Question-217521

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

Commented by mahdipoor last updated on 16/Mar/25

ΣF=0 ⇒ { (([A_x −Tcos20=0]×g)),(([A_y −Tsin20−m=0]×g)) :} ΣM_A =0 [Tsin20×lsin45+m×(l/2)sin45=Tcos20×lcos45]×g

$$\Sigma{F}=\mathrm{0}\:\Rightarrow \\ $$$$\begin{cases}{\left[{A}_{{x}} −{Tcos}\mathrm{20}=\mathrm{0}\right]×{g}}\\{\left[{A}_{{y}} −{Tsin}\mathrm{20}−{m}=\mathrm{0}\right]×{g}}\end{cases} \\ $$$$\Sigma{M}_{{A}} =\mathrm{0}\:\: \\ $$$$\left[{Tsin}\mathrm{20}×{lsin}\mathrm{45}+{m}×\frac{{l}}{\mathrm{2}}{sin}\mathrm{45}={Tcos}\mathrm{20}×{lcos}\mathrm{45}\right]×{g} \\ $$$$ \\ $$

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

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

((DC)/(CB))=((sin 25°)/(sin (45°+25°))) ((DC)/(AC))=((sin (45°−α))/(sin α)) ((sin (45°−α))/(sin α))=((sin 25°)/(sin (45°+25°))) (1/(tan α))−1=(((√2) sin 25°)/(sin 70°)) tan α=(1/(1+(((√2) sin 25°)/(sin 70°)))) ⇒α≈31.434° α+β=45°+25°=70° (R/(sin (α+β)))=(T/(sin α))=((mg)/(sin β)) mg=10×10=100 N T=((sin α mg)/(sin β))=((sin α mg)/(sin (70°−α)))≈83.66 N R=((sin (α+β) mg)/(sin β))=((sin 70° mg)/(sin (70°−α)))≈150.74 N

$$\frac{{DC}}{{CB}}=\frac{\mathrm{sin}\:\mathrm{25}°}{\mathrm{sin}\:\left(\mathrm{45}°+\mathrm{25}°\right)} \\ $$$$\frac{{DC}}{{AC}}=\frac{\mathrm{sin}\:\left(\mathrm{45}°−\alpha\right)}{\mathrm{sin}\:\alpha} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{45}°−\alpha\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{sin}\:\mathrm{25}°}{\mathrm{sin}\:\left(\mathrm{45}°+\mathrm{25}°\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\mathrm{1}=\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{25}°}{\mathrm{sin}\:\mathrm{70}°} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{1}+\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{25}°}{\mathrm{sin}\:\mathrm{70}°}}\:\Rightarrow\alpha\approx\mathrm{31}.\mathrm{434}° \\ $$$$\alpha+\beta=\mathrm{45}°+\mathrm{25}°=\mathrm{70}° \\ $$$$\frac{{R}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{{T}}{\mathrm{sin}\:\alpha}=\frac{{mg}}{\mathrm{sin}\:\beta} \\ $$$${mg}=\mathrm{10}×\mathrm{10}=\mathrm{100}\:{N} \\ $$$${T}=\frac{\mathrm{sin}\:\alpha\:{mg}}{\mathrm{sin}\:\beta}=\frac{\mathrm{sin}\:\alpha\:{mg}}{\mathrm{sin}\:\left(\mathrm{70}°−\alpha\right)}\approx\mathrm{83}.\mathrm{66}\:{N} \\ $$$${R}=\frac{\mathrm{sin}\:\left(\alpha+\beta\right)\:{mg}}{\mathrm{sin}\:\beta}=\frac{\mathrm{sin}\:\mathrm{70}°\:{mg}}{\mathrm{sin}\:\left(\mathrm{70}°−\alpha\right)}\approx\mathrm{150}.\mathrm{74}\:{N} \\ $$

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

$${is}\:{the}\:{answer}\:{correct}? \\ $$

Commented by Tawa11 last updated on 15/Mar/25

Commented by Tawa11 last updated on 15/Mar/25

They drew this. T = 5.9Kg R_(AH) = 5.54 kg R_(AV) = 15.89 kg

$$\mathrm{They}\:\mathrm{drew}\:\mathrm{this}. \\ $$$$\mathrm{T}\:\:=\:\:\mathrm{5}.\mathrm{9Kg} \\ $$$$\:\mathrm{R}_{\mathrm{AH}} \:=\:\:\mathrm{5}.\mathrm{54}\:\mathrm{kg} \\ $$$$\:\mathrm{R}_{\mathrm{AV}} \:\:=\:\:\mathrm{15}.\mathrm{89}\:\mathrm{kg} \\ $$

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

$${you}\:{can}\:{not}\:{follow}\:{my}\:{solution}\:{and} \\ $$$${judge}\:{if}\:{my}\:{answer}\:{is}\:{correct}\overset{} {?} \\ $$

Commented by Tawa11 last updated on 15/Mar/25

I can sir. I thought you only want to see what they did sir.

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

Commented by Tawa11 last updated on 15/Mar/25

I mean I am only showing you ehat they wrote since you asked sir

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

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

$${from}\:{your}\:{respond}\:{i}\:{can}\:{not}\:{know} \\ $$$${if}\:{my}\:{answer}\:{is}\:{correct}\:{or}\:{if}\:{you} \\ $$$${have}\:{checked}\:{it}\:{at}\:{all},\:{therefore}\:{i} \\ $$$${asked}.\: \\ $$

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

$${their}\:{answer}\:{differs}\:{very}\:{much} \\ $$$${from}\:{mine}.\:{hopefully}\:{you}\:{can}\:{check} \\ $$$${which}\:{is}\:{correct}. \\ $$

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

$${i}\:{asked}\:{you}\:{to}\:{check}.\:{when}\:{i}\:{post}\:{a} \\ $$$${solution},\:{certainly}\:{i}\:{think}\:{it}\:{is}\:{not} \\ $$$${wrong}.\:{when}\:{you}\:{can}\:{follow}\:{my} \\ $$$${solution}\:{and}\:{find}\:{my}\:{answer}\:{is} \\ $$$${correct},\:{then}\:{the}\:{answer}\:{in}\:{the}\:{book} \\ $$$${must}\:{be}\:{wrong}. \\ $$

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

$${i}\:{thought}\:{you}\:{might}\:{see}\:{the}\:{silly} \\ $$$${mistake}\:{in}\:{the}\:{book}: \\ $$

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

Commented by Tawa11 last updated on 16/Mar/25

sin45 = 0.7071 ≈ 0.71 I saw it sir, I thought the approximation doesn′t change their answer. When I used 0.7071, the result is still T ≈ 5.9

$$\mathrm{sin45}\:\:=\:\:\mathrm{0}.\mathrm{7071}\:\:\:\approx\:\:\:\mathrm{0}.\mathrm{71} \\ $$$$\mathrm{I}\:\mathrm{saw}\:\mathrm{it}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{thought}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$\mathrm{doesn}'\mathrm{t}\:\mathrm{change}\:\mathrm{their}\:\mathrm{answer}. \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{used}\:\:\mathrm{0}.\mathrm{7071},\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{still}\:\:\mathrm{T}\:\:\approx\:\:\mathrm{5}.\mathrm{9} \\ $$

Commented by Tawa11 last updated on 16/Mar/25

Sir, or you saw error that will make their answers as yours.

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

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

i didn′t mean that sin 45°=0.71 is the error, but sin 45° itself is the error. AE is already the arm length of the weight force. sin 45° ×AE is then totally wrong. just think!

$${i}\:{didn}'{t}\:{mean}\:{that}\:\mathrm{sin}\:\mathrm{45}°=\mathrm{0}.\mathrm{71}\:{is}\:{the} \\ $$$${error},\:{but}\:\mathrm{sin}\:\mathrm{45}°\:{itself}\:{is}\:{the}\:{error}.\: \\ $$$${AE}\:{is}\:{already}\:{the}\:{arm}\:{length}\:{of}\:{the}\: \\ $$$${weight}\:{force}.\:\underline{\mathrm{sin}\:\mathrm{45}°\:×{AE}}\:{is}\:{then}\: \\ $$$${totally}\:{wrong}.\:{just}\:{think}! \\ $$

Commented by Tawa11 last updated on 16/Mar/25

Seen sir. They don′t need to resolve it again. It should be 10 × 1.41 only.

$$\mathrm{Seen}\:\mathrm{sir}. \\ $$$$\mathrm{They}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{to}\:\mathrm{resolve}\:\mathrm{it}\:\mathrm{again}. \\ $$$$\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\:\:\:\mathrm{10}\:\:×\:\:\mathrm{1}.\mathrm{41}\:\:\:\:\mathrm{only}. \\ $$

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