Question Number 40928 by rahul 19 last updated on 29/Jul/18
Answered by MrW3 last updated on 29/Jul/18
$${length}\:{of}\:{rope}\:{l}=\mathrm{2}\pi{r}=\pi{R} \\ $$$${let}\:\lambda=\frac{{W}}{{l}}=\frac{{W}}{\pi{R}} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{3}}=\mathrm{60}° \\ $$$${p}={pressure}\:{between}\:{rope}\:{and}\:{sphere} \\ $$$${in}\:{the}\:{plane}\:{of}\:{rope} \\ $$$${p}=\frac{\lambda}{\mathrm{tan}\:\theta} \\ $$$${T}={tension}\:{in}\:{rope} \\ $$$${T}={rp}=\frac{{r}\lambda}{\mathrm{tan}\:\theta}=\frac{{RW}}{\mathrm{2tan}\:\theta\:\pi{R}}=\frac{{W}}{\mathrm{2}\pi\sqrt{\mathrm{3}}}=\frac{{W}}{\pi\sqrt{\mathrm{12}}}=\frac{{W}}{\pi\sqrt{{K}}} \\ $$$$\Rightarrow{K}=\mathrm{12} \\ $$$$\Rightarrow\frac{{K}}{\mathrm{6}}=\mathrm{2} \\ $$
Commented by MrW3 last updated on 29/Jul/18
Commented by MrW3 last updated on 30/Jul/18
![at each contact point between rope and sphere two extern forces are acting: weight of rope: λ [N/m] normal force: n [N/m] the resultant of both is p [N/m] which presses the rope outwards and causes a tension force in the rope since the rope is a closed circle.](https://www.tinkutara.com/question/Q40959.png)
$${at}\:{each}\:{contact}\:{point}\:{between}\:{rope}\:{and} \\ $$$${sphere}\:{two}\:{extern}\:{forces}\:{are}\:{acting}: \\ $$$${weight}\:{of}\:{rope}:\:\lambda\:\left[{N}/{m}\right] \\ $$$${normal}\:{force}:\:{n}\:\left[{N}/{m}\right] \\ $$$${the}\:{resultant}\:{of}\:{both}\:{is}\:{p}\:\left[{N}/{m}\right]\:{which} \\ $$$${presses}\:{the}\:{rope}\:{outwards}\:{and}\:{causes} \\ $$$${a}\:{tension}\:{force}\:{in}\:{the}\:{rope}\:{since}\:{the} \\ $$$${rope}\:{is}\:{a}\:{closed}\:{circle}. \\ $$
Commented by rahul 19 last updated on 30/Jul/18
thank u so much sir��
Commented by MrW3 last updated on 30/Jul/18
