11.3 Motion on accelerated incline plane(application)  (Page 3/4)

Let the mass of the block be “m”. The net component of force parallel to incline is acting downward and is given by :

$$F=mg\sin \theta -\mu mg\cos \theta$$

The incline moves with a certain acceleration due to forces acting on it. We need to apply a pseudo force to apply Newton's second law of motion. It is clear that pseudo force should have a component along the incline in upward direction so that net component of force (without friction) is zero. This will ensure that there is no friction between interface of block and the incline.

Now, for the pseudo force to have a component along the incline upward direction, the acceleration of the incline should be towards left as indicated in the figure below.

The free body diagram of the block with pseudo force is shown here.

The components of gravitational force and pseudo force form balanced force system so that there is no net component of forces along the incline and hence there is no friction between the interface.

$$mg\sin \theta =ma\cos \theta$$

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

$a=g\tan \theta$ towards left

Problem 4 : A block of mass "m" is placed on an incline of angle "θ" and mass "M", which is placed on a horizontal surface. All surfaces are friction - less. Find the acceleration of the block with respect to incline.

Solution : Here, block and incline both have different accelerations. The block moves down along the incline. The incline, on the other hand, moves right as shown in the figure above. As such, we can not treat two elements as one body and find the their common accelerations with respect to ground.

Let " $a_B$ " , " $a_I$ " and " $a_{\mathrm{BI}}$ " respectively denote acceleration of block w.r.t ground, acceleration of the incline w.r.t ground and acceleration of the block w.r.t incline.

We first carry out force analysis on the "block" in the non-inertial frame of incline. Here, the forces on the block are (i) weight of the block, "mg", (ii) normal force due to underlying incline, " $N_1$ ", and (iii) Pseudo force, " $m{a}_I$ " applied in the direction opposite to the acceleration of the incline. The free body diagram of the incline in non-inertial reference of incline is shown here.

Force analysis on the block yields two relation,

$$\sum F_x=m{a}_I\cos \theta +mg\sin \theta =m{a}_{BI}$$

$$\Rightarrow a_{BI}=a_I\cos \theta +g\sin \theta$$

$$\sum F_y=N_1+m{a}_I\sin \theta =mg\cos \theta$$

It is clear from the first equation that we need to know " $a_I$ " i.e. the acceleration of incline with respect to ground to determine the value of " $a_{\mathrm{BI}}$ ". It is imperative that we carry out force analysis on the "incline" in the inertial frame of reference of ground.

The forces on the incline are (i) weight of the incline, "Mg", (ii) normal force due to overlying block, " $N_1$ ", and (iii) Normal force due to underneath horizontal surface, " $N_2$ ". The free body diagram of the incline in inertial ground reference is shown here.