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Chapter Notes: Friction Physics Class 11


Notes for Friction chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

 

If we push a book on a horizontal table with a certain initial velocity, it eventually comes to rest. This shows that there is some force opposing the motion of the book on the table. This is called the ‘frictional force’. It arises due to interaction between the molecules of the book and of the table.
Actually, whenever the surface of a body slides over the surface of another body, each body exerts a frictional force on the other which is parallel to the surface in contact. The frictional force on each body is opposite to the direction of its motion relative to the other. Thus, when a book slides along the surface of a table from left to right, the frictional force acting on the book is directed to the left while an equal frictional force acts on the table directed to the right. Frictional force may also exist between surfaces even before the relative motion starts.

Static Friction

Let us consider a block at rest on a horizontal table surface. The block is acted upon by two forces: its weight mg acting vertically downward at its centre of gravity, and the reactionary-force N exerted on it by the table which is directed vertically upward and passes through its centre of gravity. Since the block is in equilibrium, N = mg. In the Figure the lines of action of mg and N are shown slightly displaced for clarity.
When we apply a small horizontal force F, say towards right, the block does not move. The force R exerted on the block by the table surface is now so inclined (figure (a)) toward left that R, mg and F may form a ‘closed’ triangle (since the block is still in equilibrium). Figure (b).
The force R can be resolved into two components; parallel and perpendicular to the contact-surfaces. The component parallel to the contact surface is called the ‘force of static frictionf, which balances the applied force F (F = f ). The component perpendicular to the contact surfaces is the ‘normal reactionN exerted on the block which balances the weight mg of the block (N = mg). Figure (c).
Now, if the applied force F is slightly increased, the block still does not begin to move. This means that the force R is further inclined towards left so that the force of static friction f also increases to become equal to the new value of F. Thus, as the applied force F is increased, the force of static friction f also increases, but after a certain limit f cannot increase any more. At this moment the block is just about to move.
This maximum value of the static frictional force {f_s} is called ‘limiting frictional force’ (it is equal to the smallest force required to start motion). Now, as the applied force is further increased, the block begins to move.

Law of Limiting Friction

The limiting (maximum) static frictional force depends upon the nature of the surfaces in contact. It does not depend upon the size or area of the surfaces. For the given surfaces, the limiting frictional force {f_s} is directly proportional to the normal reaction N.
{f_s} \propto N
or {f_s} = {\rm{ }}{\mu _s}N (for limiting frictional force)
where the constant of proportionality ms is called the 'coefficient of static friction'. The above formula holds only when {f_s} has its maximum (limiting) value. Before this stage, force of friction {f_s} < {\mu _s}N.
If the direction of the applied force F is reversed, the direction of f also reverses, while the direction of N remains unchanged. In fact f is always opposite to F.

Angle of Friction

In the case of limiting friction, the angle which the resultant R of the limiting frictional force {f_s} and the normal reaction N makes with the normal reaction N is called the angle of friction. If this angle is {\theta _s}, then
tan{\theta _s} = {{{f_s}} \over N}
But {{{f_s}} \over N} = {\mu _s}

tan{\theta _s} = {\mu _s}
Thus, the coefficient of static friction is equal to the tangent of the angle of friction.

Kinetic Friction

Once the motion starts, the frictional force acting between the surfaces decreases, so that a smaller force F is required to maintain uniform motion. The force acting between the surfaces in relative motion is called the ‘kinetic frictional force{f_k} which is less than the limiting force of static friction {f_s}. We know from daily experiences that a lesser force is required to maintain the motion of a body than the force required to start the body from rest.
Thus, when the block is in uniform motion, the force of kinetic friction is
{f_k} = {\mu _k}N
where {\mu _k} is the coefficient of kinetic friction and its value is less than {\mu _s}.

Friction on an Inclined Plane
In the figure(a), a block is placed on a rough inclined plane making an angle \theta with the horizontal. Initially, the block is stationary. The F.B.D. of the block is shown in figure (b).
The weight mg can be resolved into two components, mg\,\sin \theta  parallel to the inclined plane and mg\,\cos \theta  perpendicular to it. Due to the parallel component mg\,\sin \theta , the block has a tendency to slide down the plane, but the static frictional force f exerted on the block by the plane is preventing it from sliding.
Since the block is in equilibrium, the net force on the block parallel to the plane and also perpendicular to the plane is zero. That is
f - mg\,\sin \theta = 0 and {\rm N} - mg\,\cos \theta = 0
On increasing the angle of inclination \theta , the component mg\,\sin \theta  increases, but simultaneously f also increases. When f increases to its limiting (maximum)) value, the block is just about  to slide. Suppose in this position the angle of inclination of the plane is \theta_s .
Then 
{f_s} = {\mu _s}N
Substituting this value of fs in the above expression, we get
{\mu _s}N = mgsin{\theta _s} and
{\rm N} = mg\,\cos \theta
Dividing, we get
{\mu _s} = {\rm{ }}tan{\theta _s}
Hence, measuring the angle of inclination at which the block just starts sliding, the coefficient of static friction can be determined.

 Application 1

A block of mass M = 10 kg is placed at rest on a horizontal surface as shown in the figure. The coefficient of friction between the block and the surface is µ = 0.3. It is pulled with a horizontal force  F.
Find the magnitude of the force of friction if
(i) F = 20 N
(ii)
F = 40 N

Solution

The maximum value of friction force is
{{f_{max}} = \mu N = {\rm{ }}\mu Mg}
or {{f_{max}} = {\rm{ }}\left( {0.3} \right){\rm{ }}\left( {10} \right){\rm{ }}\left( {10} \right){\rm{ }} = {\rm{ }}30{\rm{ }}N}
(i) To keep the block stationary the magnitude of friction force should be f = F = 20 N since F < fmax . Therefore the force of friction is f =20 N
(ii) To keep the block stationary the magnitude of friction force should be f = F = 40 N. Since f cannot be more than fmax = 30 N, therefore, the force of friction is f = fmax = 30 N

Note  that in this case friction force is unable to keep the block stationary and the block accelerators with
a = {{F\, - f} \over M}\, = {{40\, - \,30} \over {10}} = {\rm{ }}1{\rm{ }}m/{s^2}\,
Also note that friction force is not always equal to µN. It is the limiting or maximum value of friction. At any stage friction force may attain any value between 0 and µ N.
i.e. 0 \le f \le \mu N

Application 2

For the system shown in the figure, the coefficient of static and kinetic friction between the block m2 and the horizontal surface is µs and µk.
If m2 = 10 kg, µs= 0.4 and µk = 0.35
(i) Find the maximum value of m1 so that the block m2 does not move.
(ii) If m1 = 5 kg, find the acceleration of the system.

Solution

(i) The force which tries to accelerate the system is m1g, while the opposing force is fs. For maximum value of munder equilibrium condition.
(m1g)max = fs max =µsm2g
or   m1max  = µsm2
here µs = 0.4, m2  = 10 kg
Thus, m1max  = (0.4) (10) = 4 kg
(ii) When m1 = 5 kg, the system starts accelerating; m1 accelerates downward and m accelerates rightward. Let a be the acceleration of the system, then
a = {{{F_{net}}} \over {{M_{system}}}}\, = {{{m_1}g - \,{f_k}} \over {{m_1}\, + {m_2}}}             or           a = \left[ {{{{m_1}\, - \,{\mu _k}\,{m_2}} \over {{m_1}\, + {m_2}}}} \right]\,g
Putting m1 = 5kg; m2 = 10 kg; µk = 0.35; g = 10 m/s2
We get, a = \left[ {{{5 - \left( {0.35} \right)\,(10)} \over {5\, + \,10}}} \right]\,(10) = 1 m/s2

Application 3

A block is placed at rest on a horizontal surface. The coefficient of friction between the block and the surface is µ. It is pulled with a force F at an angle q with the horizontal as shown in the figure. Find the value of q at which minimum force is required to move the block. Also find the magnitude of this minimum force.

Solution

The free body diagram of the block is shown in the figure. The friction force is maximum because the block is just about to move.
Applying Newton’s Second Law
N + F\,\sin \theta = mg             (1)
F\,\cos \theta = {f_{\max }} = \mu N         (2)
or     F\,\cos \theta = \mu (mg - F\,\sin \theta )
or     F(\cos \theta + \mu \sin \theta ) = \mu mg   (3)
For minimum value of F,    {{dF} \over {d\theta }}\, = 0
Thus, differentiating equation (3) w.r.t. \theta ,  by applying chain rule, we get
{{dF} \over {d\theta }}\left( {\cos \,\theta + \mu \sin \,\theta } \right) + F{d \over {d\theta }}\left( {\cos \theta + \mu \sin \theta } \right)=0
{{dF} \over {d\theta }}(cos\theta + \mu sin\theta ) + F( - sin\theta + \mu cos\theta ) = 0
Since {{dF} \over {d\theta }}=0, therefore ( - sin\theta + \mu \cos \theta ) = 0
or tan \theta = µ  \Rightarrow \theta = tan-1 µ
Note that minimum force required to move the block is at the angle of friction.
Now from equation (3), {F_{min}} = {{\mu mg} \over {\sqrt {1 + {\mu ^2}} }} with  q = tan-1(\mu )

ROUGH PULLEYS

We know that when a string passes over a smooth pulley, the tension on each side of the string is same. But if the surface of the pulley is rough then the tensions on either side of the string are different. The difference in tension depends on the coefficient of friction (µ) and the angle (\theta ) through which the string is wrapped over the pulley.

Application 4

Show that the tension on two sides of the string wrapped over a rough pulley are related as
T2 = T1 e \mu \theta     (T2 > T1)
where µ is the coefficient of friction and \theta
 is the angle subtended by the string over the pulley.

Solution

Consider a small element of string, which subtends an angle d\theta at the centre.
For the equilibrium of the element
dN = T\sin \left( {{{d\theta } \over 2}} \right) + (T + dT)sin\left( {{{d\theta } \over 2}} \right)
or dN = 2T\sin \left( {{{d\theta } \over 2}} \right) = T\,d\theta       (1)
\sin \left( {{{d\theta } \over 2}} \right) \approx \left( {{{d\theta } \over 2}} \right)
(T + dT)cos\left( {{{d\theta } \over 2}} \right) - Tcos\left( {{{d\theta } \over 2}} \right) = dF = \mu \,dN  (cos\left( {{{d\theta } \over 2}} \right)  \approx  1)
or   dT = \mu dN            (2)
From equations (1) and (2)
dT = \mu T\,d\theta      or      {{dT} \over T}  = dT = \mu d\theta
Integrating both sides, we get
\int\limits_{{T_1}}^{{T_2}} {{{dT} \over T}} \, = \,\int\limits_0^\theta {\mu d\theta }    or    n\,\left| {{{{T_2}} \over {{T_1}}}} \right|\, = \mu \theta
or T2 = T1. This shows that T2 > T1.

Application 5

Find the minimum and maximum values of M2 for which the system remains in equilibrium. Take M1 = 10 kg and µ = 0.2.

Solution

For minimum value of M2
{M_1}g = {M_{2\min }}g{e^{\mu \pi }} (\theta = \pi )
or {M_{2\min }} = {{{M_1}} \over {{e^{\mu \pi }}}} = {M_1}{e^{ - \mu \pi }}
Putting M1 = 10 kg; µ = 0.2, after solving
{M_{2\min }} = (10){e^{ - (0.2)(3.14)}}= 5.33 kg
For maximum value of M2
{M_{2\max }}g = {M_1}g{e^{\mu \pi }}
or {M_{2\max }}g = (10){e^{(0.2)(3.14)}} = 18.73 kg



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