Solutions to Exercises on Newton's Laws of Motion HC Verma's Concepts of Physics Part 1 (29-33)

 Q#29

In the previous problem, suppose m₂= 2.0 kg and m₃ = 3.0 kg. What should be the mass m so that it remains at rest?  

Answer:
From (3),   
2a = 2g – 5T/12 and T = m₁(g + a), For m₁ to be in rest a = 0.  It gives,

T = m₁g. Putting these in (3)

0 = 2g – 5m₁g/12

m₁ = 24/5 = 4.8 kg

Q#30
Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all the surfaces are friction-less. Take g = 10 m/s². 

Answer:
Let the accelerations of blocks be 'a' m/s² and tension in the string be 'T' N. See the Free Body Diagram for both the blocks below (Forces are taken in the direction of acceleration only):
For the block on the surface, we get, T = ma, 

For the hanging block mg – T = ma     

mg – T = ma (Replace ma = T)

mg – T = T

2T = mg 

T = mg/2   (given m = 1 kg and g = 10 m/s²)   

T = 10/2 = 5 N     

Q#31
Consider the situation shown in figure (5-E14). Both the pulleys and string are light and all the surfaces are friction-less. (a) Find the acceleration of the mass M. (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on the pulley A in the figure.

Answer:
From the arrangement it is clear that pulley B moves half the distance the block M moves. So, acceleration of B is half than M. Let acceleration of block M be 'a' and that of pulley B be a/2. If tension in the string be T, foe block M we have, 

Mg – T = Ma
T = Mg – Ma                                                   (1) 

and for block 2M,

2T = 2M(a/2) = Ma                                       (2)

2Mg – 2Ma = Ma (Replacing T)

2g – 2a = a 

3a = 2g

a = 2g/3 

(a) So, acceleration of block M is 2g/3 

(b) From (B), Tension in the string 2T = Ma = M(2g/3)

T = Mg/3

(c) First calculate the force on the pulley by the strings. As shown in the figure below forces on pulley are T in horizontal and T downwards by strings.

Resultant = √(T²+T²) = √(2T²) = √2T = √2Mg/3

The strings exert this force on the pulley and the pulley on the clamp. As per third law of motion equal and opposite force = √2Mg/3 is applied by the clamp on the pulley.

The direction of this force will be tanθ = T/T = 1 → θ = 45° (From the horizontal downward by the pulley on the clamp) and 45° from the horizontal upward by the clamp on the pulley.

Q#32      
Find the acceleration of the block of mass M in the situation shown in figure (5-E15). All the surfaces are friction-less and the pulleys and string are light.

Answer:
Let us first find out the tension in the string. Let the tension be T. Consider the mass 2M, it is hanging from a moving pulley and the tension in string on both side is T. So, the upward force on this block is 2T and the downward force is its weight 2Mg. It is clear from the arrangement that the displacement of block 2M will always be half the displacement of block M. So acceleration of block 2M will also be half the acceleration of block M. Let the acceleration of block M be 'a' (up the plane) and that of 2M be 'a/2' (Downward). See the 'Free body diagram' of these two blocks in the figure below (Normal force on the block not shown because its component along the slope is zero):

Net force in the direction of acceleration for the block 2M is 2Mg – 2T and the acceleration is 'a/2'. It gives, 

2Mg – 2T = 2M(a/2) = Ma

2T = 2Mg – Ma

T = Mg – Ma/2                               (1) 

Similarly net force in the direction of acceleration for block M is T – Mgcos60°  = T – Mg/2 and the acceleration is 'a'. It gives,

T – Mg/2 = Ma
T = Mg/2 + Ma                               (2)

Equating (A) and (B) we get,

Mg/2 + Ma = Mg – Ma/2
g/2 + a = g – a/2
a + a/2 = g – g/2
3a/2 = g/2

a = g/3 (Up along the plane as assumed initially)              

Q#33    
Find the mass M of the hanging block in figure (5-E16) which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction-less and the pulleys and string are light.

Answer:
Smaller block tends to slip down under the force that is the component of its weight along the plane = mgsinθ.
In order to prevent the block from slipping, this block along with the triangular block needs to be accelerated by the system so that the component of pseudo force on the block balances the component of weight which is trying to move the block. Let the required acceleration of the triangular block be 'a'. In order to apply Newton's Laws of Motion with respect to the non-inertial frame of the triangular block we add a pseudo force 'ma' on the smaller block opposite to the direction of 'a'. Component of this force along the plane is 'macosθ'. See the diagram below. (Normal force on the block has zero component along the plane hence not shown in the diagram):

So, in order to prevent the smaller block from slipping,

macosθ = mgsinθ  

a = gtanθ                                          (1)

Let the tension in the string be T. Downward force on the hanging block M is 'Mg – T' and its acceleration is 'a'. It gives, 

Mg – T = Ma
T = Mg – Ma                                   (2)

Consider the forces on the triangular block along the acceleration 'a'. Only horizontal force is T which is causing the movement of both blocks having mass M' and m. It gives,

T = (M' + m)a                                  (3) 

Equating (2) and (3).

Mg – Ma = (M' + m)a

M = (M' + m)a/(g – a)                   {Put the value of 'a' from (1)}

M = (M' + m)gtanθ/(g – gtanθ) 

M = (M' + m)tanθ/(1 – tanθ)     

 (Divide numerator and denominator by tanθ)

M = (M' + m)/(cotθ – 1)

It is the required mass of the hanging block to prevent the smaller block from slipping over the triangular block.

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