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Frictional force opposes relative motion between two objects in contact. Its magnitude depends on the surface properties of the objects and the normal force between them.


Static friction is the force that must be overcome in order to start moving an object at rest. It has a limiting value given by \begin{align} f_\text{max}=\mu_s N, \end{align} where $\mu_s$ is the coefficient of static friction and $N$ is the normal reaction.

Dynamic (or kineatic or sliding) friction is the force that acts on an object that is already in motion. Its value is given by \begin{align} f_k=\mu_k N, \end{align} where $\mu_k$ is the coefficient of static friction.

Rolling friction is the force that opposes the motion of a rolling object, such as a wheel or a ball. Rolling friction is generally weaker than sliding friction.

Problems from IIT JEE

Problem (IIT JEE 1980): A block of mass 2 kg rests on a rough inclined plane making an angle of ${30}^\mathrm{o}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is,

  1. 9.8 N
  2. $0.7\times 9.8\times\sqrt{3}\;\mathrm{N}$
  3. $9.8\times \sqrt{3}\;\mathrm{N}$
  4. $0.7\times 9.8\;\mathrm{N}$

Solution: The forces on the block of mass $m={2}\;\mathrm{kg}$ are its weight $mg$, normal reaction $N$, and the frictional force $f$ (see figure).


The net force on the block is zero because it is at rest. Resolve $mg$ in the directions parallel and perpendicular to the inclined plane. Apply Newton's second law in these directions to get, \begin{align} N&=mg\cos30 \\ &=(2)(9.8)(0.866)={16.97}\;\mathrm{N},\\ f&=mg\sin30\\ &=(2)(9.8)(0.5)={9.8}\;\mathrm{N}. \end{align} Note that $f$ is less than $f_\text{max}=\mu N=(0.7)(16.97)={11.88}\;\mathrm{N}$.


  1. Newton's Laws of Motion
  2. Rolling Friction (Demo)
  3. Direction of the Frictional Forces on the Bicycle Wheels during Pedaling
JEE Physics Solved Problems in Mechanics