Question Number 22499 by Tinkutara last updated on 19/Oct/17
$$\mathrm{A}\:\mathrm{cylinder}\:\mathrm{of}\:\mathrm{weight}\:\mathrm{200}\:\mathrm{N}\:\mathrm{is}\:\mathrm{supported} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light} \\ $$$$\mathrm{cord}\:{AC}\:\mathrm{and}\:\mathrm{pulled}\:\mathrm{with}\:\mathrm{force}\:\mathrm{of}\:\mathrm{400}\:\mathrm{N}. \\ $$$$\mathrm{The}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{at}\:{B}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by Tinkutara last updated on 19/Oct/17
Answered by ajfour last updated on 19/Oct/17
$${T}\mathrm{cos}\:\mathrm{37}\:°=\mathrm{400}{N} \\ $$$$\Rightarrow\:{T}=\mathrm{500}{N} \\ $$$${Let}\:{normal}\:{reaction}\:{at}\:{B}\:{is}\:{R}. \\ $$$${R}={Mg}+{T}\mathrm{sin}\:\mathrm{37}\:° \\ $$$$\:\:\:=\mathrm{200}{N}+\mathrm{300}{N}\:=\:\mathrm{500}{N}\:=\frac{\mathrm{1}}{\mathrm{2}}{kN}\:. \\ $$
Commented by Tinkutara last updated on 19/Oct/17
$${Why}\:{T}\mathrm{cos}\:\mathrm{37}°=\mathrm{400}{N}? \\ $$
Commented by ajfour last updated on 19/Oct/17
$${Cord}\:{is}\:{pivoted}\:{at}\:{A}.\:{cylinder} \\ $$$${doesn}'\:{t}\:{move}\:{forward}. \\ $$$$\Sigma{F}_{{x}} ={ma}_{{x}} =\mathrm{0} \\ $$$$\mathrm{400}{N}−{T}\mathrm{cos}\:\mathrm{37}\:°=\mathrm{0} \\ $$
Commented by Tinkutara last updated on 19/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$