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Question Number 216827 by Tawa11 last updated on 22/Feb/25

Commented by Tawa11 last updated on 22/Feb/25

$$\mathrm{Reaction}\mathrm{of}\mathrm{the}7\mathrm{Kg}\mathrm{block}.$$

Answered by mr W last updated on 22/Feb/25

Commented by mr W last updated on 23/Feb/25

$$g=10m/s^2$$$$T_1=8\times 10=80N$$$$T_2=1.6\times 10=16N$$$$mg=7\times 10=70N$$$$\theta =37°$$$$\varphi =16+37=53°$$$$N=mg\cos \theta -T_1\cos \varphi$$$$=70\times 0.8-80\times 0.6=8N$$$$f=T_2+mg\sin \theta -T_1\sin \varphi$$$$=16+70\times 0.6-80\times 0.8=-6N$$$$R=\sqrt{8^2+{(-6)}^2}=10N$$

Commented by Tawa11 last updated on 22/Feb/25

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

Commented by mr W last updated on 22/Feb/25

$$rightorwrong?$$

Commented by Tawa11 last updated on 22/Feb/25

$$\mathrm{I}\mathrm{will}\mathrm{get}\mathrm{back}\mathrm{sir}.$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{answer}\mathrm{for}\mathrm{now}.$$$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by mr W last updated on 23/Feb/25

$$basicallyiknowthatmyansweris$$$$right.whatimeantiswhatdoyou$$$$thinkifitisright,inotherwords$$$$ifyoucanfollowanduseit.$$

Commented by Tawa11 last updated on 23/Feb/25

$$\mathrm{What}\mathrm{he}\mathrm{showed}\mathrm{me}\mathrm{is}\mathrm{saying}9.8\mathrm{sir}$$

Commented by Tawa11 last updated on 23/Feb/25

Commented by Tawa11 last updated on 23/Feb/25

$$\mathrm{Is}\mathrm{this}\mathrm{correct}\mathrm{sir}?$$

Commented by mr W last updated on 23/Feb/25

$$ijustcalculatedN.similarlywe$$$$cancalculatef.thenwecan$$$$calculatetheresultantR.$$$$ifhisresultisright,youshouldbe$$$$abletojudgeifyouhavefollowedmy$$$$solution.$$

Commented by Tawa11 last updated on 25/Feb/25

$$\mathrm{Yes}\mathrm{sir}.$$

Commented by Tawa11 last updated on 25/Feb/25

Commented by Tawa11 last updated on 25/Feb/25

$$\mathrm{Sir},\mathrm{you}\mathrm{gave}\mathrm{this}\mathrm{formula}\mathrm{long}\mathrm{time}\mathrm{ago}.$$$$\mathrm{I}\mathrm{am}\mathrm{trying}\mathrm{to}\mathrm{use}\mathrm{it}\mathrm{for}:$$$${2\mathrm{x}}^3-{7\mathrm{x}}^2-3\mathrm{x}+18=0$$$$\mathrm{That}\mathrm{is},{\mathrm{x}}^3-\frac{7}{2}{\mathrm{x}}^2-\frac{3}{2}\mathrm{x}+9=0$$$$\mathrm{I}\mathrm{am}\mathrm{not}\mathrm{getting}:\mathrm{x}=2,\mathrm{x}=3\mathrm{or}\mathrm{x}=-\frac{3}{2}$$$$\mathrm{Is}\mathrm{there}\mathrm{anything}\mathrm{missing}\mathrm{in}\mathrm{the}\mathrm{formular}?$$

Commented by mr W last updated on 26/Feb/25

$$no!$$

Commented by Tawa11 last updated on 26/Feb/25

$$\mathrm{I}\mathrm{am}\mathrm{getting}\frac{5}{6}\mathrm{sir}??$$

Commented by Tawa11 last updated on 26/Feb/25

$$\mathrm{x}=2\sqrt{\frac{67}{36}}\sin [\frac{1}{3}{\sin }^{-1}\left(\frac{\frac{55}{27}}{\sqrt{{\left(\frac{67}{36}\right)}^3}}\right)+0]=\frac{5}{6}$$$$\mathrm{p}=-\frac{67}{36}\mathrm{and}\mathrm{q}=\frac{55}{27}$$

Commented by mr W last updated on 26/Feb/25

$$youshouldknowforwhichequation$$$$yoursolutionis!$$$$whenyoufoundx=\frac{5}{6},thisisthe$$$$rootofthetransformedequation,$$$$thentherootoftheoriginal$$$$equationis$$$$\frac{5}{6}-(-\frac{7}{3\times 2})=2.$$

Commented by Tawa11 last updated on 26/Feb/25

$$\mathrm{Ohh},\mathrm{thanks}\mathrm{sir}.\mathrm{It}\mathrm{is}\mathrm{not}\mathrm{mention}\mathrm{in}$$$$\mathrm{the}\mathrm{formula}.$$$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by Tawa11 last updated on 26/Feb/25

$$\mathrm{Sorry}\mathrm{for}\mathrm{disturbing}\mathrm{sir}.$$$$\left(\frac{-7}{3\times 2}\right)??$$

Commented by mr W last updated on 26/Feb/25

$$thenyouhavenotunderstoodthe$$$$formulaoryouappliedtheformula$$$$wrongly.showhowyoucalculated!$$

Commented by Tawa11 last updated on 26/Feb/25

$$\mathrm{Ohhh},\mathrm{I}\mathrm{understand}\mathrm{sir}.$$$$\mathrm{x}=\mathrm{s}-\frac{\mathrm{a}}{3}$$

Commented by Tawa11 last updated on 26/Feb/25

$$\mathrm{So},\mathrm{my}\frac{5}{6}\mathrm{is}\mathrm{s},\mathrm{to}\mathrm{find}\mathrm{x}$$$$\mathrm{x}=\frac{5}{6}-(-\frac{7}{3\times 2})$$$$\mathrm{I}\mathrm{understand}\mathrm{very}\mathrm{well}\mathrm{now}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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