# A box with mass 15 kg lies stationary on an inclined plane at 20 degrees to the horizontal. What has to be the coefficient of friction for this to happen.

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The box with mass 15 kg lies stationary on an inclined plane at 20 degrees to the horizontal.

Let the coefficient of friction between the box and the plane be `mu` . The mass of the box is 15 kg. There is gravitational force equal to 15*9.8 = 147 N acting in a vertically downward direction on the box. This can be divided into components perpendicular and parallel to the plane. The normal force is equal to `147*cos 20` and the component of the force parallel to the plane is `147*sin 20` .

The frictional force acting on the box is `147*cos 20*mu` . As the box is stationary, the net force on the box is 0. This gives:

`147*cos 20*mu = 147*sin 20`

=> `mu = tan 20`

=> `mu = 0.3639`

**The coefficient of friction between the box and the inclined plane it is placed on is 0.3639**