Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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) ]

$$\mathrm{The}\:\mathrm{ends}\:\boldsymbol{\mathrm{X}}\:\mathrm{and}\:\boldsymbol{\mathrm{Y}}\:\mathrm{of}\:\mathrm{an}\:\mathrm{inextensible}\:\mathrm{strings}\:\mathrm{27m} \\ $$$$\mathrm{long}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{at}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{horizontal}\:\mathrm{line}\:\mathrm{which}\:\mathrm{are}\:\mathrm{20}\:\mathrm{m}\:\mathrm{apart}. \\ $$$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{7}.\mathrm{5}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{suspended} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\boldsymbol{\mathrm{P}}\:\mathrm{on}\:\mathrm{the}\:\mathrm{string}\:\mathrm{12}\:\mathrm{m}\:\mathrm{from}\:\boldsymbol{\mathrm{X}}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Illustrate}\:\mathrm{this}\:\mathrm{information}\:\mathrm{in}\:\mathrm{a}\:\mathrm{diagram}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{calculate},\:\mathrm{correct}\:\mathrm{to}\:\boldsymbol{\mathrm{two}}\:\mathrm{decimal} \\ $$$$\mathrm{places},\:<\mathrm{YXP}\:\mathrm{and}\:<\mathrm{XYP}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{Find},\:\mathrm{correct}\:\mathrm{to}\:\mathrm{the}\:\boldsymbol{\mathrm{nearest}}\:\mathrm{hundredth}, \\ $$$$\mathrm{the}\:\mathrm{magnitudes}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tensions}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{string}.\:\:\left[\mathrm{take}\:\boldsymbol{\mathrm{g}}=\mathrm{10}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$

Answered by mr W last updated on 09/Jul/22

Commented by mr W last updated on 09/Jul/22

cos α=((20^2 +12^2 −15^2 )/(2×20×12))=((319)/(480)) α=cos^(−1) ((319)/(480))≈48.35° cos β=((20^2 +15^2 −12^2 )/(2×20×15))=((481)/(600)) β=cos^(−1) ((481)/(600))≈36.71° (T_1 /(cos β))=(T_2 /(cos α))=((mg)/(sin (α+β))) mg=7.5×10=75 N ⇒T_1 =((cos β mg)/(sin (α+β)))=60.35 N ⇒T_2 =((cos α mg)/(sin (α+β)))=50.03 N

$$\mathrm{cos}\:\alpha=\frac{\mathrm{20}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }{\mathrm{2}×\mathrm{20}×\mathrm{12}}=\frac{\mathrm{319}}{\mathrm{480}} \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{319}}{\mathrm{480}}\approx\mathrm{48}.\mathrm{35}° \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{20}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{2}×\mathrm{20}×\mathrm{15}}=\frac{\mathrm{481}}{\mathrm{600}} \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{481}}{\mathrm{600}}\approx\mathrm{36}.\mathrm{71}° \\ $$$$ \\ $$$$\frac{{T}_{\mathrm{1}} }{\mathrm{cos}\:\beta}=\frac{{T}_{\mathrm{2}} }{\mathrm{cos}\:\alpha}=\frac{{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$${mg}=\mathrm{7}.\mathrm{5}×\mathrm{10}=\mathrm{75}\:{N} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{\mathrm{cos}\:\beta\:{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\mathrm{60}.\mathrm{35}\:{N} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{\mathrm{cos}\:\alpha\:{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\mathrm{50}.\mathrm{03}\:{N} \\ $$

Commented by pete last updated on 09/Jul/22

Thank Sir W

$$\mathrm{Thank}\:\mathrm{Sir}\:\boldsymbol{\mathrm{W}} \\ $$

Commented by Tawa11 last updated on 11/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com