## How Friction acts on Inclined Plane?

*Friction on an inclined plane depends upon the angle of inclination of the plane. As the angle of inclination increases, the friction force decreases.*

To understand this, consider about a solid block resting on an adjustable *inclined plane* AB as shown in figure. Let, *mass* of the block is ( m ) and inclination of the plane to the horizontal is ( \alpha ) .

At this situation, *rectangular components* of weight of body are –

- Parallel to the inclined plane is ( mg \sin \alpha ) .
- Perpendicular to the inclined plane is ( mg \cos \alpha ) .

The component ( mg \sin \alpha ) acting parallel to the inclined plane helps in moving the body downwards but the body remains in rest due to developed friction ( f ) .

Now, inclination of plane ( \alpha ) is gradually increases. Due to increase in inclination, an instance achieved when the block is in a state of just to move down.

At this instant the body is is equilibrium due to the effect of following forces –

- Component of weight of body in a direction parallel to the plane is ( mg \sin \alpha ) . It is acting downwards.
- Force of friction at the contact surfaces of body and plane. Since body has a tendency to move down the plane, hence friction force will be acting upward.
- Normal reaction ( f ) in a direction perpendicular to plane.

Since the body is in rest, so for *equilibrium* of the body –

N = mg \cos \alpha

*Thus, normal reaction acting on the contact surface depends on angle of inclination of inclined plane.*

Also, \quad f = mg \sin \alpha

From geometry of the figure, we have –

- Angle between normal reaction ( N ) and weight of body ( W = mg ) is ( 180 \degree - \alpha ) .
- Angle between friction force ( f ) and weight no body ( W = mg ) is ( 90 \degree + \alpha ) .

By applying *lami’s theorem* for the forces, we get –

\left [ \frac {N}{\sin \left (90 \degree + \alpha \right )} \right ] = \left [ \frac {f}{\sin \left (180 \degree - \alpha \right )} \right ]

But, \quad f = \mu N

*In this way, as the angle of inclination increases, normal reaction ( N ) decreases so friction decreases also.*

### Angle of Inclination and Angle of Friction

By applying *lami’s theorem* for the forces acting on body, we get –

\left [ \frac {N}{\sin \left (90 \degree + \alpha \right )} \right ] = \left [ \frac {\mu N}{\sin \left (180 \degree - \alpha \right )} \right ]

So, \quad \left ( \frac {N}{\cos \alpha} \right ) = \left ( \frac {\mu N}{\sin \alpha} \right )

Or, \quad \left ( \frac {\sin \alpha}{\cos \alpha} \right ) = \left ( \frac {\mu N}{N} \right )

Therefore, \quad \tan \alpha = \mu

But, from the definition of *coefficient of friction*, we have –

\mu = \tan \phi

Therefore, \quad \tan \alpha = \tan \phi

Or, \quad \alpha = \phi

*Therefore, at limiting condition, the angle of inclination of plane is just equal to the angle of friction for the inclined plane.*

__Friction during motion down an Inclined Plane__

Consider about a body as shown in figure. Let, the angle of inclination of plane increases and a pull ( P ) is applied at an angle ( \theta ) to the plane, to keep the body in *equilibrium*.

Now resolving the pulling force in rectangular components, we get –

- Effect of pull in the direction perpendicular to the plane is ( P \sin \theta )
- Effect of pull in the direction parallel to the plane is ( P \cos \theta )

Consider that, the pull is not sufficient and the body has a tendency to slide down the plane.

At this situation, from geometry of figure, we get –

( P \cos \theta + f ) = mg \sin \alpha

So, \quad ( P \cos \theta + \mu \ N ) = mg \sin \alpha

And, \quad ( N + P \sin \theta ) = mg \cos \alpha

So, \quad N = ( mg \cos \alpha - P \sin \theta )

Eliminating ( N ) from above equations, we get –

P \cos \theta + \mu \left (mg \cos \alpha - P \sin \theta \right ) ] = mg \sin \alpha .

Or, \quad [ P \left ( \cos \theta - \mu \sin \theta \right ) ] = [ mg \left ( \sin \alpha - \mu \cos \alpha \right ) ]

So, \quad P = \left [ \frac { mg \left ( \sin \alpha - \mu \cos \alpha \right )}{ \left ( \cos \theta - \mu \sin \theta \right )} \right ]

But, \quad \mu = \tan \phi \quad

Therefore, \quad P = \left [ \frac {mg \left ( \sin \alpha - \tan \phi \cos \alpha \right )}{\left ( \cos \theta - \tan \phi \sin \theta \right )} \right ]

Multiplying numerator and denominator by ( \cos \phi ) , we get –

P = \left [ \frac {mg \left ( \sin \alpha \cos \phi - \sin \phi \cos \alpha \right )}{\left ( \cos \theta \cos \phi - \sin \phi \sin \theta \right )} \right ]

= \left [ \frac {mg \sin \left ( \alpha - \phi \right )}{\cos \left ( \theta + \phi \right )} \right ]

*Solving this equation, we can get the minimum value of pulling force ( P ) which is necessary to hold the body in equilibrium.*

__Friction during motion up an Inclined Plane__

Now consider that, the pulling force ( P ) is just sufficient and the body is just at point of motion up the plane. Hence the force of friction will act downward.

In this condition, for *equilibrium* of body –

P \cos \theta = ( f + m g \sin \alpha )

= ( \mu \ N + m g \sin \alpha )

And \quad ( N + P \sin \theta ) = m g \cos \alpha

So, \quad N = ( m g \cos \alpha - P \sin \theta )

Eliminating ( N ) from above equations we get –

P \cos \theta = [ \mu \left ( m g \cos \alpha - P \sin \theta \right ) + m g \sin \alpha ]

So, \quad P \left ( \cos \theta + \mu \sin \theta \right ) = mg \left ( \sin \alpha + \mu \cos \alpha \right )

Or, \quad P = \frac { m g \left ( \sin \alpha + \mu \cos \alpha \right )}{ \left ( \cos \theta + \mu \sin \theta \right )}

But, \quad \mu = \tan \phi

Therefore, \quad P = \left [ \frac { m g \left ( \sin \alpha + \tan \phi \cos \alpha \right )}{ \left ( \cos \theta + \tan \phi \sin \theta \right )} \right ]

Multiplying numerator and denominator by ( \cos \phi ) we get –

P = \left [ \frac { m g \left ( \sin \alpha \cos \phi + \sin \phi \cos \alpha \right )}{ \left ( \cos \theta \cos \phi + \sin \phi \sin \theta \right )} \right ]

= \left [ \frac {mg \sin \left ( \alpha + \phi \right )}{\cos \left ( \theta - \phi \right )} \right ]

In this expression ( P ) will be minimum when numerator will be minimum or denominator will be maximum. But ( mg ), \ ( \alpha ) \ \text {and} \ ( \phi ) , all are constants. Hence, numerator \left [ mg \sin ( \alpha + \phi ) \right ] is also a constant term.

Therefore, for ( P ) to be minimum, denominator [ \cos ( \theta – \phi ) ] , should be maximum.

We know that, maximum value of ( \cos \theta ) is 1 . But ( 1 = \cos 0 \degree ) . This maximum value of \left [ \cos ( \theta – \phi ) \right ] = 1 will arise when [ ( \theta – \phi ) = 0 ]

Therefore, \quad [ ( \theta – \phi ) = 0 \degree ]

So, \quad \theta = \phi

Putting the value of ( \theta = \phi ) in the expression for ( P ) , we get –

P_{Minimum} = mg \sin \left ( \alpha + \phi \right )

*Therefore, minimum pull required to draw a body up an inclined plane is \left [ mg \sin ( \alpha + \phi ) \right ] which is required to apply in a direction inclined at an angle equal to the angle of friction, i.e. ( \theta = \phi ) *

__Friction in Inclined Plane (when force act horizontal)__

Consider about the figure as shown. Let, the pull ( P ) is applied horizontally.

**When the body is just to move upward –**

By, *parallelogram law of forces*,* *we have –

P \cos \alpha = ( mg \sin \alpha + \mu N )

And, \quad N = ( P \sin \alpha + mg \cos \alpha )

Eliminating ( N ) from above equations, we get –

P \cos \alpha = \left [ mg \sin \alpha + \mu \left ( mg \cos \alpha + P \sin \alpha \right ) \right ]

Or, \quad P \left ( \cos \alpha - \mu \sin \alpha \right ) = mg \left ( \sin \alpha + \mu \cos \alpha \right )

So, \quad P = \left [ \frac {mg \left ( \sin \alpha + \mu \cos \alpha \right )}{ \left ( \cos \alpha - \mu \sin \alpha \right )} \right ]

= \left [ \frac {mg \left ( \sin \alpha + \tan \phi \cos \alpha \right )}{ \left ( \cos \alpha - \tan \phi \sin \alpha \right )} \right ]

Multiplying numerator and denominator by ( \cos \phi ) , we get –

P = \left [ \frac {mg \left ( \sin \alpha \cos \phi + \sin \phi \cos \alpha \right )}{ \left ( \cos \alpha \cos \phi - \sin \phi \sin \alpha \right )} \right ]

= \left [ \frac {mg \sin \left ( \alpha + \phi \right )}{ \cos \left ( \alpha + \phi \right )} \right ] = \left [ mg \tan \left ( \alpha + \phi \right ) \right ]

**When the body is just to move downward –**

In this case, \quad P \cos \alpha = ( mg \sin \alpha - \mu N )

And, \quad N = ( P \sin \alpha + mg \cos \alpha )

Solving the equations by the procedures discussed above, we get –

P = \left [ mg \tan \left ( \alpha - \phi \right ) \right ]

Therefore, following conclusions are made from above equations –

*When ( \alpha > \phi ) ( i.e. angle of inclination is greater than the angle of friction ) then, the required pull ( P ) is positive i.e. it is in the direction shown in the figure above. Therefore, an effort of magnitude [mg \tan ( \alpha + \phi ) ] will require to move the block up the plane.*

*When ( \alpha < \phi ) ( i.e. angle of inclination is lower than the angle of friction ) then, the required pull ( P ) is negative i.e. it is in the opposite direction of the direction shown in figure above. Therefore, an effort of magnitude [mg \tan ( \alpha - \phi ) ] will be required to move the block down the plane.*