 |
| Angle of Friction |
|
| |
 |
| |
Consider a block placed on a rough floor. Now, the reaction force is 
because it is equal and opposite to the weight
. Now apply a horizontal force
 so that the block just begins to
slide, i.e., the frictional force is equal to the limiting friction. When this condition is satisfied, the angle which the resultant (between the normal and external force) makes with the vertical is called the angle of friction. |
| |
 |
| |
 |
| |
| The tangent of the angle of friction is equal to the coefficient of static friction. |
| |
|
| Consider a body placed on a platform whose angle with the horizontal can be changed for e.g., a tipper with stones at the back. |
| |
 |
| |
| Only when the platform is raised to a certain angle do the stones begin to slide down. This is because the component of gravitational force along the platform just overcomes the frictional force. This angle to which the platform is inclined with the horizontal, is called the angle of sliding or the angle of repose. |
| |
| Let us see the mathematical aspect of it. Consider the figure . |
| |
 |
| |
| The weight W can be resolved into two
rectangular components. W cos f and W sin
f. The component W cos f
balances the normal reaction R while the component W sin
f is equal to the limiting friction flimiting. |
| |
 |
| |
 |
| |
 |
| |
 |
| |
 We notice that the angle of repose or sliding and the angle of friction are both equal. |
| |
 |
| |
|
 |
| |
| Consider the body to be sliding down the plane. You must take all force components along and perpendicular to the surface of the block. |
| |
| Balancing components along the perpendicular of the inclined plane |
| |
 |
| |
Balancing components along the surface of the inclined plane  |
| |
| [a is the acceleration of the block along the inclined surface] |
| |
 |
| |
 |
| |
 |
| |
| This equation gives the acceleration of a body sliding down a rough inclined plane. |
| |
|
| Work is done in displacing a body with constant velocity. For e.g., your mother asks you to shift an almirah by pushing it to the other room. You may feel tired after pushing it. But if she now says that the third room is more suitable and you push it to the third room you feel more tired than before. This is because you have done work against friction by shifting the almirah. |
| |
| Similarly, when any body is pushed on a horizontal floor, work is done against friction. |
| |
 |
| |
| Work done against friction = force . displacement |
| |
| = fkinetic . S |
| |
 |
| |
 |
| |
 |
| |
|
| The situation here is almost the same as the situation in the figure, just that the body here is sliding up instead of down, the friction acts downward. |
| |
 |
| |
| Component perpendicular to the surface |
| |
 |
| |
| Component parallel to the surface |
| |
 |
| |
| (Note that the body is being pulled with a constant velocity) |
| |
 |
| |
 |
| |
| Let the body be pulled through a distance S. |
| |
| Hence, the work done will be P.S |
| |
 |
| |
| In the case of moving a block down, mg sin q will help it to come down, but friction will oppose it. Hence, work done in moving a block down the incline: |
| |
| Work = mg (m cos q - sin q)S. |
| |
