Class 11-science H C VERMA Solutions Physics Chapter 6 - Friction

Vertical equilibrium

N=mg

For horizontal; ma

Net force = max acceleration

Now,  

s=50m

Since no driving force is present to move the block. So, frictional force will be zero.

Now,

From (ii) and (iii)

0.11

Along inclined plane

Now,

s=10 m

a)

Applied force must be greater than net force which is acting downwards to make to move up.

F_req = µN + mgsin30° 

N = mgcos30° 

M = 2kg , g = 9.8m/s² , μ = 0.2

On substituting, F_req = 13N.

b)

Net force acting down the incline is given by

F_net = 2gsin30° - μN 

= 2×9.8×1/2 - (0.2)[mgcos30°]

= 9.8 - 0.2[2×9.8×√3/2]

= 6.41N

6.41 is the force acting down the inclined plane.

This is enough for the body to slide down. No need to exert extra force. So, force required is zero.

Since the block is just to move up the incline so frictional force will act in downward direction.

Since, block is in equilibrium.

 ………(perpendicular to incline)

 ……..(along the incline)

Now,

F=17.5 N

 ……(perpendicular to incline)

 ….(along the incline)

In the first half metre,

u=0m/s, s=0.5m, t=0.5s

v=u+at

v= 0+(0.5×4) = 2m/s

s=ut+1/2at²

0.5=0+12(a)(0.5)²

a=4m/s²

For the next half metre, u=2m/s , a=4m/s², s=0.5m

0.5 = 2t+(1/2)(4)t²

4t²+4t-1=0

On solving, t=0.2027 sec

Time taken to cover the next half metre is 0.21s.

Angle of friction,  

The value of friction force depends upon external force applied. If the body does not move then. 

When body is about to move or moves then ff== μN 

So,

ff<=μN 

λ≤tan^-1(μ)

For mass 0.5 Kg

for 1 Kg night mass

For 1 Kg left mass

Solving (i), (ii) and (iii)

For 2Kg block

For 4Kg block

Adding (i), (ii) and (iii)

15g-g-5g=15a+5a+50

9g=25a

Now, put in eq.(i)

T= 96 N

From eq.(3)

=68N 

Now, by translatory motion equation

Squaring

On solving,

(a) frictional force exerted by ground will be in forward direction.

sec

(b) While stopping, frictional force will act in opposite direction of motion.

After covering 50m, the velocity of athelete is

 Now,

t=3.33 sec 

When driver applies brakes frictional force will be opposite to motion of car.

s=12.8

We know that,

  Kmph

v=36 Kmph 

The maximum acceleration for car is given by,

t=10 sec 

P is the contact force.

N

for 4Kg block

For 2Kg block

On solving (i) and (ii)

Both blocks are connected by rod.

So, there acceleration must be same.

 -----(1)

 ------(2)

 ------(3)

 -------(4)

Add (1) and (3)

Put in equation (1)

T=0  

Let minimum force required be P at an angle θ.

 (vertical equilibrium) -(i) 

 (Body is about to move) -(ii)

From (i)

 -(iii)

To minimize P, we have to maximize  

Maximum value of  

Put in (3)

So,  at  

For man

N+T=Mg

For board

 (vertical equilibrium)

 (Horizontal equilibrium)

It is maximum force exerted by man. 

(a)

Normal contact for 2Kg = 2g =20 N.

Maximum friction force between 2Kg and 4Kg=  

= 4 N

Since F=12 N >ff bond will break

for 2Kg block,

12-4=2a

8=2a

for 4Kg block,

4-0=4 

(b)

Individual acc.

for 4Kg block,

12-4=4a

for 2Kg block,

4-0=2 

So, acceleration of both blocks are. 

For 2Kg;

For 3Kg;

For 7Kg;

(a)

The bond between 2Kg and 3Kg breaks as applied force is more than limiting friction.

Acceleration of 2Kg block  

Bond between 3Kg and 7Kg will not be broken by 4N force so both will move together.

Acceleration of 3Kg and 7Kg block  

(b) The bond between 3Kg and 7Kg will not be broken so they will definitely move together.

Now individual acceleration

For 2Kg;

For 3Kg and 7Kg;

Acceleration of 2Kg block cannot be greater than 3Kg block.

So, all will move together.

So,  

(c) The bond between 3Kg and 7Kg will not be broken as applied force (10N) is less than . So, both of them moves together.

Now, individual acceleration

For 2Kg;

For 3Kg and 7Kg;

Acceleration of 2Kg block cannot be greater than 3Kg block.

So, all will move together.

So,  .

(a)For block m;

N=mg

ff=μN=μmg 

F=T + ff (∴ acceleration = μ) -(i)

For block M;

T=ff (∴ acceleration = 0) -(ii)

Solving (i)and(ii),

F=ff + ft

F=2μmg

(b) Now applied force is F=2(2μmg)=4μmg

For block m;

F-T-ff=ma -(i)

For block M;

T-ff=Ma -(ii)

Adding (i)and(ii),

F-2ff =(m + M)a

4μmg-2μmg = (m + M)a

 in opposite directions. 

In elevator, moving down with acceleration a, effective acceleration will be

So, replace g by  in above answers.

For block m,

QE+N=mg (vertical equilibrium)

N=mg-QE

F=T + ff (Horizontal equilibrium) -(i)

For block M,

T=ff (Horizontal Equilibrium) -(ii)

Solving (i)and(ii),

F=ff + ft

F=2ff = 2μ(mg-QE)

When block slips, the limiting friction force acts.

Since the table remains at rest.

When a block M, moves with acceleration a towards right block m moves downwards and rightwards with acceleration 2a and a respectively.

Drawing FBD of M mass

 (Vertical) -(i)

 (Horizontal) -(ii)

FBD of mass m

N=ma (Horizontal) -(iii)

 (vertical) -(iv)

From (iii) and (iv)

 -(v)

Solving equation (i), (ii), (iii) and (iv)

Net driving force on the block

Limiting friction force= 

 block will move.

For acceleration,

Direction w.r.t 15N force

with 15N force.

(a) Since the man is at rest horizontally. So, both by both walls on man will be equal and opposite in direction.

(b) In vertical equilibrium condition

N=250 N

Initial velocity of both blocks is same. So,  

When m separates from M.  

Let and be the acceleration of the blocks m and M respectively with respect to ground.

Here, > as both blocks separates from each other.

So,  

 -(i)

Now,

FBD of mass m,

 -(ii)

FBD of mass M

 -(iii)

Subtract (ii) and (iii)

Put in e.q. (i)