Question

the horizontal acceleration that should be given to a smooth inclined plane of inclination Cos^(-1) ½, to keep an object stationary on the plane relative to it is :-

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Answer 1

inclination of plane =cis^-1/2

so from here angle of inclination=60°

if the object will placed on the inclined plane,it will have accleration of gsin60°=g√3/2

so to keep the body at rest,there should be horizontal acceleration of g√3/2

so option1) is correct

Answer 2

Hello mate,

Here is your answer,

(For better understanding check the above diagram)

Given:

• Inclination of plane => ${cos}^{ - 1} ( \frac{1}{2} )$

To Find:

• Horizontal acceleration of the plane to keep the object stationary on the plane relative to it.

Solution:

Inclination:

$cos (\alpha ) = \frac{1}{2} \\ \\ \alpha = 60 \: \: degrees$

Now as per the diagram,

$f = ma \sin (\alpha ) = mg \cos ( \alpha ) \\ \\ a. \sin( \alpha ) = g. \cos( \alpha ) \\ \\ a = \frac{g. \cos( \alpha ) }{ \sin( \alpha ) } \\ \\ a = g. \tan( \alpha ) \\ \\ we \: \: know \: \: \alpha = 60 \: \: degrees \\ \\ a = g. \tan(60) \\ \\ we \: \: know \\ \\ \tan(60) = \sqrt{3} \\ \\ so \\ \\ a = \sqrt{3} g$

Option (3) is your required answer

Hope it helps:)