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In-the-figure-shown-below-all-surfaces-are-smooth-strings-and-pulley-are-ideal-If-the-wedge-is-moving-with-acceleration-a-towards-the-right-then-the-acceleration-of-the-block-with-respect-to-the-w

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Question Number 24301 by Tinkutara last updated on 15/Nov/17
In the figure shown below, all surfaces are smooth, strings and pulley are ideal. If the wedge is moving with acceleration a towards the right, then the acceleration of the block with respect to the wedge at that instant is
$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown}\:\mathrm{below},\:\mathrm{all}\:\mathrm{surfaces} \\ $$$$\mathrm{are}\:\mathrm{smooth},\:\mathrm{strings}\:\mathrm{and}\:\mathrm{pulley}\:\mathrm{are}\:\mathrm{ideal}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{acceleration} \\ $$$${a}\:\mathrm{towards}\:\mathrm{the}\:\mathrm{right},\:\mathrm{then}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{block}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wedge} \\ $$$$\mathrm{at}\:\mathrm{that}\:\mathrm{instant}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 15/Nov/17
Commented by ajfour last updated on 15/Nov/17
let length of string from top of wedge to block m be y and from wall pulley to wedge be x. Then y+x+(√(h^2 +x^2 )) =l ⇒ v_(rel) +v_M +((xv_M )/( (√(h^2 +x^2 )))) =0 or a_(rel) +a+(((v_M ^2 +xa)(√(h^2 +x^2 ))−xv_M (((xv_M )/( (√(h^2 +x^2 ))))))/(h^2 +x^2 )) =0 so a_(rel) =−a−(((v_M ^2 +xa)(h^2 +x^2 )−(xv_M )^2 )/((h^2 +x^2 )(√(h^2 +x^2 )))) =−a−((h^2 (v_M ^2 +xa)+x^3 a)/((h^2 +x^2 )(√(h^2 +x^2 )))) =−a−((h^2 v_M ^2 )/((h^2 +x^2 )^(3/2) ))−((xa)/( (√(h^2 +x^2 )))) if only system has started now from rest and initial relation is sought for, we can have v_M =0 a_(rel) = −a(1+cos θ_0 ) the minus sign is there because ((dx/dt)) is −ve here, and ((dy/dt)) is +ve if wedge accelerates towards right. so ((d^2 x/dt^2 )) = a (i have assumed) but according to question it is ((d^2 x/dt^2 )) =−a_(given in question) . given answer must be a_(rel) =a(1+cos θ) .
$${let}\:{length}\:{of}\:{string}\:{from}\:{top}\:{of} \\ $$$${wedge}\:{to}\:{block}\:{m}\:{be}\:{y}\:{and}\:{from} \\ $$$${wall}\:{pulley}\:{to}\:{wedge}\:{be}\:{x}.\:{Then} \\ $$$${y}+{x}+\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:={l} \\ $$$$\Rightarrow\:\:\:{v}_{{rel}} +{v}_{{M}} +\frac{{xv}_{{M}} }{\:\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }}\:=\mathrm{0} \\ $$$${or}\:\:{a}_{{rel}} +{a}+\frac{\left({v}_{{M}} ^{\mathrm{2}} +{xa}\right)\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }−{xv}_{{M}} \left(\frac{{xv}_{{M}} }{\:\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }}\right)}{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${so}\:\:{a}_{{rel}} =−{a}−\frac{\left({v}_{{M}} ^{\mathrm{2}} +{xa}\right)\left({h}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)−\left({xv}_{{M}} \right)^{\mathrm{2}} }{\left({h}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:=−{a}−\frac{{h}^{\mathrm{2}} \left({v}_{{M}} ^{\mathrm{2}} +{xa}\right)+{x}^{\mathrm{3}} {a}}{\left({h}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:=−{a}−\frac{{h}^{\mathrm{2}} {v}_{{M}} ^{\mathrm{2}} }{\left({h}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }−\frac{{xa}}{\:\sqrt{{h}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$${if}\:{only}\:{system}\:{has}\:{started}\:{now} \\ $$$${from}\:{rest}\:{and}\:{initial}\:{relation} \\ $$$${is}\:{sought}\:{for},\:{we}\:{can}\:{have} \\ $$$$\:\:{v}_{{M}} =\mathrm{0} \\ $$$$\:\:\:\:\:{a}_{{rel}} \:=\:−{a}\left(\mathrm{1}+\mathrm{cos}\:\theta_{\mathrm{0}} \right) \\ $$$${the}\:{minus}\:{sign}\:{is}\:{there}\:{because} \\ $$$$\:\left(\frac{{dx}}{{dt}}\right)\:{is}\:−{ve}\:{here},\:{and}\:\left(\frac{{dy}}{{dt}}\right)\:{is}\:+{ve} \\ $$$${if}\:{wedge}\:{accelerates}\:{towards}\:{right}. \\ $$$${so}\:\:\left(\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\right)\:=\:{a}\:\:\:\:\left({i}\:{have}\:{assumed}\right) \\ $$$${but}\:{according}\:{to}\:{question}\:{it} \\ $$$${is}\:\left(\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\right)\:=−{a}_{{given}\:{in}\:{question}} \:. \\ $$$${given}\:{answer}\:{must}\:{be}\: \\ $$$$\:\:\:\:\:{a}_{{rel}} \:={a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\:. \\ $$
Commented by Tinkutara last updated on 15/Nov/17
Thank you very much Sir! Your solution is right.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$$$\mathrm{Your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{right}. \\ $$
Commented by Tinkutara last updated on 15/Nov/17

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