note : that tension in the string is equal to the force by which he pulls the rope. (i) applying Newton s second law 2T (M +m) g = (M +m) a Or a = Here M = 60 kg; m = 15 kg; T = 400 N g = 10 m/s 2 a = = 0.67 m/s 2 (ii) to attain a speed of 1m/s in one second, the acceleration a must be 1 m/s 2 Thus, the applied force is (iii) When the painter and the platform move (upward) together will a constant speed, it is state of dynamic equilibrium thus, 2F (M + m) g = 0 or F = = = 5

PHYS 101 Previous Exam Problems. Force & Motion I

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