11.3 Motion on accelerated incline plane(application)  (Page 2/4)

The component of acceleration of the lift in horizontal direction is zero. The motion of the block in horizontal direction, therefore, is not in accelerated frame of reference. For the condition of balanced net component of forces in horizontal direction :

$$\begin{array}{l}\sum F_x=F_s\cos \theta -N\sin \theta =0\\ \Rightarrow {\mu }_{s}N\cos \theta -N\sin \theta =0\end{array}$$

$$\begin{array}{l}\Rightarrow \tan \theta ={\mu }_s\end{array}$$

Thus, we see that angle of repose in the accelerated lift is same as that when the incline is placed on the ground. This result is expected as there is no component of acceleration of the lift in horizontal direction and application of Newton's law remains same as far as force analysis in horizontal direction is concerned.

Incline on a smooth horizontal surface

Problem 2 : A block of mass "m" is kept on a smooth incline of angle "θ" and mass "M". The incline is placed on a smooth horizontal surface. Find the horizontal force "F" required such that block is stationary with respect to incline.

Solution : Since block is stationary with respect to incline, both elements of the system (incline and block) have same acceleration. As such, we can consider "block - incline" to constitute a combined system and treat them as one. As all surfaces are smooth, friction force is zero. Hence, acceleration of the system is given by :

$$a=\frac{F}{\left(m+M\right)}$$

We can now analyze forces on the block to find force "F" for the given condition of block being stationary with respect to incline. The block is stationary with reference to incline, but it is accelerated with respect to ground. The condition of "being stationary" is, therefore, given in the context of accelerated frame of incline. Hence, it is more intuitive to carry out analysis in the non-inertial frame of reference.

Three forces on the block, including pseudo force, are shown here in the figure. Further, since two of the three forces i.e weight and pseudo forces are in horizontal and vertical directions, we choose our coordinates accordingly along these directions.

The free body diagram of the block with components of forces along the axes is shown here.

$$\sum F_x=ma-N\sin \theta =0$$

$$\Rightarrow N\sin \theta =ma$$

$$\sum F_y=N\cos \theta -mg=0$$

$$\Rightarrow N\cos \theta =mg$$

Taking ratio, we have :

$$\Rightarrow \frac{a}{g}=\tan \theta$$

Substituting for "a", we have :

$$\frac{F}{\left(m+M\right)g}=\tan \theta$$

$$\Rightarrow F=\left(m+M\right)g\tan \theta$$

Problem 3 : A block of mass is placed on a rough incline of angle “θ”. The angle of incline is greater than the angle of repose. If coefficient of friction between block and incline is “μ”, then what should be acceleration of the incline in horizontal direction such that no friction operates between block and incline.

Solution : In order to find the acceleration of the incline for “no friction”, we need to know the net component of forces parallel to the incline. The pulling force here is the component of gravitational force parallel to incline in the downward direction. It is given that the angle of incline is greater than angle of repose. It means that block has started moving downward and force of friction is equal to kinetic friction (about equal to limiting friction).