Question Number 111282 by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{500g}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\: \\ $$$$\mathrm{plane}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{30}° \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{What}\:\mathrm{force}\:\mathrm{is} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{acting}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{acting}\:\mathrm{horizontally}\:\mathrm{is}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{hold}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:\mathrm{rest}\:\left(\mathrm{g}=\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$
Answered by ajfour last updated on 03/Sep/20
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Solution}\:\mathrm{please}? \\ $$
Commented by ajfour last updated on 03/Sep/20
$$\left({i}\right)\:\:\:{F}={mg}\mathrm{sin}\:\mathrm{30}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{2}.\mathrm{5}{N} \\ $$$$\left({ii}\right)\:\:\:{F}=\:{N}\mathrm{sin}\:\mathrm{30}° \\ $$$$\:\:\:\:\:\:\:\:\:\:{N}\mathrm{cos}\:\mathrm{30}°\:=\:{mg}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}\:=\:\mathrm{5} \\ $$$$\Rightarrow\:\:{F}=\frac{\mathrm{5}}{\mathrm{cos}\:\mathrm{30}°}×\mathrm{sin}\:\mathrm{30}°\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\:{newtons}. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Kindly}\:\mathrm{explain}\:\mathrm{no}.\:\mathrm{2}\:\mathrm{in}\:\mathrm{details}\:\mathrm{please}. \\ $$$$\mathrm{Thanks}. \\ $$
Commented by ajfour last updated on 03/Sep/20
$${force}\:{balance}\:{along}\:{vertical}\:{and} \\ $$$${horizontal}\:{gives}: \\ $$$${N}\mathrm{cos}\:\mathrm{30}°={mg} \\ $$$${N}\mathrm{sin}\:\mathrm{30}°={F} \\ $$$${dividing} \\ $$$$\frac{{F}}{{mg}}\:=\:\mathrm{tan}\:\mathrm{30}°\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{F}=\frac{{mg}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\left(\mathrm{1}/\mathrm{2}\right)×\mathrm{10}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}}\:. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{diagram} \\ $$$$\mathrm{for}\:\left(\mathrm{ii}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Is}\:\mathrm{that}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{position}\:\mathrm{for}\:\mathrm{F}? \\ $$
Commented by ajfour last updated on 03/Sep/20
$${yes}\:{it}'{ll}\:{be}\:{like}\:{i}'{ve}\:{drawn}. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Why}\:\mathrm{is}\:\mathrm{that}\:\mathrm{so}? \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{horizontal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{like}\:\mathrm{that}? \\ $$
Commented by ajfour last updated on 03/Sep/20
$${yes}\:{the}\:{force}\:{that}\:{is}\:{horizontal} \\ $$$${and}\:{needed}\:{to}\:{hold}\:{the}\:{particle} \\ $$$${against}\:{the}\:{frictionless}\:{incline} \\ $$$${will}\:{be}\:{directed}\:{suchwise}\:{and} \\ $$$${F}={mg}\mathrm{tan}\:\theta. \\ $$$${horizontal}\:{means}\:{parallel}\:{to}\:{flat} \\ $$$${ground},\:{you}\:{should}\:{not}\:{say}\:{horizontal} \\ $$$${to}\:{plane}\:\left({inclined}\right)! \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{What}\:\mathrm{about}\:\mathrm{the}\:\mathrm{dotted}\:\mathrm{line}\:\mathrm{opposite} \\ $$$$\mathrm{to}\:\mathrm{it}? \\ $$
Commented by ajfour last updated on 03/Sep/20
$${the}\:{component}\:{of}\:{Normal}\:{reaction} \\ $$$${along}\:{that}\:{dotted}\:{line}\:{is}\:{N}\mathrm{sin}\:\mathrm{30}° \\ $$$${which}\:{the}\:{horizontal}\:{forcw}\:{F}\:{is} \\ $$$${balancing}. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{So}\:\mathrm{which}\:\mathrm{angle}\:\mathrm{are}\:\mathrm{we}\:\mathrm{using}? \\ $$
Commented by ajfour last updated on 03/Sep/20
Commented by ajfour last updated on 03/Sep/20
$${N}\mathrm{cos}\:\left(\mathrm{90}°−\mathrm{30}°\right)={F} \\ $$$$\Rightarrow\:{N}\mathrm{sin}\:\mathrm{30}°={F} \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{that}\:\mathrm{30}°\:\mathrm{close}\:\mathrm{to}\:\mathrm{N}? \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{Thanks}. \\ $$
Commented by ajfour last updated on 03/Sep/20
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Yes}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}.\:\mathrm{I}\:\mathrm{really} \\ $$$$\mathrm{appreciate}\:\mathrm{your}\:\mathrm{efforts}. \\ $$