Solutions to Concepts of Physics by HC Verma Chapter 5

5

Newton's Laws of Motion

1.

A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in first two seconds. Find the magnitude of F. Given that m = 2kg, displacement s = 10m in time t = 2s and u = 0 ms–1. Putting required values in 1 𝑠 = 𝑢𝑡 + 𝑎𝑡 2 2

⇒

2.

4

10 = 0 × 2 + 𝑎 2 𝑎 = 5 𝑚/𝑠 2 𝐹 = 𝑚𝑎 = 2 × 5 = 10𝑁

A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it? Here 𝑢 = 40 𝑘𝑚ℎ−1 = 40 × 𝑣 = 0 𝑠 = 4𝑚

1000 60

× 60 = 11.1 𝑚𝑠 −1

But

3.

⇒

𝑣 2 = 𝑢2 + 2𝑎𝑠 123.46 𝑎 = − 2×4 = 15.43 𝑚𝑠 −2

⇒

𝐹 = 𝑚𝑎 = 2000 𝑘𝑔 × 15.43

𝑚 𝑠2

= 30864 ≅ 3.1x104 N

In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5 x 106 m/s in travelling one cm. Assuming straight line motion, find the constant force exerted on the electron. The mass of electron is 9.1 x 10-31 kg. Given that 𝑢 = 0, 𝑣 = 5 × 106 𝑚/𝑠 2 , 𝑠 = 1 𝑐𝑚 = 0.01 𝑚

4.

25×10 12 2×0.01

= 1.25 × 1015 𝑚𝑠 −2

⇒

𝑎=

⇒

𝐹 = 𝑚𝑎 = 9.1 × 10−31 × 1.25 × 1015 = 1.1 × 10−15 N

A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings.

5

Newton's Laws of Motion The tension T1 is the effect of gravitational force of both blocks. ∴ T1 = (0.2+0.3)g = 0.5(10) N = 5 N The tension T2 is due to gravitational force of the lower block. ∴ T2 = 0.3g = 3 N

5.

T1

T2 0.3 kg

Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the two blocks. Let the blocks accelerate by 𝑎. Then by Newton's scond law we have 𝐹 = 2𝑚𝑎 ⇒ a = F/2m Now considering motion of the block other than on which the force F is applied. This block moves because of the tension force T applied on it. Therefore T = ma = m(F/2m) = F/2

6.

A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure. Find the force acting on the particle at t =2, 4 and 6 seconds. Let 𝑣2 , 𝑣4 𝑎𝑛𝑑 𝑣6 be velocities at t = 2, 4 and 6 respectively. Then from the graph, at t=2 𝑣2 = 10 ⇒ 10 = 𝑢 + 𝑎2 𝑡, where u = 0 and t = 2 10 ⇒ 𝑎2 = = 5 𝑚𝑠 −2 20 2

⇒ 𝐹 = 𝑚𝑎2 = 0.25 N

50 1000

5 =

15 10 5

The graph clearly shows that velocity 2 𝑣2 is constant when t = 4 and therefore acceleration and consequently force is zero. Again when t = 6 𝑣6 = 10 = 𝑢 + 𝑎6 𝑡

4

[∵ 𝑢 = 15 𝑎𝑛𝑑 𝑡 = 1]

6

8

5

7.

Newton's Laws of Motion −5 1

= −5 𝑚𝑠 −2

⇒

𝑎6 =

⇒

𝐹 = 𝑚𝑎 =

50 1000

– 5 = – 0.25 𝑁

Two blocks of mass 𝑚𝐴 𝑎𝑛𝑑 𝑚𝐵 respectively are kept in contact on a frictionless table. The experimenter pushes the block a from behind so that the blocks accelerate . If the block A exerts a force F on the block B, what is the force exerted by the experimenter on block A ? Let 𝐹𝐴𝐵 be the force exerted by the experimenter on the two blocks. The acceleration of the two blocks due this force and that of the block B due to force exerted by block A on it are same. Therefore 𝐹𝐴𝐵 𝐹 = 𝑚 𝑚 +𝑚 𝐴

⇒

8.

𝐵

𝐵

𝐹𝐴𝐵 = 𝐹 1 +

𝑚𝐴 𝑚𝐵

Raindrops of radius 1 mm and mass 4 mg are falling with a speed 30 m/s on the head of a bald person. The drops splash on the head of the person and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head. Let 𝑣0 , 𝑚 and s be the initial velocity, mass and the distance covered after striking the head and before coming to rest. Then 𝑣 = 0, 𝑣0 = 30 𝑚/𝑠 𝑚 = 4 mg = 4 x 10–6 kg 𝑠 = 1 mm = 10–3 m ⇒ 𝑣 2 = 𝑣0 2 + 2𝑎𝑠 ⇒ 0 = 900 + 2𝑎 10–3 ⇒ ⇒

9.

𝑎 = – 900 1000/2 = – 4.5 x 105 𝑚𝑠 –1 𝐹 = 𝑚𝑎 = 4 x 10−6 x 4.5 x 105 = 1.8 N

A particle of mass 0.3 kg is subjected to a force 𝐹 = – 𝑘𝑥 with k = 15 N/m. What will be the initial acceleration if it is released from a point 𝑥 = 20 𝑐𝑚? Given that 𝑚 = 0.3 kg, 𝑘 = 15 Nm−1 , 𝑥 = 20 cm = 0.2 m. We now have F = −𝑘𝑥 = –15(0.2) = –3N

5

Newton's Laws of Motion ⇒

a = F/m = 3/0.3 = 10N k1

10. Both the springs shown in the m figure are unstretched. If the block is displaced by a distance 𝑥 and released, what will be the initial acceleration?

k2

Let 𝑥 be the initial displacement of the block. Then the forces on the block are – 𝑘1 𝑥 and −𝑘2 𝑥. ⇒ F = – 𝑘1 𝑥 − 𝑘2 𝑥 = − 𝑘1 + 𝑘2 𝑥 𝑘 +𝑘 𝑥 ⇒ a = F/m = − 1 2 𝑚

11. A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially the block B is near the right end of block A. A constant horizontal force of 10 N is applied to the block A. All surfaces are assumed frictionless. Find the time elapsed before the block B separates from block A. The block A will accelerate under the force 10 N applied on it. But the surfaces being frictionless, block B will remain stationary. So block B will separate from block A when block A has travelled a distance s equal to its length i.e. 20 B cm. 10 N ⇒ 𝑠 = 0.2𝑚. A Also we have 𝑎 = 𝐹/𝑚 = 10/5 = 2𝑚/𝑠 2 . Also u = 0. Putting values in 𝑣 2 = 𝑢2 + 2𝑎𝑠 we get 2 2 × 2 × 0.2 = 0.894 𝑚/𝑠 0.894 = 𝑢 + 𝑎𝑡 = 2𝑡 𝑡 = 0.45𝑠

𝑣 = ⇒ ⇒

12. A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in the figure. Show that the force (assumed equal for both friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth h.

5

Newton's Laws of Motion (a)

(b)

d/2 Let tension in the either rope be T when the friends are pulling the man in the T T h ditch out. The forces acting T T on the man in the ditch are T by the first friend, T by mg θ the second friend and mg the gravitational force. Both the tension forces make angle θ with the vertical. Thus 2 𝑇 𝑐𝑜𝑠𝜃 − 𝑚𝑔 = 0 𝑚𝑔 or 𝑇= 2 cos 𝜃 But as man comes up θ increases to π/2 and accordingly T increases. When man in the ditch is at a height h, let the rope make an angle θ with the vertical. Let l be the length of the rope between the man in the ditch and one of the friends pulling him out. Therefore 𝑙 = ℎ2 + 𝑑2 /4. ℎ ℎ cos 𝜃 = = 𝑙 𝑑2 ℎ2 + 4 𝑚𝑔 𝑚𝑔 𝑚𝑔 𝑇 = 2cos 𝜃 = 2ℎ = 4ℎ 4ℎ2 + 𝑑2 𝑑2 ℎ 2+ 4

13. The elevator shown in the figure is descending with a acceleration of 2 m/s2. The mass of block A is 0.5 kg. What force is exerted by the block A o the block B?

2m/s2 A

B

Force exerted by block A on block B is due to its acceleration. ∴ 𝐹 = 𝑚𝑎 = 2 x 2 = 4 N

14. A pendulum bob of mass 50 g is suspended from the ceiling of the elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s2, (b) goes up with deceleration 1.2 m/s2, (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s2, (e) goes down with deceleration 1.2 m/s2,(a) goes down

5

Newton's Laws of Motion with uniform velocity. Taking the earth as the inertial frame of reference, 𝑎 as the acceleration of the elevator and noting that the tension T in the string of the pendulum is the force applied by it on the bob, we have 𝑇 – 𝑚𝑔 = 𝑚𝑎 ⇒ 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚(𝑚 + 𝑎) (a) Here acceleration is upward, i.e. 𝑎 = 1.2 𝑚𝑠 −2 . 50 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 1000 9.8 + 1.2 = 0.55 N (b) Here acceleration is upward but negative, i.e. 𝑎 = −1.2 𝑚𝑠 −2 . 50 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 9.8 − 1.2 = 1000 0.43 N (c) Here velocity is constant and therefore 𝑎 = 0 𝑚𝑠 −2 . 50 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 1000 9.8 + 0 = 0.49 N (d) Here acceleration is downward i.e. 𝑎 = −1.2 𝑚𝑠 −2 . 50 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 1000 9.8 − 1.2 = 0.43 N (e) Here acceleration is downward but negative, i.e. 𝑎 = 1.2 𝑚𝑠 −2 . 50 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 9.8 + 1.2 = 1000 0.55 N (f) Here velocity is constant and therefore 𝑎 = 0 𝑚𝑠 −2 . 50 𝑇 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 1000 9.8 + 0 = 0.49 N

15. A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and deceleration are the same, find (a) the true weight of the person and (b) the magnitude of acceleration. Take 𝑔 = 9.9 𝑚𝑠 −2 .

5

Newton's Laws of Motion Since the acceleration of the elevator first upward and then downward is same in magnitude let it be 𝑎. Let m be the mass of the person standing on the scale. Then 𝑚𝑔 + 𝑚𝑎 = 72𝑔 and 𝑚𝑔 – 𝑚𝑎 = 60𝑔 Solving we have 𝑚 = 132𝑔/2𝑔 = 66 kg and ⇒

2𝑚𝑎 = 12𝑔 𝑎 = 6𝑔/𝑚 = 6 x 9.9/66 = 0.9 ms−2

16. Find the reading of the spring balance shown in the figure . The elevator is going up with an acceleration of g/10, the pulley and the string are light and the pulley is smooth. Here the spring balance and the blocks are moving in a non-inertial frame of reference, the accelerating elevator. Since Newton’s laws g/10 don’t hold in non-inertial frames, we choose earth as the inertial frame of reference and measure all motion parameters with reference to earth. Let the acceleration of the 1.5 kg block w.r.t elevator be 𝑎. Then acceleration of 1.5 kg this block w.r.t earth will be 𝑎 + 𝑔/10. Let T be 3 kg the tension in the string going round the pulley on either side. Noting that the only forces acting on 1.5 kg block are tension T in the upward direction and gravitational force g in the downward direction, the motion of 1.5 kg block gives 𝑔 𝑇 – 1.5𝑔 = 1.5 𝑎 + 10 …I But the acceleration of 3.0 kg block is equal and in the opposite direction to the acceleration of 1.5 kg block and therefore is equal to – 𝑎. Thus from the motion of the 3 kg block we have 𝑔 𝑇 – 3.0𝑔 = 3.0 −𝑎 + …II 10

Adding twice I to II we get 3T – 6.0g = 3.0(g/5) ⇒ T = 2.2g

5

Newton's Laws of Motion If M is the reading on the spring balance then Mg = 2T ⇒ M = 2T/g =2(2.2) = 4.4 kg

17. A block of 2 kg is suspended from the ceiling through a massless spring of spring constant 𝑘 = 100 𝑁𝑚−1 . What is the elongation of the spring? If another 1 kg is added to the block what would be the further elongation ? Let elongation of the spring be 𝑥. Then the equilibrium of the spring mass system ⇒ 𝑚𝑔 = 𝑘𝑥. ⇒ 𝑥 = 𝑚𝑔/𝑘 = 2x10/100 = 0.2 m If another 1 kg mass is added to the block then 𝑚𝑔 3x10 𝑥 = 𝑘 = 100 = 0.3 m ⇒ additional elongation = 0.3 – 0.2 = 0.1 m

18. Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 𝑚𝑠 −2 . Find the elongations. In this case the force exerted by the block on the spring is 𝑚𝑔 + 𝑚𝑎 where a is acceleration of the elevator. Also 𝑔 and 𝑎 act in opposite directions. ⇒ 𝑚(𝑔 + 𝑎) = 𝑘𝑥 ⇒ 𝑥 = 𝑚(𝑔 + 𝑎)/𝑘 = 2(10 + 2)/100 = 0.24 m and after adding 1 kg to the block 𝑥 = 𝑚(𝑔 + 𝑎)/𝑘 = 3(10 + 2)/100 = 0.36 m Thus additional elongation = 0.12 m

19. The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall down near earth's surface with a constant velocity 𝑣 . How much mass should be removed from the balloon so that it may rise with a constant velocity ?

5

Newton's Laws of Motion Since velocity 𝑣 near earth's surface is constant, acceleration 𝑎 = 0. The net force on the balloon being 𝑚𝑎 = 0 we have Fnet = 𝑀𝑔 – 𝐵 + 𝑐𝑣 = 0 ⇒ 𝑐𝑣 = (𝐵 – 𝑀𝑔) After mass m is removed from the balloon, the velocity becomes – 𝑣 which is constant and therefore the acceleration and the net force on balloon is again zero ⇒ F′net = 𝑀 − 𝑚 𝑔 – 𝐵 − 𝑐𝑣 = 𝑀 − 𝑚 𝑔 − 𝐵 − 𝐵 − 𝑀𝑔 = 2(𝑀𝑔 − 𝐵) − 𝑚𝑔 = 0 B ⇒ m= 2 M– g

20. An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6. The mass of the empty plastic box is 𝑚 and its acceleration 𝑎 = 𝑔/6 is upward. If sand of mass M is put into box the mass becomes m+M and acceleration becomes 𝑎 = −𝑔/6. Let B be the force of buoyancy of water. Then −𝑚𝑔 + 𝐵 = 𝑚𝑔/6 and − 𝑚 + 𝑀 𝑔 + 𝐵 = −(𝑚 + 𝑀)𝑔/6 ⇒ –Mg + mg/6 = –(m+M)g/6 ⇒ M(–1+1/6) = –m(1/6 +1/6) ⇒ M(–5/6) = –m(1/3) ⇒ M = 2m/5

21. A force 𝑭 = 𝒗 𝑥 𝑨is exerted on a particle in addition to the force of gravity, where 𝑣 is the velocity of the particle and 𝑨 is a constant vector in the horizontal direction. With what minimum speed a particle of mass m be projected so that it continues to move undeflected with a constant velocity ? Let 𝒊, 𝒋 and 𝒌 be a right handed triad of unit vectors along x, y, z axes, such that 𝒊 is in horizontal direction parallel to 𝑨 and 𝒋 in the vertical direction. ∴ 𝑨 = 𝐴𝒊 …I The particle, when projected with velocity 𝑣, continues to move undeflected with constant velocity 𝑣 so that the acceleration and

5

Newton's Laws of Motion y consequently the net force acting on the particle both vanish. ∴ 𝑭net = 𝑭 − 𝑚𝑔𝒋 = 𝟎 F=vxA ⇒ 𝒗 x 𝑨 = 𝑭 = 𝑚𝑔𝒋 … II x Thus by definition of vector cross v = mg/A mg product 𝒋 is perpendicular to 𝒗 and therefore y-component of 𝒗 z vanishes. ∴ 𝒗 = 𝑣𝑥 𝒊 + 𝑣𝑧 𝒌 … III Equations I, II and III together give 𝑚𝑔𝒋 = 𝒗 x 𝑨 = (𝑣𝑥 𝒊 + 𝑣𝑧 𝒌) x 𝐴𝒊 = 𝑣𝑧 𝐴𝒋 ⇒ 𝑚𝑔 = 𝑣𝑧 𝐴 ⇒ 𝑣𝑧 = 𝑚𝑔/𝐴 which is a constant and therefore velocity 𝑣 = 𝑣𝑥2 + 𝑣𝑧2 is minimum when 𝑣𝑥 is zero. 𝑚𝑔 ⇒ 𝑣min = 𝑣𝑧 = 𝐴

22. In a simple Atwood machine, two unequal masses 𝑚1 𝑎𝑛𝑑 𝑚2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement 𝑚1 = 300 𝑔 𝑎𝑛𝑑 𝑚2 = 600 𝑔. The system s released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley. (a)

Let 𝑢, 𝑣, 𝑎 and 𝑠 be the initial velocity, velocity, acceleration and distance covered at time t. If T be the tension in the string on both sides of the pulley, then T – 0.3𝑔 = 0.3𝑎 T – 0.6𝑔 = −0.6𝑎 ⇒ 0.3𝑔 = 0.9𝑎 ⇒ 𝑎 = 𝑔/3 Then 𝑢=0 𝑣 = 𝑢 + 𝑎𝑡 = 2𝑔/3

m1 m2

5

Newton's Laws of Motion

(b)

⇒

𝑣 2 = 𝑢2 + 2𝑎𝑠

⇒

𝑠 =

3 2𝑔

=

2𝑔 3

=

2x9.8 3

= 6.5 m

From part (a) 𝑇 = 0.3𝑎 + 0.3𝑔 = 0.3 𝑎 + 𝑔 = 0.3 3 10

(c)

4𝑔 2 9

4𝑔 3

𝑔 3

+𝑔 =

= 3.92 N 𝐹 = 𝑇 + 𝑇 = 2 3.92 = 7.84 N

23. Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again. As 𝑚2 stops T becomes zero and only force acting on 𝑚1 is – 𝑚1 𝑔 and acceleration −𝑔. At 𝑡 = 2 the velocity of 𝑚1 is 𝑢1 given by 2𝑔 𝑢1 = 𝑢 + 𝑎𝑡 = 3 When the block comes to stop after further time t we have 0 = 2𝑔/3 − 𝑡𝑔 ⇒ t = 2/3 s The distance covered by 𝑚1 is 1 2 2𝑔 1 4 2𝑔 𝑠1 = 𝑢1 𝑡 + 𝑎𝑡 2 = + −𝑔 = 2 3 3 2 9 9 After 𝑚2 stops for a moment it again start downward journey starting with zero initial velocity and acceleration – 𝑔. So distance covered by 𝑚2 is 1 1 4 2𝑔 𝑠2 = 𝑢2 𝑡 + 2 𝑎𝑡 2 = 2 −𝑔 9 = − 9 However at this instant the velocity of 𝑚1 is zero, while velocity of 𝑚2 is non zero and away from 𝑚1 , the block shall move further and hence the string will be tight again. Therefore the required time is 2/3 s. 10 cm

20 cm

24. Figure shows a uniform rod of length 30 cm having mass 3.0 kg. The 20 N 32 N strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part on the 10 cm part. All the surfaces are smooth and the strings

5

Newton's Laws of Motion and the pulleys are light. Clearly the 10 cm part of the rod has mass 1 kg and the 20 cm part 2kg. Since both parts move together, let their acceleration due to the two force be 𝑎. If F be the force extered by the 20 cm part on 10 cm part then from the motions of 10 cm and 20 cm parts we get 𝐹 – 20 = 𝑎 −𝐹 + 32 = 2𝑎 72 ⇒ 𝐹 = 3 = 24 N

25. Consider the situation shown in the figure. All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks. Let T be the tension in the string and 𝑎 be the acceleration of the two blocks. Taking component of the gravitational force along the slants respectively 𝑇 − g sin 𝑐𝑜𝑠 −1

3 5

𝑇 − g sin⁡ (cos−1

4 5

=

−1𝑎 =

4m

3m

1 kg

1 kg

1𝑎 4𝑔 5 3𝑔 – 5

𝑇–

= −𝑎

𝑇

=𝑎

𝑔 5

5m

= 2𝑎

𝑎=

𝑔 10

ms −2

26. A constant force 𝐹 = 𝑚2 𝑔 is applied on the block of mass 𝑚1 as shown in the figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of 𝑚1 . Let T be the tension in the string. Then motion of 𝑚1 and m2 respectively give 𝑇 − 𝐹 = 𝑚1 𝑎 𝑇 − 𝑚2 𝑔 = −𝑚2 𝑎 −𝑚2 𝑎 + 𝑚2 𝑔 − 𝐹 = 𝑚1 𝑎

F

m1

m2

5

Newton's Laws of Motion ⇒

𝑚 𝑔−𝐹

𝑎 = 𝑚 2+𝑚 = 2

1

𝑚 2𝑔 2(𝑚 2 +𝑚 1 )

27. In the figure 𝑚1 = 5 𝑘𝑔, 𝑚2 = 2 𝑘𝑔 𝑎𝑛𝑑 𝐹 = 1 𝑁. Find the acceleration of the either block. Describe the motion of 𝑚1 if the string breaks but F continues to act. Let T be the tension in the string and a be the acceleration of either block. Then motion of the blocks give T 𝑇 − 𝐹 − 𝑚1 𝑔 = −𝑚1 𝑎 𝑇 − 𝐹 − 𝑚2 𝑔 = 𝑚2 𝑎 m1 𝑚 −𝑚 𝑔 3𝑔 ⇒ 𝑎 = 𝑚1 +𝑚2 = 7 = 4.3 𝑚𝑠 −2 1

2

If the string breaks 𝑚1 being the lighter among the two will continue to move up with gravitational acceleration 𝐹 𝑎 = − 𝑔 + 𝑚 = − 𝑔 + 0.2 ms−2

T m2

F

F

1

28. Let 𝑚1 = 1 𝑘𝑔, 𝑚2 = 2 𝑘𝑔 𝑎𝑛𝑑 𝑚3 = 3 𝑘𝑔 in the figure. Find the accelerations of 𝑚1 , 𝑚2 𝑎𝑛𝑑 𝑚3 . The string from the upper pulley to 𝑚1 is 20 cm when the syatem is released from rest. How long will it take before 𝑚1 strikes the pulley. Let 𝑇1 and 𝑇2 be the tensions in the strings as shown in the figure. Block 𝑚1 being lighter will move up and let its acceleration be 𝑎1 . T1 Let 𝑎 be the acceleration of 𝑚2 relative to the pulley and 𝑎2 and 𝑎3 accelerations of 𝑚2 and m1 𝑚3 . So the equation of motion of 𝑚1 is 𝑇_1 − 𝑚_1 𝑔 = 𝑚_1 𝑎_1 T2 …I Equation of motion of 𝑚2 and 𝑚3 are m2 𝑇2 − 𝑚2 𝑔 = 𝑚2 𝑎 − 𝑎1 … II 𝑇2 − 𝑚3 𝑔 = −𝑚3 𝑎 + 𝑎1 … III Also 𝑇1 = 2𝑇2 … IV 𝑚3 + 𝑚2 𝑇2 − 2𝑚2 𝑚3 𝑔 = −2𝑚2 𝑚3 𝑎1

T1

T2 m3

5

Newton's Laws of Motion 𝑇

( 𝑚3 + 𝑚2 1 − 2𝑚2 𝑚3 𝑔 = −2𝑚2 𝑚3 𝑎1 2 𝑚3 + 𝑚2 𝑚1 𝑎1 + 𝑔 − 4𝑚2 𝑚3 𝑔 = −4𝑚2 𝑚3 𝑎1 𝑚3 + 𝑚2 𝑚1 𝑔 − 4𝑚2 𝑚3 𝑔 = − 4𝑚2 𝑚3 + 𝑚1 𝑚2 + 𝑚1 𝑚3 𝑎1 𝑚1 𝑚3 + 𝑚1 𝑚2 − 4𝑚2 𝑚3 𝑔 = − 4𝑚2 𝑚3 + 𝑚1 𝑚2 + 𝑚1 𝑚3 𝑎1 …V 19𝑔 ⇒ 𝑎1 = 29 𝑢𝑝 Again 𝑇2 = 𝑚2 𝑔 + 𝑎 − 𝑎1 = 𝑚3 (𝑔 − 𝑎 − 𝑎1 ) 𝑚 −𝑚 𝑔−𝑎 1 19 2𝑔 ⇒ 𝑎 = 3 𝑚 2+𝑚 1 = 5 𝑔 − 29 𝑔 = 29 2

3

𝑎2 = 𝑎 − 𝑎1 = And

𝑎3 = 𝑎 + 𝑎1 =

2𝑔 29 2𝑔 29

− +

19𝑔 29 19𝑔 29

17𝑔 17𝑔 = 𝑑𝑜𝑤𝑛 29 29 21𝑔 𝑢𝑝 29

=− =

29. In the previous problem, suppose 𝑚2 = 2 𝑘𝑔 𝑎𝑛𝑑 𝑚3 = 3 𝑘𝑔. What should be the mass 𝑚1 so that it remains at rest ? For 𝑚1 to remain at rest 𝑎1 should vanish in equation V of part (a) ∴ 3𝑚1 + 2𝑚 − 24 = 0 ⇒ 𝑚1 = 24/5 = 4.8 𝑘𝑔

30. Calculate the tension in the string shown in the figure. The pulley and the string are light and all the surfaces are frictionless. Let 𝑎 be the acceleration of either block and T the tension in the string, then from the motion of the hanging block and the other one 𝑇 − 𝑔 = −𝑎 & 𝑇=𝑎 ⇒ T = g/2 = 5 N

T T

1 kg T 1 kg

B

T

A

2T 31. Consider the situation T 12M shown in the figure. Both the pulleys and T the string are light and M the surfaces are frictionless. (a) Find the acceleration of mass M. (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on the pulley A in the figure.

5

Newton's Laws of Motion (a)

(b) (c)

If the 2M block moves d towards right the block M moves 2d downwards. Therefore if the acceleration of block 2M is 𝑎 the acceleration of block M is 2𝑎. If T is the tension in the string then from the motions of 2M and M blocks 𝑇 – 𝑀𝑔 = −2𝑀𝑎 & 2𝑇 = 2𝑀𝑎 ⇒ 2𝑀𝑎 = 2𝑇 = 2𝑀𝑔 − 4𝑀𝑎 ⇒ 2𝑎 = 2𝑔/3 T = Mg/3 If F is the force of the pulley on the clamp then F = Ti + Tj ⇒

𝐹 = 𝑇2 + 𝑇2 = 𝑇 2 =

&

𝜃 = 𝑡𝑎𝑛−1

𝑇 𝑇

=

𝜋 4

𝑀𝑔 2 3

32. Find the acceleration of the block of mass M in the situation shown in the figure. All the surfaces are frictionless and the pulleys and the string are light. If the 2M block goes down by distance d, the block M goes up by distance 2d. So the acceleration 𝑎1 of M block is twice the acceleration M 𝑎2 of block 2M and opposite in sign. The motion of 2M block gives 2𝑇 − 2𝑀𝑔 = 2𝑀𝑎2 The motion of block M gives 𝑇 − 𝑀𝑔 sin 30 ° = 𝑀𝑎1 = −2𝑀𝑎2 𝑔 ⇒ 𝑀(−2𝑎2 + 2 ) = 𝑇 = 𝑀(𝑔 + 𝑎2 ) ⇒ 𝑎2 = −𝑔/6 𝑔 ⇒ 𝑎1 = −2𝑎2 = 3

T

T

30°

2T 2M

33. Find the mass M of the hanging block in the figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

5

Newton's Laws of Motion Let acceleration of block M m be 𝑎. The motion of block M T gives us mg M' 𝑇 – 𝑀𝑔 = −𝑀𝑎 θ …I T And from motion of block m along the slant we get M −𝑚𝑔 sin 𝜃 = −𝑚𝑎 cos 𝜃 … II Lastly since m does not slip over M' we can take m+M' as the system and the motion of blocks m+M' gives 𝑇 = 𝑚 + 𝑀′ 𝑎 … III From equations I, II and III we get 𝑇 = 𝑀 𝑔 − 𝑎 = 𝑚 + 𝑀′ (𝑔 tan 𝜃) ⇒ 𝑀 = 𝑚 + 𝑀′ (𝑔 tan 𝜃)/(𝑔(1 − tan 𝜃)) ⇒ 𝑀 = (𝑚 + 𝑀′ )/(cot 𝜃 − 1)

34. Find the acceleration of the block A and B in the three situations shown in the figure. (a)

2 kg A

A 4 kg

B

A

6 kg

B

5 kg

B

2 kg

1 kg

Let T be tension in the string supporting block A. Then tension the string supporting block B is 2T. Now if block B moves distance d downward then block A moves 2d upward. Thus acceleration 𝑎𝐴 of block A is twice the acceleration 𝑎𝐵 of block B. From the motion of A and B blocks respectively we have 𝑇 − 4𝑔 = 4𝑎𝐴 = −8𝑎𝐵 & 2𝑇 − 5𝑔 = 5𝑎𝐵 3𝑔 𝑔 ⇒ 𝑎𝐵 = = 𝑚𝑠 −2 21

&

7

𝑎𝐴 = −2𝑎𝐵 = −

2𝑔 𝑚𝑠 −2 7

5

Newton's Laws of Motion (b)

Let T be tension in the string supporting block A. Then tension the string supporting block B is 2T. As in part (a) if block B moves distance d downward then block A moves 2d upward. Thus acceleration 𝑎𝐴 of block A is twice the acceleration 𝑎𝐵 of block B. From the motion of A and B blocks respectively we have 𝑇 = 2𝑎𝐴 = −4𝑎𝐵 & 2𝑇 − 5𝑔 = 5𝑎𝐵 5𝑔 ⇒ 𝑎𝐵 = − 𝑚𝑠 −2 13

(c)

10𝑔

& 𝑎𝐴 = −2𝑎𝐵 = 13 𝑚𝑠 −2 Let T be tension in the string supporting block A. Then tension the string supporting block B is 2T. Again if block B moves distance d downward then block A moves 2d upward. Thus acceleration 𝑎𝐴 of block A is twice the acceleration 𝑎𝐵 of block B. From the motion of A and B blocks respectively we have 𝑇 − 2𝑔 = 2𝑎𝐴 = −4𝑎𝐵 & 2𝑇 − 𝑔 = 𝑎𝐵 𝑔 ⇒ 𝑎𝐵 = 3 𝑚𝑠 −2 &

𝑎𝐴 = −2𝑎𝐵 = −

2𝑔 𝑚𝑠 −2 3

35. Find the acceleration of the 500 g block in the figure. Let 𝑇1 be the tension in the string supporting 500 g block and 𝑇2 be the tension in the string supporting 50 g block. Let 𝑎 be the acceleration of 500 g block. Then from the motions of 500 g, 100g and 50 g blocks we get 𝑇1 − 0.5𝑔 = −0.5𝑎 𝑇1 − 𝑇2 − 0.1𝑔 sin 30° = 0.1𝑎 100 g 𝑇2 − 0.05𝑔 = 0.05 𝑎 ⇒ 0.40𝑔 = 0.65𝑎 500 g ⇒ 30° 𝑎 = 8𝑔 −2 𝑚𝑠 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 13

50 g

36. A monkey of mass 15 kg is climbing on a rope with one end fixed to

5

Newton's Laws of Motion the ceiling. If it wishes to go up with an acceleration of 1 𝑚𝑠 −2 how much force should it apply to the rope ? If the rope is 5 m long and the monkey starts from rest, how much time will it take to reach the ceiling ? The mass 𝑚 of the monkey is 15 kg. The required acceleration 𝑎 is 1 𝑚𝑠 −2 and the height h to be climbed is 5 meter. If F is the required force then 𝐹 – 15𝑔 = 15 ⇒ F = 165 N ℎ = 𝑢𝑡 + ½ 𝑎𝑡 2 ⇒

𝑡 =

2ℎ 𝑎

=

10 s

37. A monkey is climbing on a rope that goes over a light smooth pulley and supports a block of equal mass at the other end. Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes. Suppose the monkey applies force F on the rope. Consequently the monkey will accelerate and due to change in tension in the rope the block will also accelerate. Let the acceleration of monkey be 𝑎𝑀 and that of the block be 𝑎𝐵 . We note that ony forces acting on the monkey are the tension T in the rope and the gravitational force 𝑚𝑔. From motions of the monkey and the block we have 𝑇 − 𝑚𝑔 = 𝑚𝑎𝑀 𝑇 – 𝑚𝑔 = 𝑚𝑎𝐵 ⇒ 𝑎𝑀 = 𝑎𝐵 [Q. E. D.] If the monkey and the block start from rest, their accelerations being equal and in the same direction, both will cover same distance and therefore will remain separated by a constant distance.

38. The monkey B shown in the figure id holding on to the tail of the monkey A which is climbing up the rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it ?. Take 𝑔 = 10 𝑚𝑠 −2

5

Newton's Laws of Motion Suppose monkey A applies force F on the rope and accelerates at the rate 𝑎. Consequently let the tension in the rope be T and let T' be the tension in the tail of monkey A. Then equation of motion of monkey A and B are 𝑇 − 𝑇′ − 5𝑔 = 5𝑎 …I 𝑇 ′ − 2𝑔 = 2𝑎 … II ′ ′ ⇒ 𝑇 = 2 𝑎 + 𝑔 = 2(𝑇 − 𝑇 )/5 7 ⇒ 𝑇 = 𝑇′ 2 Now the monkeys will continue to move up as long as T is greater than zero. The force applied F = T is maximum when T' is maximum, that is 30 N 7 ⇒ 𝑇𝑚𝑎𝑥 = 2 30 = 105 N The force is minimum when acceleration of the monkeys is zero and consequently from equation II 𝑇 ′ = 2𝑔 ⇒ 𝑇𝑚𝑖𝑛 = 𝑇 ′ + 5𝑔 = 7𝑔 = 70 𝑁

39. Figure shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine ? What force should he exert on the rope to get his correct weight on the machine ? Let us assume that the man exerts force F on the rope and that T is the tension in the rope. Let the man's weight shown on the scale be 𝑊𝑅 . Then let us write the equation of the motion of the box which is staionary. 𝑇 − 30𝑔 − 𝑊𝑅 𝑔 = 0 …I Again let us write equation of man who is also stationary 𝑇 − 60𝑔 + 𝑊𝑅 𝑔 = 0 … II Solving I and II we get 2𝑊𝑅 𝑔 = 30g ⇒ 𝑊𝑅 = 15 kg Now suppose the man changes force applied on the rope and consequently accelerates by 𝑎 and the scale shows correct weight, i.e. 60 kg. Now the equation of motion are 𝑇 − 30𝑔 − 60𝑔 = 30𝑎 … III

5

Newton's Laws of Motion 𝑇 − 60𝑔 + 60𝑔 = 60𝑎 which gives 𝑇 = 60𝑎 = 2𝑇 − 180𝑔 𝑇 = 180𝑔 = 1800 N

… IV

40. A block can slide on a frictionless incline of angle θ and length 𝑙 , kept inside an elevator going up with a uniform velocity 𝑣 . Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline. Since the elevator is moving with a uniform velocity its acceleration is zero. Let the block move on the incline. The forces on the block are the gravitational force and the normal force. Therefore the equation of motion of the block along the incline is 𝑚𝑔 sin 𝜃 = 𝑚𝑎 ⇒ 𝑎 = 𝑔 sin 𝜃 which is constant. Therefore from the motion of block 1 1 𝑙 = 𝑢𝑡 + 2 𝑎𝑡 2 = 2 𝑎𝑡 2 ⇒

𝑡=

2𝑙 𝑎

=

2𝑙 g sin 𝜃

41. A car is speeding up a horizontal road with an acceleration 𝑎. Consider the following situations in the car. (i) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal. (i)

The motion of the ball along the line of the string gives 𝑚𝑔 sin 𝜃 = 𝑚𝑎 cos 𝜃 𝑎 ⇒ tan 𝜃 = 𝑔 ⇒

𝜃 = 𝑡𝑎𝑛−1

𝑎 𝑔

(ii) The motion of the ball along the incline gives 𝑚𝑔 sin 𝜃 = 𝑚𝑎 cos 𝜃 𝑎 ⇒ tan 𝜃 = 𝑔 ⇒

𝜃 = 𝑡𝑎𝑛−1

𝑎 𝑔

5

Newton's Laws of Motion

42. A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 𝑚𝑠 −2 . Find the displacement of the block during the first 0.2 s after the start. Take 𝑔 = 10 𝑚𝑠 −2 . Since the elevator and block both start at zero velocity and the downward acceleration of the elevator is greater than g the block separates from the floor of the elevator at the start and starts descending with acceleration g. 1 1 10 4 ∴ 𝑥 = 𝑢 − 2 𝑔𝑡 2 = 0 − 2 𝑔(0.2)2 = 2 100 = −20 cm

∎