Learn static friction definition physics and types of friction with examples, Practice to calculate angle of repose and friction problems.
where, \(\mu_s\) is coefficient of static friction
N is normal force
The block does not move even the applied force is increasing as a counteracting force of friction is also increasing till it reaches its maximum value and balances the applied force.
The block does not move upto the maximum limit of f_{s} i.e., \(f_s = \mu_ s\,N\).
Wheel of moving car
A Static Friction
B Kinetic Friction
C Rolling Friction
D None of these
Case 1: When there is no external force applied on the block
\(f_s = 0\)
Case 2: When external force is applied horizontally and the block is still at rest
Maximum friction force = \(\mu _s N\)
\(f_s - F \,cos\theta = 0 \) (As body is at rest)
\(N+F\,sin\theta-mg=0\)
Since, \(\Sigma F_y = 0\) (As, there is no acceleration along y-axis)
\(N=mg-F\,sin\theta\)
\( F_{max} = \mu_s N \)
\(f_{s_{max}} = \mu_s (mg - F sin \theta)\)
At this value of \(\theta\), \(f_s= f_{max} = \mu_s \,N\)
Net force in \(x- \text{direction} \,\,{(\sum f_x)}\),
\(= -mg\,sin\, \theta + \mu_s \,N =0 \) ..........(i)
Net force in \(y- \text{direction} \,\,{(\sum f_y)}\),
\(= -mg \,cos \,\theta +N =0 \) ...........(ii)
Putting the value of N from equitation (ii) in (i)
\(tan\,\theta = \mu_s\)
\(\theta = \theta _R = tan^{-1} \, \mu_s\)
This angle \(\theta _R\) is known as angle of Repose.
A tan -1 (4)
B tan -1 (2)
C tan -1 (0.5)
D tan -1 (0.4)
Case 1:
If m_{2} is smaller than m_{1}, then m_{1} will tends to move downward on the inclined plane for the smallest possible value of m_{2}.
To remain at equilibrium, m_{1} will experience maximum friction in upward direction.
For the block of mass m_{2,}
T = m_{2 }g ..............(i)
For equilibrium of m_{1,}
\(T + \mu_s N - m_1 g \,sin\theta =0\) ............(ii)
\(m_1 \,g \,cos \theta - N =0\)
\( N = m_1 g \,cos\theta \) ..............(iii)
Putting the value of N and T in equation (ii) from (iii) and (i)
\(m_2 g + \mu_s (m_1 g \,cos\theta) - m_1 g \, sin \theta = 0 \)
\(m_2 = m_1 (sin \theta - \mu_s cos\theta)\)
\(\dfrac{m_2}{m_1} = sin \theta - \mu_s cos\theta\)
Case 2:
Now if m_{2} is increased, then m_{1} will tends to move in upward direction on the inclined plane. The force of friction will be directed in downward direction on the inclined plane.
For equilibrium of m_{2},
T = m_{2} g ...........(i)
For equilibrium of m_{1,}
\(T - \mu_s N - m_1 g \,sin\theta = 0 \) ...............(ii)
\(m_1\, g \,cos \theta - N = 0\)
\(N = m_1 \,g \, cos\theta\) ...............(iii)
Putting the value of N and T in equation (ii) from (iii) and (i)
\(m_2 = m_1 (sin\theta + \mu_s cos\theta)\)
\(\dfrac{m_2}{m_1} = sin\theta+ \mu_s cos\theta\)