Learn kinetic friction definition and formula with examples, calculate acceleration with force applied at an angle.
It always acts in opposite direction to the relative motion between the two surfaces in contact.
\(f= \mu_k\,N\)
\(a= \dfrac{\sum\,F_{net}}{m}\)
\(a= \dfrac{\mu_k\,N}{m}\)
\(a = \dfrac{\mu_k\,mg}{m}\) (\(\because N=mg\))
\(a= \mu_k\,g\)
A \(2\,m/s^2\)
B \(3\,m/s^2\)
C \(4\,m/s^2\)
D \(5\,m/s^2\)
FBD of the Block
\(N + F\,sin \,\theta = mg\)
\(N = mg - F\,sin \,\theta\) ....(1)
\(F\,cos\,\theta - f_k = m\,a\)
\(F\,cos\,\theta - \mu_k \,N = m\,a\)
\(F\,cos\,\theta - \mu_k (mg-F\,sin\,\theta) =ma\) [from (1)]
\(a= \dfrac{F\,cos\,\theta-\mu_k(mg-F\,sin\,\theta)}{m}\)
A \(5.8\,m/s^2\)
B \(6\,m/s^2\)
C \(7.2\ m/s^2\)
D \(8\,m/s^2\)
\(N =mg\) ........(1)
Applying Newton's second law in horizontal direction
\(- f_k = m\,a\) [taking right side positive]
\(a = \dfrac{-f_k}{m}\)
\(a= \dfrac{-\mu_k\,mg}{m}\)
\(a= -\mu_k\,g\)
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.
It is the maximum value of static friction i.e., \(f_{max} = \mu_s \,N\)
where \(\mu_s\) is the coefficient of static friction.
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.
A \(4\,m/s^2\)
B \(5\,m/s^2\)
C \(3\,m/s^2\)
D \(2\,m/s^2\)
FBD of Block A
\(N = mg \,cos\,\theta\)
FBD of Block B
\(T = mg\) .......(1)
\(f_s = f_{max} = \mu_s\,N= \mu_s \,mg\,cos\,\theta\)
\(f_k = \mu_kN = \mu_k \,mg \,cos\,\theta\) ......(2)
\(a_A =0\)
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.
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.
Thus, kinetic frictional force will act.
\(T- f_k - mg\,sin\,\theta = m\,a\)
\(a= \dfrac{T- f_k -mg\,sin\,\theta}{m}\)
\(a= \dfrac{mg- \mu_k \,mg \,cos\,\theta - mg \,sin\,\theta }{m}\)
[ From (1) and (2) ]
\(a= g[1 -\mu_k \,cos\,\theta -sin\,\theta ]\)
A \(4\,m/s^2\)
B \(2.3\,m/s^2\)
C \(3.2\,m/s^2\)
D \(6\,m/s^2\)
Limiting frictional force, \(f_{\ell} = \mu\,N\)
\(f_{\ell } = \mu\,m_1\,g\) ........(1)
\([\because N= m_1 \,g]\)
Let the acceleration of both the blocks are \(a_{\ell}\).
Applying Newton's Second Law on block m_{2}
\(f_{\ell} = m_2 \,a_{\ell}\)
\( a_{\ell} = \dfrac{f_{\ell}}{m_2} \,\)
\(a_{\ell} = \dfrac{\mu \,m_1\,g}{m_2} \,\)
[from equation (1)]
\(a_{\ell} = \mu \left(\dfrac{\,m_1}{m_2}\right)\,g\) .......(2)
Applying Newton's Second law on block m_{1}
\(F_{max} - f_{\ell} = m_1\,a_{\ell}\)
\(F_{max} = m_1\left(\dfrac{\mu\,m_1\,g}{m_2}\right) +\mu\,m_1 \,g\)
[from (1)and (2)]
\(F_{max} =\dfrac{\mu\,{m_1}^2\,g}{m_2} +\mu\,m_1 \,g\)
\(F_{max} = \mu\,m_1 \,g \left[\dfrac{m_1+m_2}{m_2}\right] \)
A \(40 \,N\)
B \(37.5\,N\)
C \(42 \,N\)
D \(44 \,N\)