Solutions to Concepts of Physics by HC Verma Chapter 6

6 1.

Friction A body slipping on a rough horizontal plane moves with a deceleration of the coefficient of kinectic friction between the block and the plane ? Let be the frictional force, N the normal normal force and friction. Then If

. What is

be the coefficient of kinetic

be the deceleration then

⇒

2.

A block is projected along a rough horizontal road with a speed of coefficient of kinectic friction is 0.10, how far will it travel before it comes to rest ? If

be the frictional force then with usual notation

⇒ ⇒ ⇒ ⇒

3.

If the

x

A block of mass m is kept on a horizontal table If the static friction coefficient is μ, find the frictional force acting on the block. Since the block is kept on the horizontal table, it does not move and thus the acceleration and consequently net force acting on the block is zero. Since no force other than friction is acting on the block, the force of friction is also zero.

4.

A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two. Let be the mass of the block and the frictional force acting the block. Then net force acting on the block along the incline is contributed by frictional force and the gravitational force only as normal force, being perpendicular to the incline, has zero component along the incline. ∴

f

N

mg sin 30°

mg 30°

To find N let us consider forces acting on the block perpendicular to the incline. These include the normal force N and the component of gravitational force mg parallel to N, i.e. cos( – ) But there is no motion and hence no acceleration perpendicular to the incline and therefore the net force parallel to N is zero. Hence . This implies cos If

( ) be the acceleration along the incline then

Also from the given data;

,

6

Friction ⇒ ⇒

( )

⇒

5.

Suppose the block of previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest ? The mass of the block is 4 kg. In this case cos (

)

⇒ ⇒

6.

x

A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2. (a)

When the block moves up, the direction of the frictional force is downward. Therefore the required minimum force F is given by cos

(

x x

)

( ) (b) Let be F the net force acting on the block. Then cos (

(

x x

)

)

As the net downward force is greater than zero, the block will slip down the incline with acceleration without any additional force and therefore the required force is zero.

7.

Repeat part (a) of the problem 6 if the push is applied horizontally and not parallel to incline. If F is the horizontal force applied on the block such that the block moves up, then cos ⇒

8.

cos

cos cos

sin sin

In a children park an inclined plane is constructed with an angle of incline at 45° in the middle part. Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and . The forces acting on the boy are the gravitational force and the friction. If force acting on boy then

is the net

6

Friction sin ⇒

9.

A boy starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter ? Given that

and

, the acceleration of the body is given by

i.e. If t is the time taken to move next half meter, then body moves 1 m in t+0.5 s after starting from rest. ∴ ⇒

10. The angle between the resultant contact force and the normal force exerted by a body on the other body is called the angle of friction Show that, if λ be the angle of friction and μ the coefficient of static friction, . From the figure we have ⇒ ⇒

N

R λ

tan

f

11. Consider the situation shown in the figure. Calculate (a) the acceleration of the 1 kg blocks, (b) the tension in the string connecting the 1 kg blocks and (c) the tension in the string attached to the 0.5 kg block. (a)

Let the tension in the string between the 1 kg blocks be T' and that in other string be T. Let the acceleration of the blocks be . Then the motion of 0.5 kg block gives

N

N' 1 kg

f' μ

T'

1 kg

f μ

T

T

…I 0.5 Let the kinetic frictional force acting on kg the right and the left 1 kg blocks be and respectively. The motion of right and left 1 kg blocks give … II and … III But denoting the normal forces acting on the right and left 1kg blocks with N and N' respectively we have … IV and …V Solving for gives …VI (b) Equations III, V and VI of part (a) gives

6

Friction (c)

Again from I and VI of part (a)

12. If the tension in the string in figure is 16 N and the acceleration of each block is find the friction coefficient at the two contacts with the blocks. From the motion of 2 kg block we have – ⇒ From the motion of 4 kg block we have cos ⇒

T=16N

,

4 kg

μ2

2 kg

30°

μ1

13. The friction coefficient between the table and the block shown figure is 0.2. Find the tensions in the two strings. Let T and T' be the tensions in the two strings T and a be the acceleration towards left. Then by 5 kg motion of 15 kg block we get – …I Similarly by motion of 5 kg hanging block, we 15 kg get – … II By the motion of the 5 kg block on the table we have … III Where f is the kinetic friction acting on the 5 kg block and is given by … IV By multiplying I with 2/3 and adding it to the sum of II and III we get

T'

5 kg

⇒ and

14. The friction coefficient between the road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped in 5 m. The given data is

To find acceleration, we put values of ,

which gives us

Let θ be the angle of incline, the mass of the vehicle. The net force acting the vehicle, is provided by the gravitational force and the force of friction. ∴ sin sin cos ⇒ cos or sin …I To find the value of θ, let us define angle ϕ such that cos

6

Friction ⇒

√

sin

( )

∴ from equation I we have sin cos sin cos cos or sin cos sin ( ) ⇒

cos

15. The friction coefficient between an athlete's shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time that he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop ? (a)

Given that . Let m be the mass of the athlete. The force for the motion of the athlete is provided by the friction between the athlete's shoes and the ground and is given by where N is normal force exerted by ground on the athlete and therefore by Newton's third law is mg. ∴ x Since acceleration is constant, using we have

or ⇒ s and . Since the acceleration (b) For stopping in the minimum time is constant, for the motion of the athlete covering next 50 meters, we have ( ) Using we get t ⇒ s

16. A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30°. The friction coefficient between the road and the tyre is Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take . The car is moving with velocity 21.6 km/hr along horizontal. When it encounters the slope the initial velocity along the slope will be ( ) The acceleration is provided by the gravitational force and the force of friction between the tyre and the road. Both these forces act along the slope. The hardest brake will produce maximum friction force. ∴ sin sin ⇒

( )

Also

. To find √

we use ms

(

)

kph

6

Friction

17. A car starts from rest on a half km long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one can not drive through the bridge in less than 10 s. The given data is , , The maximum frictional force is where m is the mass of the car. Since the only force on the car is the force of friction, we have The minimum time is the time taken to move the length of the bridge with above acceleration. ∴ ⇒ t = 10 s

18. Figure shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is and that between the block of mass 4.0 kg and the incline is . Calculate the acceleration of the 2.0 kg block if (a) , (b) . (a)

Since the coefficient of friction of the 4 kg block with the incline is less than that of the 2 kg 2 kg block with the incline, both blocks will 4 kg move together. The forces on the 2 kg block along the incline are downward gravitational 30° force, the contact F due to its contact with 4 kg block and upward frictional force. If is the net force acting on 2 kg block and is its acceleration, then cos …I Also the forces on the 4 kg block along the incline are downward gravitational force, the contact force F due to its contact with the 2 kg block and the upward frictional force. Since the blocks are supposedly moving together both move with same acceleration. If is the net force on 4 kg block, then cos … II Adding I and II sin cos (

)

⇒ (b) Since the coefficient of friction of the 4 kg block with the incline is greater than that of the 2 kg block with the incline, they instantly separate after starting off and therefore there is no force exerted by 4 kg block on the 2 kg block. The forces on the 2 kg block along the incline are downward gravitational force and upward frictional force. If is the net force acting on 2 kg block and is its acceleration, then cos ⇒ sin cos ( (

) )

19. The masses are connected with a light rod and the system is slipping down a rough incline of angle θ with the horizontal The friction coefficient at both the contacts is μ Find the acceleration of the system and the force by the rod on one of the blocks

6

Friction Let T be the tension in the rod, the frictional force acting on and the acceleration of the blocks. The sin or cos …I or

M2

M1 θ

cos

… II Adding I and II ⇒ and

T

sin

sin sin

cos

cos cos

cos

sin

20. A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is μ The block is to be pulled by applying a force to it What minimum force is needed to slide the block ? In which direction should this force act ? Let the force , applied to the block at an angle θ with the horizontal, be the minimum force required to move the block. Let N be the normal force acting on the block and be the frictional force. Then as a limiting case the normal force is given by sin …I and … II Now the net force acting on the block along the horizontal is zero as long as the block does not accelerate. ∴ cos sin ⇒ cos sin … III To find minimum value of F, we equate to zero. i.e. cos sin sin cos ⇒ tan … IV Putting this value of μ in III, we get cos √

The direction of the minimum applied force is θ such that

21. The friction coefficient between the board and the floor shown in the figure is μ Find the maximum force that the man can exert so that the board does not slip on the floor. Let T be the tension in the string consequent to the force applied by the man. Let us consider the motion of the board. The forces acting on the board are the gravitational force due to the masses of the man and the board, the tension T in the string attached to the board and the frictional force. Since the board does not slip we have ⇒ ⇒

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 the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take .

6

Friction (a)

Let be accelerations of the 2 kg and 4 kg blocks. From the motion of the 2 kg block we have ⇒ From the motion of the 4 kg block we have

⇒ (b) In this case if be accelerations of the 2 kg and 4 kg blocks, we have from the motion of the 2 kg block

2 kg 4 kg

⇒ From the motion of the 4 kg block we get

23. Find the accelerations , , of the three blocks shown in figure if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take . (a)

Let , be the friction forces between 2 and 3 kg blocks and the 3 and 7 kg blocks. Let the normal force on the 2 kg and 3 kg blocks be and Consider the motion of 2 kg block.

2 kg 3 kg 7 kg

But ⇒ The only horizontal forces acting on the 3 kg block are the frictional force due to motion of 2 kg block and frictional force . But maximum value of and only a force greater than this can accelerate the 3 kg block. Thus the force can not move the 3 kg block relative to 7 kg block. But this force also acts on the combined mass of 3 kg and 7 kg blocks and since there is no friction between the 7 kg block and the floor, both these blocks will move together. ⇒ (b) When the 10 N force acts on the 3 kg block, the maximum frictional force that can arise from the 2 kg and 7 kg blocks is (4+15 =) 19 N and thus the 10 N force applied to it can not move this block relative to these blocks. But the 10 N force when acts on the three blocks with mass 12 kg they will move together as there is no friction between the 7 kg block and the floor. ⇒ (c) When 10 N force is applied to the 7 kg block, the maximum frictional force that 2 and 3 kg blocks can apply to it is . Thus it does not move relative to the 2 kg and 3 kg blocks. But there being no friction between the floor and the 7 kg block, all three blocks will move together and

24. The friction coefficient between the two blocks shown in the figure 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.

6

Friction (a)

Let T be the tension in the string. Then the equilibrium of the block M gives

F

m M

⇒ where is the normal force exerted by M on m. Similarly equilibrium of block m gives (b) In this case F μ mg Let be the acceleration of the block M. Clearly acceleration of the block m is – Then ⇒ and ⇒ ⇒ and

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

Let T be the tension in the string. Then the equilibrium of the block M gives ⇒ where is the normal force exerted by M on m. Similarly equilibrium of block m gives

(b) In this case F μ m g-a). Let acceleration of the block m is –

be the acceleration of the block M. Clearly Then

⇒ and ⇒ ⇒ and

26. Consider the situation shown in figure. 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 floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium ? The force due to charge/field is equilibrium of the block M gives

. Let T be the tension in the string. Then the

⇒ where is the normal force exerted by M on m. Similarly equilibrium of block m gives

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

6

Friction the table ? Since the table does not move the frictional force applied by the floor on the legs of the table is equal to the kinetic frictional force applied by the block on the table. But You do not need the coefficient of friction between the table and the floor or the mass of the table to calculate the frictional force applied by the floor on the legs of the table.

28. Find the acceleration of the block of mass M in the situation of the figure. The coefficient of friction between the two blocks is and that between the bigger block and the ground is . Let T be the tension in the string and be the acceleration of the block M. Then block moves with acceleration downward and to the T right. Let be the frictional force between the M m blocks and be the normal force exerted by bigger block on the smaller block. is parallel to the floor and is due to the motion of the bigger block towards right. Thus …I ⇒ … II Let be the frictional force between the bigger block and the floor and be the normal force exerted by the floor on the bigger block. But is contributed by the gravitational force on block M, the tension T pulling down on the clamp attached to M and the downward frictional force exerted by the motion of blokc m. Thus … III … IV The downward motion of m gives ⇒ …V The rightward motion of the block M gives Substituting values of [

… VI from IV and I and then value of T from V we get

,

] (

)

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. There will be no motion perpendicular to wall, since by Newton's third law, wall will exert equal and opposite force on the block. The resultant of the 15 N force and the gravitational force acting on the block is Thus without frictional force the block will move in the direction tan and its magnitude is the opposite direction. Its magnitude will be

tan

. The frictional force will act in μ

N

6

Friction Thus the block will move with an acceleration horizontal.

at angle 53° downwards with the

30. A person (40 kg) is managing to be at rest between two walls by pressing against wall A by his hands and feet and wall B by his back. Assume that the static friction coefficient between the walls and the body is 0.8 and the 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 the either wall on the person. Take . (a)

Let be the normal forces exerted by the walls A and B on the body. Since limiting friction acts at all contact points, force ⇒ [Q.E.D] (b) Since the gravitational force on the person is balanced by the frictional force by the walls on the body, we have

31. The figure shows a small block of mass kept at the left end of a larger block of mass M and length . The system can slide on a horizontal road. The system is started towards right with an initial velocity The friction coefficient between the road and the bigger block is μ and that between the blocks is . Find the time elapsed before the smaller block separates from the bigger block. The system starts with velocity . Let be the frictional force acting on smaller block. Then ( ) ( ) If is the frictional force acting on the bigger block from the road, then The acceleration of the block m relative to block M is ( )( ) The relative initial velocity of the blocks is . The smaller block on which smaller frictional force acts will move faster and separate from the bigger block when its left end moves ahead of the right end of the bigger block. If is the time elapsed before the blocks separate, the length will be given by ⇒

√

√ ∎