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Question Number 20038 by Tinkutara last updated on 20/Aug/17

$$\mathrm{In}\mathrm{the}\mathrm{figure}\mathrm{shown},m\mathrm{slides}\mathrm{on}$$$$\mathrm{inclined}\mathrm{surface}\mathrm{of}\mathrm{wedge}M.\mathrm{If}\mathrm{velocity}$$$$\mathrm{of}\mathrm{wedge}\mathrm{at}\mathrm{any}\mathrm{instant}\mathrm{be}v,\mathrm{find}$$$$\mathrm{velocity}\mathrm{of}m\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{ground}.$$

Commented by Tinkutara last updated on 20/Aug/17

Answered by ajfour last updated on 21/Aug/17

$$v=-\frac{dy}{dt},u_{rel}=\frac{dx}{dt}$$$$x+y=l\left(isconstant\right)$$$$so\frac{dx}{dt}=-\frac{dy}{dt}\Rightarrow u_{rel}=v$$$$u=\sqrt{u_x^2+u_y^2}$$$$u_x=v-u_{rel}\cos 60°=v-\frac{v}{2}=\frac{v}{2}$$$$u_y=v_{rel}\sin 60°=\frac{v\sqrt{3}}{2}$$$$Hence{block}^{\prime }sspeedwithrespect$$$$togroundu=\sqrt{{\left(\frac{v}{2}\right)}^2+{\left(\frac{v\sqrt{3}}{2}\right)}^2}=v.$$$$$$

Commented by Tinkutara last updated on 21/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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