Question Number 20035 by Tinkutara last updated on 20/Aug/17
$$\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases},\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{and}\:\mathrm{the}\:\mathrm{block}, \\ $$$$\mathrm{if}\:\mathrm{an}\:\mathrm{external}\:\mathrm{force}\:{F}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{as} \\ $$$$\mathrm{shown}.\:\left(\mathrm{Both}\:\mathrm{pulleys}\:\mathrm{and}\:\mathrm{strings}\:\mathrm{are}\right. \\ $$$$\left.\mathrm{ideal}\right) \\ $$
Commented by Tinkutara last updated on 20/Aug/17
Answered by ajfour last updated on 21/Aug/17
$${let}\:\boldsymbol{{A}}\:{be}\:{acceleration}\:{of}\:{wedge}, \\ $$$${and}\:\boldsymbol{{a}}\:{acceleeation}\:{of}\:{block}. \\ $$$${Tension}\:{in}\:\left({light}\right)\:{string}\:={F} \\ $$$${Case}\:{II}: \\ $$$$\:\:\:\:\:{F}−{F}\mathrm{cos}\:\theta={MA} \\ $$$$\Rightarrow\:\:\:\:\:{A}=\frac{{F}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{{M}} \\ $$$$\:−{F}={ma}\:\:\:\Rightarrow\:\:{a}=−\frac{{F}}{{m}}\:\: \\ $$$$\:\mid{a}\mid=\frac{{F}}{{m}}\:\:{towards}\:{left}\:{for}\:{all}\:{cases}. \\ $$$${Case}\:{I}:\:\:\:\:\theta=\mathrm{0}°\:\:{so}\:\:\:{A}=\mathrm{0} \\ $$$${Case}\:{III}:\:\:\theta=\mathrm{90}°\:\:{so}\:\:{A}=\frac{{F}}{{M}}\:. \\ $$
Commented by Tinkutara last updated on 21/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$