Solutions to Exercises on Newton's Laws of Motion HC Verma's Concepts of Physics Part 1 (34-37)

 Q#34

Find the acceleration of the A and B in the three situations shown in the figure (5-E17).

Answer:
In each of the cases block B is hanging from a moving pulley and the systems are such that for each displacement of block A, the displacement of block B is its half. So acceleration of B is half the acceleration of A. Let the acceleration of block A be 'a' and that of block B be 'a/2' and the tension in the string is T. 

Case (a)
See the F.B.D. of blocks in the diagram below:

Suppose that the acceleration 'a' of block A is downward and of block B is 'a/2' upward. We get, 
For block A,

mg - T = ma
4g – T = 4a
T = 4g – 4a 

For block B,   
2T – 5g = 5a/2
2(4g – 4a) – 5g = 5a/2 
3g – 8a = 5a/2 
21a = 6g
a = 2g/7 

So acceleration of block A is 2g/7 (Downward as assumed) and that of B is g/7  (upward as assumed).

Case (b)
Let the acceleration of block A is 'a' m/s² towards right and that of block B is 'a/2' m/s² downward. See the free body diagram of the blocks below:

Consider block A. The only force along the direction of acceleration is force of tension T. It gives,
T = ma
T = 2a   

Consider block B. Upward force is 2T by strings on both side of the pulley and downward force is weight 5g. Net downward force is 5g – 2T that accelerates a mass of 5 kg by 'a/2'. It gives, 

5g – 2T = 5(a/2) 
5g – 2 x 2a = 5a/2       (replacing T)
5g – 4a = 5a/2 
10g – 8a = 5a
13a = 10g
a = 10g/13

So the acceleration of block A is 10g/13  towards right (as assumed earlier) and that of block B is a/2 = 10g/26 = 5g/13 downward as assumed.

Case (c)
Let the acceleration of block A (mass 2 kg) is 'a' m/s² downwards and that of block B (mass 1 kg) is 'a/2' m/s² upwards. See the Free Body Diagrams of both blocks below:

Consider block A. Forces on it are weight 2g downward and tension of one string T upward. Net downward force along acceleration 'a' is 2g – T. It gives,

2g – T = 2a
T = 2g – 2a
Consider block B. Upward force is by two strings = 2T and downward force is weight 1 x g = g. Net upward force is 2T-g along the direction of acceleration 'a/2'. It gives,

2T – g = 1 x a/2 = a/2
4g – 4a – g = a/2                  (Replacing T)
3g – 4a = a/2
a = 2g/3

So, acceleration of block A is 2g/3 m/s² downward as assumed and that of block B is a/2 = g/3  upward as assumed earlier.         

Q#35                
Find the acceleration of the 500 g block in figure (5-E18).

Answer:
Let tension in the string connected to 500 g block be 'T' Newton and its acceleration is 'a' m/s² downward. Forces acting on it are its weight 0.5g N downward and 'T' N upward. See the Free Body Diagram below. Net downward force along acceleration is '0.5g – T' N. It gives,

0.5g – T = 0.5a
T = 0.5g – 0.5a

Consider the 50 g block. Let tension in the string connected to 50 g block be 'F' Newton. Its acceleration will also be 'a' m/s² but upward. Its weight will be 0.05g N downward. Net force upward along 'a' is F – 0.05g. It gives,

F – 0.05g = 0.05a
F = 0.05g + 0.05a

Consider 100 g block. It will also have acceleration 'a' along the slope upward. Tension in upper string is T and lower string is F. Component of weight (0.10g N) along the slope will be 0.10g x sin30° = 0.05g downward. Normal force on the block by the plane will have zero component along the slope (Hence not shown). So net force along the direction of acceleration is 'T – F – 0.05g'  N. It gives,

T – F – 0.05g = 0.10a

(In these three equations we have three unknowns T, F and a. We eliminate T and F in last equation by replacing them to find the value of 'a'.)

0.5g – 0.5a – 0.05g – 0.05a – 0.05g = 0.1a
0.65a = 0.4g
a = 8g/13  m/s²

So, acceleration of 500 g block will be 8g/13 downward as assumed earlier.                

Q#36
A monkey of mass 15 kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1 m/s², 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?     

Answer:
Let the force applied by the monkey be T. It will also be the tension in the rope. Mass of monkey = m = 15 kg and acceleration = a = 1 m/s². Weight of monkey = mg = 15g. Net upward force = 'T – 15g'  N. It gives,

T – 15g = 15 kg x 1 m/s² = 15 
T = 15g + 15 = 162 N

Let the time taken to reach the ceiling be 't' s. Distance s = 5 m, Initial velocity u = 0, acceleration a = 1 m/s². From the relation,

s = ut + ½at² 
5 = 0 + ½(1)t²
t² = 10
t = √10 s    

Q#37                
A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5-E19). 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.

Answer:
The weight of the monkey is balanced by the weight of equal mass on the other side of the pulley.Whatever force the monkey exerts on the rope equal and opposite force is exerted by the rope on the monkey and this force is the tension in the rope. Tension in the rope is equal on both side of the smooth pulley. Let this tension be T and masses of monkey and the counterweight be m. If the monkey goes up that means net force T – mg is upward ie T > mg. It is also true for the counterweight for which net force T – mg is upward. So its acceleration will also be upward. For both monkey and counterweight acceleration = Force/mass, 

= T – mg/m.
Same is the case if the monkey goes downward. In that case net force downward is mg – T for both, and acceleration = g – T/m. So whatever force the monkey exerts on the rope the monkey and the block move in the same direction with the equal acceleration. 

So if the monkey starts from the rest both will travel in same direction equal to a distance

s = ut + ½at² = 0 +½at² = ½at²

So, their separation will not change as the time passes.                       

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