Physics - NLM Question with Solution | TestHub

Let the wedge be moving in the horizontal direction with acceleration $a$ as shown in the figure, and the block be moving along the wedge with acceleration $b$ with respect to the wedge.

On drawing the FBD of the wedge in the horizontal direction:

$$\mathrm{N}\cos 30^{\circ }+\mathrm{T}-\mathrm{T}\cos 60^{\circ }=\mathrm{Ma}$$

$$\mathrm{N}\cos 30^{\circ }+\mathrm{T}\cos 60^{\circ }=\mathrm{Ma}\dots \dots \dots \left(i\right)$$

On drawing the FBD of the block with respect to the wedge:

$$\mathrm{N}+\mathrm{ma}\sin 60^{\circ }=\mathrm{mg}\cos 60^{\circ }..\dots \dots \left(ii\right)$$

$$\mathrm{ma}\cos 60^{\circ }+\mathrm{mg}\sin 60^{\circ }-\mathrm{T}=\mathrm{mb}\dots \dots \left(iii\right)$$

Now, from the wedge constraint:

$$\mathrm{a}=\mathrm{b}\text{ (relative to wedge) }$$

For the block:

$$b_x=a-a\cos \theta =a-a\times \frac{1}{2}=\frac{a}{2}$$

$$b_y=a\sin \theta =a\times \frac{\sqrt{3}}{2}$$

So, $b=\sqrt{b_x^2+b_y^2}=\sqrt{(a/2{)}^2+{\left(\frac{\sqrt{3}a}{2}\right)}^2}$

$$=b=a\dots \dots \left(iv\right)$$

Now, on solving the above equations, we have:

$$\mathrm{a}=\mathrm{b}=\frac{\sqrt{3}\mathrm{g}}{4},\mathrm{T}=\frac{3\sqrt{3}\mathrm{mg}}{8},\mathrm{N}=\frac{\mathrm{mg}}{8}$$