• Consider a block of mass \(m\) which is moving with initial velocity \(v\) on a rough surface, as shown in figure.

• As the body is in motion, the kinetic friction will act on it.
• The coefficient of kinetic friction is \(\mu_k\).

Direction of kinetic frictional force 

It always acts in opposite direction to the relative motion between the two surfaces in contact.

Net Horizontal Force 

       \(f= \mu_k\,N\)

• Using Newton's Second Law 

     \(a= \dfrac{\sum\,F_{net}}{m}\)

 \(a= \dfrac{\mu_k\,N}{m}\)

 \(a = \dfrac{\mu_k\,mg}{m}\)        (\(\because N=mg\))

\(a= \mu_k\,g\)

• Consider a block of mass \(m\) which is pulled by a force \(F\) at an angle \(\theta\) with the horizontal.

FBD of the Block 

• Here, normal force will not be equal to mg because vertical component of force is also acting.
• Net Force in Vertical Direction 

\(N + F\,sin \,\theta = mg\)

\(N = mg - F\,sin \,\theta\)     ....(1)

• Applying Newton's Second law in horizontal direction 

\(F\,cos\,\theta - f_k = m\,a\)

\(F\,cos\,\theta - \mu_k \,N = m\,a\)

\(F\,cos\,\theta - \mu_k (mg-F\,sin\,\theta) =ma\)          [from (1)]

\(a= \dfrac{F\,cos\,\theta-\mu_k(mg-F\,sin\,\theta)}{m}\)

• Consider two blocks A and B of masses m each, are attached with a string.
• Block A lies on the rough surface of a fixed wedge with an angle of inclination \(\theta\), while block B hangs through a massless pulley, as shown in figure. 

• The coefficient of static and kinetic friction between the surface of wedge and block A are \(\mu_s\) and \(\mu_k\), respectively.
• Representation of forces on the system 

FBD of Block A   

\(N = mg \,cos\,\theta\)

FBD of Block B   

      \(T = mg\)    .......(1)

• Maximum frictional force 

    \(f_s = f_{max} = \mu_s\,N= \mu_s \,mg\,cos\,\theta\)

• Kinetic friction 

      \(f_k = \mu_kN = \mu_k \,mg \,cos\,\theta\)    ......(2) 

• As the mass of block B is equal to the mass of block A, thus, assume that there is no slipping of block A.

                \(a_A =0\)

• Direction of frictional force 

       To find the direction of frictional force, first find the tendency of relative motion.

         As  \(T > mg\,sin\,\theta\)

\([\because \,T =mg \,\,\,{\text{and}} \,\,\theta\neq 90°]\)

Thus, block A has tendency to move upwards.

Hence, the frictional force \(f\) will act downwards.

• Since, acceleration of block A is zero thus, it will not move.

Applying Newton's Second Law on block A 

      \(mg\,sin\,\theta +f -T=0\)

        \(f= T- mg\,sin\,\theta\) 

• Case 1 :   If \(f< f_{max}\)  

    Then, assumed direction of frictional force is correct.

• Case 2 :    If  \(f>f_{max}\) 

Then, it is not possible.

Hence, there must be some acceleration and the block A will move in upward direction.

• The block A will, move with some acceleration.

            Thus, kinetic frictional force will act.

• Using Newton's second law 

            \(T- f_k - mg\,sin\,\theta = m\,a\)

            \(a= \dfrac{T- f_k -mg\,sin\,\theta}{m}\)

    \(a= \dfrac{mg- \mu_k \,mg \,cos\,\theta - mg \,sin\,\theta }{m}\)

[ From (1) and (2) ]

 \(a= g[1 -\mu_k \,cos\,\theta -sin\,\theta ]\)

• Consider two blocks of masses m₁ and m₂, placed on each other, as shown in figure.

• The coefficient of friction between the blocks is \(\mu\).
• An external force \(F_{ext}\) is applied on the block of mass m₁.
• To calculate the maximum value of F such that blocks of mass m₁ and m₂ could move together, firstly we have to calculate the frictional force acting between the blocks.
• At the verge of slipping, the limiting frictional force \(f_{\ell}\) acts between the blocks.

FBD of block of mass m₁

Limiting frictional force,  \(f_{\ell} = \mu\,N\)

\(f_{\ell } = \mu\,m_1\,g\)    ........(1) 

\([\because N= m_1 \,g]\)

FBD of block of mass m₂

Let the acceleration of both the blocks are \(a_{\ell}\).

Applying Newton's Second Law on block m₂ 

   \(f_{\ell} = m_2 \,a_{\ell}\)

    \( a_{\ell} = \dfrac{f_{\ell}}{m_2} \,\)

    \(a_{\ell} = \dfrac{\mu \,m_1\,g}{m_2} \,\)

[from equation (1)]

     \(a_{\ell} = \mu \left(\dfrac{\,m_1}{m_2}\right)\,g\)       .......(2)  

• Applying Newton's Second law on block m₁ 

\(F_{max} - f_{\ell} = m_1\,a_{\ell}\)

\(F_{max} = m_1\left(\dfrac{\mu\,m_1\,g}{m_2}\right) +\mu\,m_1 \,g\)

[from (1)and (2)]

 \(F_{max} =\dfrac{\mu\,{m_1}^2\,g}{m_2} +\mu\,m_1 \,g\)

\(F_{max} = \mu\,m_1 \,g \left[\dfrac{m_1+m_2}{m_2}\right] \)

• Hence, \(F_{max}\) is the value of maximum external force for which both the blocks move together.
• It is also clear that  \(F_{max}\neq f_{\ell}\).

• Consider a block of mass \(m\) which is thrown with initial velocity \(v\) on a rough horizontal surface, as shown in figure.
• The coefficient of kinetic friction between rough surface and block is \(\mu_k\). 

• The  sliding of block on surface will start only when the block comes in contact with surface and kinetic friction comes into act to oppose it.
• As the block is sliding towards right, thus to oppose it, the kinetic frictional force will act towards left.
• Applying Newton's second law in vertical direction 

         \(N =mg\)  ........(1)

• Applying Newton's second law in horizontal direction 

         \(- f_k = m\,a\)         [taking right side positive]

\(a = \dfrac{-f_k}{m}\)

\(a= \dfrac{-\mu_k\,mg}{m}\)

\(a= -\mu_k\,g\)

• Finally, the block will come to rest.

           Thus,   \(v=0\) 

using second law of motion,

\(v^2 =u^2 +2\,a\,s\)

\(0 = u^2 -2\,\mu_k \,g\,s\)

\(s= \dfrac{u^2}{2\,\mu_k\,g}\)

where, \(s\) is the distance traveled by block before it comes to rest.

• Consider a flatbed truck which is standing on a horizontal road.
• A block of mass \(m\) is kept on the truck, as shown in figure.

• Now, the truck accelerates. The block will remain, as it is till maximum limit of acceleration is achieved.
• If the truck accelerates above this limits, the block may slip.
• Consider that the truck moves towards right side.
• At the verge of slipping, the acceleration of both truck and block is same and here, limiting friction must be acting.

• Limiting Friction 

         It is the maximum value of static friction i.e., \(f_{max} = \mu_s \,N\) 

where \(\mu_s\) is the  coefficient of static friction. 

• As truck accelerates towards right direction, the block will have tendency to slip towards left. So, frictional force on the block will act towards right direction.

• Applying Newton's Second law in vertical direction 

          \(N = mg\) 

{Equilibrium in vertical direction }

• Applying Newton's Second law in horizontal direction 

          \(f_{max} = m\,a_{max}\)

[At the verge of slipping \(f_{max}\) is acting]

          \(\Rightarrow \mu_s\,N = m\,a_{max}\) 

         \(\Rightarrow \mu_s\,mg=ma_{max}\)   

         \(\Rightarrow a_{max} = \mu_s\,g\)

This is the maximum value of acceleration of truck so that block does not slip.