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Physics-

General

Easy

Question

For the situation shown in figure, the block is stationary w.r.t. incline fixed in an elevator. The elevator is having an acceleration of $square root of 5 a subscript 0 end subscript$ whose components are shown in the figure. The surface is rough and coefficient of static friction between the incline and block is $mu subscript s end subscript$ . Determine the magnitude of force exerted by incline on the block. (take $a subscript 0 end subscript equals fraction numerator g over denominator 2 end fraction a n d blank theta equals 37 degree comma blank mu subscript s end subscript equals 0.2$ )
<img src="https://mycourses.turito.com/tokenpluginfile.php/c161933dbfaab094c54655ab71e9b8f0/1/question/questiontext/965275/1/1211145/Picture899.png" alt=""

$\frac{m g}{10}$
$\frac{9 m g}{25}$
$\frac{3 m g}{25} \times \sqrt{41}$
$\frac{\sqrt{13} m g}{2}$

hint Hint:

F=m(g+a₀)sino-2ma_ocoso

The correct answer is: $fraction numerator square root of 13 blank m g over denominator 2 end fraction$

The FBD of block from lift frame is as shown in figure. From given data, as $m open parentheses g plus a subscript 0 end subscript close parentheses sin invisible function application theta greater than 2 blank m a subscript 0 end subscript cos invisible function application theta$

So friction force acts upwards
$f equals m open parentheses g plus a subscript 0 end subscript close parentheses sin invisible function application theta minus 2 m a subscript 0 end subscript cos invisible function application theta$
$equals fraction numerator 9 g over denominator 10 end fraction minus fraction numerator 4 m g over denominator 5 end fraction equals fraction numerator m g over denominator 10 end fraction$
$N equals m open parentheses g plus a subscript 0 end subscript close parentheses cos invisible function application theta plus 2 m a subscript 0 end subscript sin invisible function application theta equals fraction numerator 9 m g over denominator 5 end fraction$
As $f subscript L end subscript equals mu subscript s end subscript N equals fraction numerator 18 m g over denominator 50 end fraction equals fraction numerator 9 m g over denominator 25 end fraction greater than f$ so static friction
Reaction force,
$R equals square root of f to the power of 2 end exponent plus N to the power of 2 end exponent end root equals fraction numerator m g over denominator 5 end fraction square root of fraction numerator 1 over denominator 4 end fraction plus 9 to the power of 2 end exponent end root equals fraction numerator m g square root of 13 over denominator 2 end fraction$
Alternative solution:
Net force $stack F with rightwards arrow on top minus m g stack i with hat on top equals m left parenthesis a subscript 0 end subscript stack j with hat on top minus 2 a subscript 0 end subscript stack i with hat on top right parenthesis$
$rightwards double arrow blank stack F with rightwards arrow on top equals m open parentheses negative 2 a subscript 0 end subscript stack i with hat on top plus open parentheses a subscript 0 end subscript plus g close parentheses stack j with hat on top close parentheses rightwards double arrow blank stack F with rightwards arrow on top equals m open parentheses negative g stack i with hat on top plus fraction numerator 3 g over denominator 2 end fraction stack j with hat on top close parentheses$
$rightwards double arrow F equals m square root of g to the power of 2 end exponent plus open parentheses fraction numerator 3 g over denominator 2 end fraction close parentheses to the power of 2 end exponent end root equals fraction numerator square root of 13 m g over denominator 2 end fraction f$

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