An adamant block

Classical Mechanics Level 3

A block of mass m kg \text{m kg} rests on an inclined plane of inclination 30 ° {30}^° . Now, the inclined plane is pulled forward in such a way that the block is relatively at rest with respect to the inclined plane's surface. What is the acceleration produced by the applied force?

Details and assumptions :-

  • Assume that the inclined plane is frictionless.
  • Take g \text{g} (acceleration due to gravity) as 9.8 m/sec 2 9.8 {\text{m/sec}}^2 .
  • Give your answer (in m/sec 2 {\text{m/sec}}^2 ) upto two decimal places.


The answer is 5.65.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Jul 23, 2016

The vertical (downwards) component of the block is given by mg sin ⁡ θ \text{mg} \sin\theta .

Now, to stop the block from moving, a pseudo force acts on te block in the direction opposite to that of the applied force which is given by ma \text{ma} .

The vertical (upwards) component of this pseudo force which keeps the block at rest = ma cos ⁡ θ \text{ma}\cos\theta .

Since the block is at rest, ma cos ⁡ θ = mg sin ⁡ θ a = g tan ⁡ θ = 9.8 × tan ⁡ ( 30 ) ° = 9.8 3 = 5.65 m/sec 2 \begin{aligned} \text{ma} \cos\theta & = \text{mg} \sin\theta\\ \text{a} & = \text{g}\tan\theta\\ & = 9.8 × \tan {(30)}^°\\ & = \dfrac{9.8}{\sqrt{3}}\\ & = \color{#3D99F6}{\boxed{5.65 \ {\text{m/sec}}^2}} \end{aligned}

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