- Download PDF
1 Answer | Add Yours
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
We’ve answered 319,233 questions. We can answer yours, too.Ask a question