Friction

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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-and-kinetic-friction

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).

a-block-of-mass-2kg

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 $f_\text{max}=\mu N=(0.7)(16.97)={11.88}\;\mathrm{N}$.

Related

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

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