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Question Number 217521 by Tawa11 last updated on 15/Mar/25

Commented by mahdipoor last updated on 16/Mar/25

$$\mathrm{\Sigma }F=0\Rightarrow$$$$\{\begin{array}{l}[A_x-Tcos20=0]\times g\\ [A_y-Tsin20-m=0]\times g\end{array}$$$$\mathrm{\Sigma }{M}_A=0$$$$[Tsin20\times lsin45+m\times \frac{l}{2}sin45=Tcos20\times lcos45]\times g$$$$$$

Answered by mr W last updated on 15/Mar/25

Commented by mr W last updated on 15/Mar/25

$$\frac{DC}{CB}=\frac{\sin 25°}{\sin (45°+25°)}$$$$\frac{DC}{AC}=\frac{\sin (45°-\alpha )}{\sin \alpha }$$$$\frac{\sin (45°-\alpha )}{\sin \alpha }=\frac{\sin 25°}{\sin (45°+25°)}$$$$\frac{1}{\tan \alpha }-1=\frac{\sqrt{2}\sin 25°}{\sin 70°}$$$$\tan \alpha =\frac{1}{1+\frac{\sqrt{2}\sin 25°}{\sin 70°}}\Rightarrow \alpha \approx 31.434°$$$$\alpha +\beta =45°+25°=70°$$$$\frac{R}{\sin (\alpha +\beta )}=\frac{T}{\sin \alpha }=\frac{mg}{\sin \beta }$$$$mg=10\times 10=100N$$$$T=\frac{\sin \alpha mg}{\sin \beta }=\frac{\sin \alpha mg}{\sin (70°-\alpha )}\approx 83.66N$$$$R=\frac{\sin (\alpha +\beta )mg}{\sin \beta }=\frac{\sin 70°mg}{\sin (70°-\alpha )}\approx 150.74N$$

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by mr W last updated on 15/Mar/25

$$istheanswercorrect?$$

Commented by Tawa11 last updated on 15/Mar/25

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{They}\mathrm{drew}\mathrm{this}.$$$$\mathrm{T}=5.9\mathrm{Kg}$$$${\mathrm{R}}_{\mathrm{AH}}=5.54\mathrm{kg}$$$${\mathrm{R}}_{\mathrm{AV}}=15.89\mathrm{kg}$$

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{They}\mathrm{did}\mathrm{not}\mathrm{show}\mathrm{any}\mathrm{workings}$$

Commented by mr W last updated on 15/Mar/25

$$youcannotfollowmysolutionand$$$$judgeifmyansweriscorrect\stackrel{}{?}$$

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{I}\mathrm{can}\mathrm{sir}.$$$$\mathrm{I}\mathrm{thought}\mathrm{you}\mathrm{only}\mathrm{want}\mathrm{to}\mathrm{see}$$$$\mathrm{what}\mathrm{they}\mathrm{did}\mathrm{sir}.$$

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{I}\mathrm{mean}\mathrm{I}\mathrm{am}\mathrm{only}\mathrm{showing}\mathrm{you}$$$$\mathrm{ehat}\mathrm{they}\mathrm{wrote}\mathrm{since}\mathrm{you}\mathrm{asked}\mathrm{sir}$$

Commented by mr W last updated on 15/Mar/25

$$fromyourrespondicannotknow$$$$ifmyansweriscorrectorifyou$$$$havecheckeditatall,thereforei$$$$asked.$$

Commented by mr W last updated on 15/Mar/25

$$theiranswerdiffersverymuch$$$$frommine.hopefullyyoucancheck$$$$whichiscorrect.$$

Commented by Tawa11 last updated on 15/Mar/25

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{Sir},\mathrm{please}\mathrm{check}\mathrm{this}.$$

Commented by mr W last updated on 15/Mar/25

$$iaskedyoutocheck.wheniposta$$$$solution,certainlyithinkitisnot$$$$wrong.whenyoucanfollowmy$$$$solutionandfindmyansweris$$$$correct,thentheanswerinthebook$$$$mustbewrong.$$

Commented by Tawa11 last updated on 15/Mar/25

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

Commented by mr W last updated on 16/Mar/25

$$ithoughtyoumightseethesilly$$$$mistakeinthebook:$$

Commented by mr W last updated on 16/Mar/25

Commented by Tawa11 last updated on 16/Mar/25

$$\sin 45=0.7071\approx 0.71$$$$\mathrm{I}\mathrm{saw}\mathrm{it}\mathrm{sir},\mathrm{I}\mathrm{thought}\mathrm{the}\mathrm{approximation}$$$${\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{change}\mathrm{their}\mathrm{answer}.$$$$\mathrm{When}\mathrm{I}\mathrm{used}0.7071,\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{still}\mathrm{T}\approx 5.9$$

Commented by Tawa11 last updated on 16/Mar/25

$$\mathrm{Sir},\mathrm{or}\mathrm{you}\mathrm{saw}\mathrm{error}\mathrm{that}\mathrm{will}\mathrm{make}\mathrm{their}\mathrm{answers}$$$$\mathrm{as}\mathrm{yours}.$$

Commented by mr W last updated on 16/Mar/25

$$i{didn}^{\prime }tmeanthat\sin 45°=0.71isthe$$$$error,but\sin 45°itselfistheerror.$$$$AEisalreadythearmlengthofthe$$$$weightforce.\underset{―}{\sin 45°\times AE}isthen$$$$totallywrong.justthink!$$

Commented by Tawa11 last updated on 16/Mar/25

$$\mathrm{Seen}\mathrm{sir}.$$$$\mathrm{They}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{need}\mathrm{to}\mathrm{resolve}\mathrm{it}\mathrm{again}.$$$$\mathrm{It}\mathrm{should}\mathrm{be}10\times 1.41\mathrm{only}.$$

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