Question Number 173245 by pete last updated on 08/Jul/22
![The ends X and Y of an inextensible strings 27m long are fixed at two points on the same horizontal line which are 20 m apart. A particle of mass 7.5 kg is suspended from a point P on the string 12 m from X. (a) Illustrate this information in a diagram. (b) calculate, correct to two decimal places, <YXP and <XYP. (c) Find, correct to the nearest hundredth, the magnitudes of the tensions in the string. [take g=10 ms^(−2) ]](https://www.tinkutara.com/question/Q173245.png)
$$\mathrm{The}\:\mathrm{ends}\:\boldsymbol{\mathrm{X}}\:\mathrm{and}\:\boldsymbol{\mathrm{Y}}\:\mathrm{of}\:\mathrm{an}\:\mathrm{inextensible}\:\mathrm{strings}\:\mathrm{27m} \\ $$$$\mathrm{long}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{at}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{horizontal}\:\mathrm{line}\:\mathrm{which}\:\mathrm{are}\:\mathrm{20}\:\mathrm{m}\:\mathrm{apart}. \\ $$$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{7}.\mathrm{5}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{suspended} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\boldsymbol{\mathrm{P}}\:\mathrm{on}\:\mathrm{the}\:\mathrm{string}\:\mathrm{12}\:\mathrm{m}\:\mathrm{from}\:\boldsymbol{\mathrm{X}}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Illustrate}\:\mathrm{this}\:\mathrm{information}\:\mathrm{in}\:\mathrm{a}\:\mathrm{diagram}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{calculate},\:\mathrm{correct}\:\mathrm{to}\:\boldsymbol{\mathrm{two}}\:\mathrm{decimal} \\ $$$$\mathrm{places},\:<\mathrm{YXP}\:\mathrm{and}\:<\mathrm{XYP}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{Find},\:\mathrm{correct}\:\mathrm{to}\:\mathrm{the}\:\boldsymbol{\mathrm{nearest}}\:\mathrm{hundredth}, \\ $$$$\mathrm{the}\:\mathrm{magnitudes}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tensions}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{string}.\:\:\left[\mathrm{take}\:\boldsymbol{\mathrm{g}}=\mathrm{10}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$
Answered by mr W last updated on 09/Jul/22
Commented by mr W last updated on 09/Jul/22
$$\mathrm{cos}\:\alpha=\frac{\mathrm{20}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }{\mathrm{2}×\mathrm{20}×\mathrm{12}}=\frac{\mathrm{319}}{\mathrm{480}} \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{319}}{\mathrm{480}}\approx\mathrm{48}.\mathrm{35}° \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{20}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{2}×\mathrm{20}×\mathrm{15}}=\frac{\mathrm{481}}{\mathrm{600}} \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{481}}{\mathrm{600}}\approx\mathrm{36}.\mathrm{71}° \\ $$$$ \\ $$$$\frac{{T}_{\mathrm{1}} }{\mathrm{cos}\:\beta}=\frac{{T}_{\mathrm{2}} }{\mathrm{cos}\:\alpha}=\frac{{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$${mg}=\mathrm{7}.\mathrm{5}×\mathrm{10}=\mathrm{75}\:{N} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{\mathrm{cos}\:\beta\:{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\mathrm{60}.\mathrm{35}\:{N} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{\mathrm{cos}\:\alpha\:{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\mathrm{50}.\mathrm{03}\:{N} \\ $$
Commented by pete last updated on 09/Jul/22
$$\mathrm{Thank}\:\mathrm{Sir}\:\boldsymbol{\mathrm{W}} \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$