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

Force analysis of the incline yields two relations,

$$\sum F_x=N_1\sin \theta =m{a}_I$$

$$\sum F_y=N_2=N_1\cos \theta +Mg$$

First of the equations is useful that can be combined with the analysis in non-inertial frame of reference. Substituting for " $N_1$ " in the earlier equation, we have :

$$\Rightarrow \frac{m{a}_I}{\sin \theta }+m{a}_I\sin \theta =mg\cos \theta$$

$$\Rightarrow a_I=\frac{\left(mg\sin \theta \cos \theta \right)}{\left(M+m{\sin }^2\theta \right)}$$

We can now determine the required acceleration of the block with respect to incline by substituting the value of " $a_I$ " in the expression of " $a_{\mathrm{BI}}$ " :

$$\Rightarrow a_{BI}=\frac{\left(mg\sin \theta \cos {}^2\theta \right)}{\left(M+m\sin {}^2\theta \right)}+g\sin \theta$$

$$\Rightarrow aBI=\frac{\left(mg\sin \theta {\cos }^2\theta +Mg\sin \theta +m{\sin }^3\theta \right)}{\left(M+m{\sin }^2\theta \right)}$$

$$\Rightarrow a_{BI}=\frac{g\sin \theta \{M+m\left({\cos }^2\theta +{\sin }^2\theta \right)\}}{\left(M+m{\sin }^2\theta \right)}$$

$$\Rightarrow a_{BI}=\frac{\left(M+m\right)g\sin \theta }{\left(M+m\sin {}^2\theta \right)}$$

Problem 5 : A block is placed on an incline, which is accelerated towards left at an acceleration equal to the acceleration due to gravity, as shown in the figure. The angle of incline is 30°. If friction is absent at all contact surfaces, then find the time taken by the block to cover a length of 4 m on the incline.

Solution : Here, initial velocity of block is zero. We need to know the acceleration of the block with respect to incline in order to find the required time to cover a displacement of 4 m along the incline.

Let “a” be the acceleration of the incline with respect to ground and “ $a_r$ ” be the acceleration of the block relative to the incline. We however do not know whether net component of forces parallel to incline is in up or down direction. The free body diagram of the block is superimposed on the block in the figure above. The forces on the block includes pseudo force (mg), acting towards right.

$$F_x=mg\cos \theta -mg\sin \theta$$

$$\Rightarrow F_x=mg\cos {30}^0-mg\sin {30}^0$$

$$\Rightarrow F_x=mgX\frac{\sqrt{3}}{2}-mgX\frac{1}{2}$$

$$\Rightarrow F_x=\frac{\left(\sqrt{3}-1\right)mg}{2}$$

Clearly, the net component of forces along the incline is positive. It means that block is moving up. The magnitude of acceleration with respect to incline in upward direction, is :

$$\Rightarrow a_r=\frac{\left(\sqrt{3}-1\right)mg}{2m}$$

$$\Rightarrow a_r=\frac{\left(\sqrt{3}-1\right)g}{2}=3.66m/s^2$$

Now, using equation of motion for constant acceleration, we have (initial velocity is zero) :

$$x=\frac{1}{2}a{t}^2$$

$$\Rightarrow 4=\frac{1}{2}X3.66X{t}^2$$

$$\Rightarrow t=\frac{\sqrt{8}}{3.66}=1.48s$$

Incline in an accelerated lift

Problem 6 : A block of 2 kg slides down a smooth inclines plane of angle 30° and length 12 m. The incline is fixed at the floor of a lift, which is acceleration up at 2 $m/s^2$ . Find the time taken by the block to travel from the top to bottom of the incline. Also find the magnitude of acceleration of the block with respect to ground. Consider g = 10 $m/s^2$ .

Solution : In order to know the time, we need to know the acceleration of the block with respect to incline (not with respect to ground). The block has to travel exactly 12 m in the reference of accelerated lift. The technique of pseudo force, therefore, has distinct advantage over analysis in ground reference as the analysis in accelerated frame will directly give the acceleration of the block with respect to incline. This will allow us to apply equation of motion to find time to cover 12 m in the lift.

Let $a_B$ , $a_L$ and $a_{\mathrm{BL}}$ be the acceleration of block w.r.t ground, acceleration of lift (incline) w.r.t ground and acceleration of block w.r.t lift (incline) respectively.

Note that block is acted upon by three forces (i) weight of the block, "mg" (ii) Normal force on the block, "N" and pseudo force, "maL". The axes are drawn parallel and perpendicular to the incline. This is advantageous as acceleration of the block with respect to incline is along the incline. As we have gained experience, free body diagram is not separately drawn. It is shown superimposed on the body system. The figure above also shows the weight of the block resolved along the axes.

$$\begin{array}{l}\sum F_x=mg\sin \theta +m{a}_L\sin \theta =m{a}_{\mathrm{BL}}\\ \Rightarrow 2x{a}_{\mathrm{BL}}=2x10x\sin {30}^0+2x2x\sin {30}^0\\ \Rightarrow 2a_{\mathrm{BL}}=2x10x\frac{1}{2}+4x\frac{1}{2}=12\\ \Rightarrow a_{\mathrm{BL}}=6m/s^2\end{array}$$

Now, applying equation of motion,

$$\begin{array}{l}x=ut+\frac{1}{2}{a_{\mathrm{BL}}}^2=\frac{1}{2}x6x{t}^2\\ \Rightarrow t^2=\frac{12x2}{6}=4\\ t=2s\end{array}$$

The relative acceleration of the block with respect to incline $a_{\mathrm{BL}}$ is related to accelerations of the block and lift w.r.t ground as :

$$\begin{array}{l}{\mathbf{a}}_{\mathrm{BL}}={\mathbf{a}}_B-{\mathbf{a}}_L\end{array}$$

The acceleration of block with respect to ground is given as :

$$\begin{array}{l}\Rightarrow {\mathbf{a}}_B={\mathbf{a}}_{\mathrm{BL}}+{\mathbf{a}}_L\end{array}$$

Evaluating the right side of the equation,

$$\begin{array}{l}\Rightarrow a_B=\surd ({a_{\mathrm{BL}}}^2+{a_L}^2+2a_{\mathrm{BL}}{a}_L\cos {120}^0\\ \Rightarrow a_B=\surd (6^2+2^2+2x2x6x-\frac{1}{2})\\ \Rightarrow a_B=\surd (36+4-12)=\surd \left(28\right)=5.29m/s^2\end{array}$$