Friction on a Rough Inclined Plane

To understand the concept of friction on a rough Inclined plane let’s consider a rough inclined plane having an inclination of angle θ with the horizontal which is greater than the angle of repose and a coefficient of friction of $ \displaystyle \mu $. Let’s first calculate the acceleration ‘$ \displaystyle a$’of a body on a rough inclined plane.

Acceleration of a body down a rough inclined plane

As it is clear from the figure,

$ \displaystyle R=mg\cos \theta $

The net force on the body down the inclined plane

$  \displaystyle \begin{array}{l}f=mg\sin \theta -F\text{ }…\text{(2)}\\f=ma=mg\sin \theta -\mu R\\\because \mu =\dfrac{F}{R}\end{array}$

$ \displaystyle \text{Using (1), we get}$

$ \displaystyle \begin{array}{l}ma=mg\sin \theta -\mu mg\cos \theta \\=mg\left( {\sin \theta -\mu \cos \theta } \right)\end{array}$

Hence,

$ \displaystyle a=g\left( {\sin \theta -\mu \cos \theta } \right)$

Clearly, $ \displaystyle a<g$

i.e. acceleration of a body down a rough inclined plane is always less than the acceleration due to gravity ($ \displaystyle g$)

Note: When a plane is inclined to the horizontal at an angle $ \displaystyle \theta $, which is less than the angle of repose then the minimum force required to move the body up the inclined plane is

$ \displaystyle \begin{array}{l}{{f}_{1}}=\left( {mg\sin \theta +F} \right)\\=mg\left( {\sin \theta +\mu \cos \theta } \right)\end{array}$

Furthe, the minimum force required to push the body down the inclined plane is

$ \displaystyle \begin{array}{l}{{f}_{2}}=\left( {F-mg\sin \theta } \right)\\=mg\left( {\mu \cos \theta -\sin \theta } \right)\end{array}$

Work done in moving a body up a rough inclined plane

Suppose m is the mass of a body that has to be moved up a rough plane AB, inclined to the horizontal at an angle $ \displaystyle \theta $. The various forces involved are shown as:

1. Weight $ \displaystyle \left( {mg} \right)$ of the body, acting vertically downwards,
2. Normal reaction, $ \displaystyle \left( R \right)$ acting perpendicular to the plane AB.
3. Force of friction $ \displaystyle \left( F \right)$, acting down the plane AB, as the body moved up the plane.

The weight $ \displaystyle mg$can be resolved into two rectangular components:

(i) $ \displaystyle mg\cos \theta $ opposite to $ \displaystyle R$

(ii) $ \displaystyle mg\sin \theta $ along the inclination or parallel to the inclination

In equilibrium,

$ \displaystyle R=mg\cos \theta \text{ }…\text{(1)}$

and $ \displaystyle P=mg\sin \theta +F\text{ }…\text{(2)}$

where $ \displaystyle P$ is the force required to be applied up the plane AB.

Under the action of this force, suppose the body moves through a distance S up the plane.

As $ \displaystyle \text{Work done = force }\times \text{distance}$

$ \displaystyle \begin{array}{l}\therefore W=P\times S\\=\left( {mg\sin \theta +F} \right)S\text{ }…\text{(using 2)}\end{array}$

$ \displaystyle W=\left( {mg\sin \theta +\mu R} \right)$

where $ \displaystyle \mu $ is the coefficient of friction between the two surfaces in contact:

$ \displaystyle \begin{array}{l}W=\left( {mg\sin \theta +\mu mg\cos \theta } \right)S\\\text{by using (1)}\\W=mg\left( {\sin \theta -\mu \cos \theta } \right)S\end{array}$