Solutions to Problems on "Friction"-'H C Verma's Concepts of Physics, Part-I, Chapter-6', EXERCISES,Q 21 -31.

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EXERCISES (21-31)

21. The friction coefficient between the board and the floor shown in figure (6-E7) is µ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Figure for problem number 21

Answer: Let the maximum force be P. At the moment when P is applied by the man his effective weight on the board reduces to Mg-P. Total force applied on the floor is equal to weight of the man plus weight of the board, i.e.

=Mg-P+mg   =Normal force by the plane on the board.

In this case maximum friction force =µ(Mg-P+mg)

This maximum friction force is equal to horizontal pull by the string which in turn is equal to P, equating these two,

µ(Mg-P+mg)=P

→µP+P=µ(M+m)g

→P= µ(M+m)g/(1+µ)

22. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of two blocks if a horizontal force of 12 N is applied to (a) the upper block , (b) the lower block. Take g=10 m/s². 

Answer:  (a) When the force of 12 N is applied to the upper 2 kg block, See the forces on the blocks in the diagrams below:--

Figure for problem no-22

Consider the upper block as in figure-1 above.

Weight of 2 kg block =2g  N=2x10  N =20 N = R, Normal force on the block by the lower block. Given µ=0.20,

Force of friction = µR = 0.20*20  N = 4 N

This force of friction will act against the motion which is in the direction of 12 N force. So net force on the block =12-4  N =8 N

If acceleration of this block be 'a' then

a=Force/mass=8 N/2 kg =4 m/s² 

Consider the lower block as in figure-2 above.  

Since the block is placed on a smooth surface so friction force from the lower surface is zero. But the friction force by the upper block surface on it will be same as the friction force applied by the lower block on the upper block but opposite in direction according to the Newton's Third Law. So the friction force 4 N will be along the 12 N force. Only force in the horizontal direction for the lower block is this 4 N which will induce acceleration to it. Let the acceleration be a'. 

a' =Force/mass = 4 N/4 kg = 1 m/s².    

(b) When the force of 12 N is applied to the lower 4 kg block, See the forces on the blocks in the diagrams below:--

Figure for Problem-22

Consider the lower block as in figure-3 above.
Lower surface is smooth so no friction force by the lower surface but for the upper surface the weight of 2g  N of upper block is normal force so friction force at this surface=µ.2g =0.20*2*10  N  
= 4 N  and its direction will be opposite to 12 N applied force. 
So acceleration of lower block =a' = Force/mass  
= (12-4) N/4 kg = 8  N/4  kg = 2 m/s²   

Consider the upper block as in figure-4 above. 

The weight of 2g N is balanced by the normal force of equal and opposite magnitude. There is no applied force on it but the force of friction applied to the lower block will have equal and opposite reaction, so the friction force of 4 N will act in the direction of 12 N force and this 4 N force of friction is the only horizontal force on the upper block that will induce acceleration to it.                   

 So the acceleration of upper block a= 4 N/2 kg =2 m/ s².      

23. Find the accelerations a₁, a2 , a3 of the three blocks shown in figure (6-E8). If a horizontal force of 10N is applied on (a) 2 kg block (b) 3 kg block , (c) 7 kg block. Take g= 10 m/s².

Figure for problem number 23

Answer:  (a) When a horizontal force of 10 N is applied to 2 kg block  

Normal force on 2 kg block = weight of 2 kg block = 2g  N =20 N 

Friction force on 2 kg block = µ₁*20 N =0.2*20 N= 4 N 

Net horizontal force on 2 kg block =10-4 N =6 N  

Acceleration a₁ =6 N/ 2 kg = 3 m/s²   

3 kg block 

Upper surface of this block will have friction force same as above but opposite in direction i.e. 4 N along applied force of 10 N.  

Normal force on the lower surface = Total weight of both blocks = 5g N = 50 N 

Friction force =µ₂*50 N =0.3*50 N =15 N but it can not be more than the 4 N. It means that the 3 kg and 7 kg blocks do not move with respect to each other.        

To calculate the acceleration of these blocks we consider both as if attached. Total mass=3 kg+7 kg = 10 kg. Total downward force on the smooth surface due to weights = 10g N+2g N =12g N =120 N =Normal force on the lower surface of 7 kg block. 

Friction force on this surface = µ₃*120 N =0*120 N = 0 

Now the net external force on this combined block is the friction force applied by the top most block = 4 N.

Acceleration of combined block = Force/mass

=4 N/10 kg = 0.40 m/s²

Hence a₂₌a₃ = 0.40 m/s²  

(b) When a horizontal force of 10 N is applied to 3 kg block  

Friction forces opposing the movement are on both surfaces of the 3 kg block. Normal force on upper surface = 2g N 

Limiting Friction force on upper surface = µ₁*2g  N =0.2*2*10 N =4 N

Normal force on the lower surface = 2g+3g =5g N =50 N

Friction force on lower surface = µ₂ *5g  N =0.3*50 N =15 N but it is the limiting value ; in this case it can not be more than applied force of 10 N meaning thereby that the lower two blocks do not move with respect to each other but move combined on the lower most surface. So consider the lower two blocks combined. As far as upper 2 kg block is concerned we do not know whether the limiting value of friction force 4 N is attained or not. So let us proceed with assumption that it is attained, then,

Mass of combined block = 7 kg+3 kg =10 kg

Total normal force on the block by the lower most surface =(2+10)g N =12*10 N =120 N 

Friction force on the block =µ_3* 120 N = 0*120 N = 0 N

So net horizontal force on this combined block = 10 N-4 N =6 N

Acceleration of this combined block = 6 N/10 kg = 0.6 m/s²

on 2 kg Block 

Only horizontal force on this block is friction force = 4 N     

Acceleration of 2 kg upper block a₁= 4 N/2 kg =2 m/s² which is greater than the acceleration of lower combined blocks 0.6 m/s² that is not possible. So our assumption that limiting friction force of 4 N is attained between 2 kg and 3 kg blocks is not true. Since this friction force is less than limiting, it means that 2 kg block do not move with respect to 3 kg block as a result all the three blocks move combined. Total mass of three combined blocks =2+3+7 kg =12 kg, Force applied = 10 N, 

Hence acceleration = 10 N/12 kg =5/6 m/s² 

So a₁ =a₂ =a₃ = 5/6 m/s²         

(c) When a horizontal force of 10 N is applied to 7 kg block    

Since the friction coefficient on lower most surface =µ₃ =0, means there is no friction force on the lower surface of the 7 kg block. On the upper surface of the 7 kg block normal force = 5g N  

Limiting Friction force=µ₂*5g  N =0.30*50 N =15 N but it can not be more than the applied force, so there is no relative movement between 3 kg and 7 kg block and they move combined. Normal force on the upper surface of 3 kg block = 2g N =20 N  

Limiting Friction force at this surface = µ₁*20 N =0.2*20 N =4 N

Same as in the above case (b) we can prove that this limiting friction force is not achieved and thus even the 2 kg block move together with the 3 kg and 7 kg block. Similarly combined acceleration of the three blocks = Total external force/total mass =10N/12kg =5/6 m/s² 

Hence accelerations a₁ =a₂ =a₃ = 5/6 m/s²                 

24. The friction coefficient between the two blocks shown in figure(6-E9) is µ but the floor is smooth (a) what maximum horizontal force F can be applied without disturbing the equilibrium of the system? (b) suppose the horizontal force applied is double of that found in part(a) , find the acceleration of the two masses.

Figure for problem 24

Answer:   (a)At maximum F friction force between the two blocks will be limiting= P =µR , where R is normal force.  

Since R=mg , P=µmg, Its direction will be opposite to F on block 'm'. 

Let the tension in the string around the pulley be T, See the Free body diagram of upper block below, Figure -1,

Diagrams for problem-24

Since the block is in equilibrium, sum of horizontal forces =0,

i.e. F-P-T=0

→F=P+T  Newton. ------------------------------- (a)  

Consider the lower block. Friction force by the upper block =P=µmg but opposite in direction that of on upper block. Lower surface is smooth, so no friction force by it. Normal force by the surface =(M+m)g  N. Horizontal force by string = T as shown in the Free body diagram Figure - 2. At equilibrium sum of all horizontal forces are zero,

i.e. T-P=0

→T=P=µmg  

Putting this value in ---(a) 

F=P+T=2P=2µmg  

(b) When the force applied is doubled,F=4µmg, limiting friction is crossed and there is net horizontal force causing acceleration. In this case accelerations of the two blocks 'a' are equal.   

So for the upper block, F-P-T=ma  ----------(b) 

And for the lower block, T-P=Ma 

T=P+Ma    

Putting this in (b) 

F-P-P-Ma=ma 

→(M+m)a=F-2P = 4µmg-2µmg =2µmg  

→a=2µmg/(M+m) 

25. Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).

Answer: Since the elevator is coming down with an acceleration 'a' the weights of the blocks will reduce to m(g-a) and M(g-a) and accordingly the Normal forces will change. So we can follow the same process as in problem number 24, and get answers with just a replacement of 'g' by 'g-a' as follows:- 

(a) F=P+T=2P=2µm(g-a)  

(b) a=2µm(g-a)/(M+m) 

26. Consider the situation shown in the figure (6-E9). Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is µ but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium?

[Hint: The force on a charge Q by the electric field E is F=QE in the direction of E.]

Answer: an additional force =QE in the upper direction on the upper block is applied due to electric field E, see the figures below, 

Figure for problem 26

For the equilibrium in vertical direction, 

R+QE=mg → R=mg-QE 

Friction force on upper block, P=µR =µ(mg-QE)  

For the equilibrium in the horizontal direction F=T+P  ............(i)  

Consider the equilibrium of lower block in vertical direction,

R=Mg+mg-QE, since the surface is smooth there is no friction on the lower surface of lower block, but the upper surface of the lower block will have friction force P by the upper block in the opposite direction. So for the equilibrium in the horizontal direction, 

P=T=µ(mg-QE), putting this value in ...(i) we get,  

F=T+P=2P=2µ(mg-QE)        

27. A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is µ. The table does not move on the floor . Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table?

Answer: Since the block moves on the table, the limiting friction has been achieved and the friction force on the block is µmg. Equal and opposite friction force will be applied by the block on the table =µmg, Since the table does not move on the floor, the limiting friction between the legs of the table and the floor has not been achieved, let this friction force be F. F will resist the force applied by the block on the table hence both will have opposite directions. Since the table is in equilibrium, sum of horizontal forces = zero, 

i.e. F-µmg=0 

→ F=µmg 

We do not need the friction coefficient between the table and the floor or the mass of the table as is obvious from the above solution.

28. Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is µ₁ and that between the bigger block and the ground is µ₂.

Figure for problem 28

Answer: Let the acceleration of block M be 'a' and the tension in the string be T. The smaller block will have acceleration 'a' towards right and acceleration '2a' downwards. The Normal force on the block m =R = ma (See FBD for smaller block in figure below),

Diagram for problem no-28

 Friction force on block m by the block M =µ₁R = µ₁.ma, in the vertical upward direction. Considering the forces in the vertical direction on block m. 

mg-T- µ₁.ma =m.2a 

→ T= mg- µ₁.ma -2ma

Now consider the vertical forces on the larger block M, See FBD.  

Downward forces are weight Mg and friction by smaller block m = µ₁.ma and Tension T at the pulley. So total downward force =Mg+µ₁.ma+T=Normal force by the ground R' 

So friction force =  µ₂(Mg+µ₁.ma+T)    its direction will be opposite to 'a'. Now consider the horizontal forces on the larger block M

2T-R-µ₂(Mg+µ₁.ma+T)= Ma 

→ 2mg-µ₁.2ma-4ma-ma-µ₂(Mg+µ₁.ma+mg- µ₁.ma -2ma) =Ma   (putting the value of T)

→ 2mg-µ₂Mg -µ_2.mg+ µ_2* 2ma  =Ma+5ma+ µ₁.2ma

→ [2m-µ₂₍M+m)]g=Ma+5ma+µ₁.2ma -µ₂.2ma   

→ [2m-µ₂₍M+m)]g=[M+m{5+2(µ₁ -µ₂)}]a 

 →a =  [2m-µ₂₍M+m)]g/[M+m{5+2(µ₁ -µ₂)}]

   

29. A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction? If no, find the frictional force exerted by the wall on the block.

Answer: Weight of the block=2g N=20 N. 

Normal force on the block by the wall when pushed by 40 N force =40 N, 

Figure for Q-29
(Please visualize in 3D)

Limiting static Friction force on the block by the wall =µR=0.5*40 N=20 N. So the weight of the block is just balanced by the friction force and block does not move if only the 40 N force is acting. When a horizontal force of 15 N parallel to the wall is applied, resultant of forces trying to move the block parallel to the wall will be resultant of weight 20 N and horizontal force 15 N both at right angles =√(20²+15²) =√625 =25 N which is more than the limiting static friction force. So the block will move and move in the direction of this resultant force. Let x be the angle between resultant and 15 N force.  

tan x =20/15 =4/3 →x=53° from the 15 N force.    

30. A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back (figure 6-E11). Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s².

Figure for problem 30

Answer: (a)The man pushes both walls, in turn the walls also push him in equal and opposite directions.Considering the horizontal forces in horizontal direction on the man. Only the normal forces by the walls on the man will act. Let normal force by walls be R and R'. Then R=R' at equilibrium. Hence the man pushes both walls with equal forces and the friction forces on both walls are also equal say F. 

(b) Weight =40g  N =400 N. 

In vertical direction 2F=400 → F=200 N     

Also F=µR  

→200=0.8 R  

→R=200/0.8 N =250 N     

31.   Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is µ and that between the blocks is µ/2. Find the time elapsed before the smaller blocks separates from the bigger block.

Figure for problem number 31

Answer:   Since the friction between the blocks is half than the friction between road and the larger block, the smaller block will slide over the larger. Normal force on the smaller block=weight of the block = mg.

Friction force = µ/2*mg =µmg/2

Retardation of the block = Friction force/mass =µg/2

See diagram below,

Diagram for problem number 31, Friction-Chapter -6

 Consider the forces on the larger block. Weights mg and Mg push the road, in turn the road applies a normal force equal and opposite in direction to the block.  

So the normal force = (m+M)g   

Since the system has been given a velocity and no other driving force acts, hence the friction force between larger block and the road will act against the motion to stop it.

This friction force =  µ(m+M)g   

Another friction force will act on the top surface applied by smaller block in the direction of motion = µmg/2  

Total force against the motion = µ(m+M)g -µmg/2 =µmg/2+µMg  =µg(m/2+M)  

Retardation of larger block = µg(m/2+M)/M 

Difference in retardation of both blocks 'a'

=  µg(m/2+M)/M-µg/2 

=  µg{(m/2+M)/M-1/2} 

=  µg{(m+2M)/2M-1/2} 

= ½µg{(m+2M)/M-1} 

= ½µg{m+2M-M/M}  

= ½µg{(m+M)/M} 

= µg(m+M)/2M 

= Relative acceleration of the smaller block with respect to larger block.  

As soon as the system is left after giving a velocity towards right, the smaller block starts to slip over the larger block. Here we consider the relative velocities of the smaller block w.r.t. larger. So initial velocity of smaller block, u=0, distance to be traveled = l, acceleration a= µg(m+M)/2M, to get the time taken to travel this distance, t,  we use the relation s=ut+½at².

Here, l=0+½* µg(m+M)/2M *t²  

t²*µg(m+M)=4lM

t²= 4lM/µg(m+M)  

t=√{4lM/µg(m+M)}   

So after this time the smaller block will separate from larger block.              

===<<O>>===

For more practice on problems on friction solve these "New Questions on Friction" .

Links for the chapters -

HC Verma's Concepts of Physics, Chapter-7, Circular Motion,

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HC Verma's Concepts of Physics, Chapter-6, Friction,

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

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HC Verma's Concepts of Physics, Chapter-4, The Forces

The Forces-

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HC Verma's Concepts of Physics, Chapter-3, Kinematics-Rest and Motion:---

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Chapter -2, "Vector related Problems"

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