Question

A point A is located above an incline plane of inclination 30o . It is possible to reach the plane by sliding under gravity down a straight smooth wire, joining A to some point A' on the plane. If AA' makes  angle with vertically downward direction. Find the value of  to minimise the time taken

Answer 1

Answer:

A point A is located above an inclined plane

61 A point A is located above an inclined plane of inclination 30°. It is possible to reach the plane by sliding under gravity down a straight smooth wire, joining A to some point A' on the plane. If AA' makes e angle with vertically downward direction.

Answer 2

Answer:

If the wire is at 15°, the time taken to slide along AA' is minimum.

Explanation:

Given the angle of the inclined plane, $\theta=30^o$

Join the point A to some arbitrary point A' on the plane, let the measure of AA' be $l$.

Draw a normal from the point A onto the plane.

It makes a right angle triangle with some angle x at A as shown in the figure. Then

$cosx=\dfrac{adj}{hyp} =\dfrac{l}{AA^'} \implies AA^{'}=\dfrac{l}{cosx}$

As the case is a sliding case, no frictional force.

The initial velocity, u = 0

And the acceleration along AA' is $gcos(30+x)$

So, applying the equation of motion,

$s=ut+\dfrac{1}{2} at^{2}$

$\implies \dfrac{l}{cosx}=0+\dfrac{1}{2} gcos(30+x)t^{2}$

$\implies t^{2}= \dfrac{2l}{gcos(30+x)cosx}\implies t= \sqrt{\dfrac{2l}{gcos(30+x)cosx}}$

For the time to be minimum, the quantity in the denominator of right hand side should be maximum.

That is let $f(x)={gcos(30+x)cosx}$ should be maximum

and hence derivative,

$\dfrac{df(x)}{dx} =0$

So, differentiating the above with respect to x

$\dfrac{df(x)}{dx} ={gcos(30+x)cosx}$

$\implies -gcos(30+x)sinx-cosxgsin(30+x)=0$

$\implies cos(30+x)sinx+cosxsin(30+x)=0$

$\implies sin(30+2x)=0$

$\implies 30+2x=0$

$\implies x=\dfrac{-30}{2} =15^o$

Therefore, if the wire is at 15°, the time taken to slide along AA' is minimum.

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