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Question Number 20035 by Tinkutara last updated on 20/Aug/17

$$\mathrm{In}\mathrm{the}\mathrm{following}\mathrm{cases},\mathrm{find}\mathrm{out}\mathrm{the}$$$$\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{wedge}\mathrm{and}\mathrm{the}\mathrm{block},$$$$\mathrm{if}\mathrm{an}\mathrm{external}\mathrm{force}F\mathrm{is}\mathrm{applied}\mathrm{as}$$$$\mathrm{shown}.(\mathrm{Both}\mathrm{pulleys}\mathrm{and}\mathrm{strings}\mathrm{are}$$$$\mathrm{ideal})$$

Commented by Tinkutara last updated on 20/Aug/17

Answered by ajfour last updated on 21/Aug/17

$$let\mathit{A}beaccelerationofwedge,$$$$and\mathit{a}acceleeationofblock.$$$$Tensionin\left(light\right)string=F$$$$CaseII:$$$$F-F\cos \theta =MA$$$$\Rightarrow A=\frac{F(1-\cos \theta )}{M}$$$$-F=ma\Rightarrow a=-\frac{F}{m}$$$$\mid a\mid =\frac{F}{m}towardsleftforallcases.$$$$CaseI:\theta =0°soA=0$$$$CaseIII:\theta =90°soA=\frac{F}{M}.$$

Commented by Tinkutara last updated on 21/Aug/17

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

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