Question-217521

Tinku Tara March 15, 2025 Others 0 Comments

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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

Σ F = 0 \Rightarrow

{\begin{cases} [A_{x} - T c o s 20 = 0] \times g \\ [A_{y} - T s i n 20 - m = 0] \times g \end{cases}

Σ M_{A} = 0

[T s i n 20 \times l s i n 45 + m \times \frac{l}{2} s i n 45 = T c o s 20 \times l c o s 45] \times 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{D C}{C B} = \frac{\sin 25 °}{\sin (45 ° + 25 °)}

\frac{D C}{A C} = \frac{\sin (45 ° - α)}{\sin α}

\frac{\sin (45 ° - α)}{\sin α} = \frac{\sin 25 °}{\sin (45 ° + 25 °)}

\frac{1}{\tan α} - 1 = \frac{\sqrt{2} \sin 25 °}{\sin 70 °}

\tan α = \frac{1}{1 + \frac{\sqrt{2} \sin 25 °}{\sin 70 °}} \Rightarrow α \approx 31.434 °

α + β = 45 ° + 25 ° = 70 °

\frac{R}{\sin (α + β)} = \frac{T}{\sin α} = \frac{m g}{\sin β}

m g = 10 \times 10 = 100 N

T = \frac{\sin α m g}{\sin β} = \frac{\sin α m g}{\sin (70 ° - α)} \approx 83.66 N

R = \frac{\sin (α + β) m g}{\sin β} = \frac{\sin 70 ° m g}{\sin (70 ° - α)} \approx 150.74 N

Commented by Tawa11 last updated on 15/Mar/25

Thanks sir. I really appreciate.

Thanks sir .

I really appreciate .

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

is the answer correct?

i s t h e a n s w e r c o r r e c t ?

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

They drew this .

T = 5 . 9 Kg

R_{AH} = 5.54 kg

R_{AV} = 15.89 kg

Commented by Tawa11 last updated on 15/Mar/25

They did not show any workings

They did not show any workings

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

you can not follow my solution and judge if my answer is correct?^

y o u c a n n o t f o l l o w m y s o l u t i o n a n d

j u d g e i f m y a n s w e r i s c o r r e c t \overset{}{?}

Commented by Tawa11 last updated on 15/Mar/25

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

I can sir .

I thought you only want to see

what they did sir .

Commented by Tawa11 last updated on 15/Mar/25

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

I mean I am only showing you

ehat they wrote since you asked 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.

f r o m y o u r r e s p o n d i c a n n o t k n o w

i f m y a n s w e r i s c o r r e c t o r i f y o u

h a v e c h e c k e d i t a t a l l, t h e r e f o r e i

a s k e d .

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

their answer differs very much from mine. hopefully you can check which is correct.

t h e i r a n s w e r d i f f e r s v e r y m u c h

f r o m m i n e . h o p e f u l l y y o u c a n c h e c k

w h i c h i s c o r r e c t .

Commented by Tawa11 last updated on 15/Mar/25

Commented by Tawa11 last updated on 15/Mar/25

Sir, please check this.

Sir, please check 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.

i a s k e d y o u t o c h e c k . w h e n i p o s t a

s o l u t i o n, c e r t a i n l y i t h i n k i t i s n o t

w r o n g . w h e n y o u c a n f o l l o w m y

s o l u t i o n a n d f i n d m y a n s w e r i s

c o r r e c t, t h e n t h e a n s w e r i n t h e b o o k

m u s t b e w r o n g .

Commented by Tawa11 last updated on 15/Mar/25

Yes sir. True.

Yes sir . True .

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

i thought you might see the silly mistake in the book:

i t h o u g h t y o u m i g h t s e e t h e s i l l y

m i s t a k e i n t h e b o o k :

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

\sin 45 = 0.7071 \approx 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 \approx 5.9

Commented by Tawa11 last updated on 16/Mar/25

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

Sir, or you saw error that will make their answers

as 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 {d i d n}^{'} t m e a n t h a t \sin 45 ° = 0.71 i s t h e

e r r o r, b u t \sin 45 ° i t s e l f i s t h e e r r o r .

A E i s a l r e a d y t h e a r m l e n g t h o f t h e

w e i g h t f o r c e . \underset{―}{\sin 45 ° \times A E} i s t h e n

t o t a l l y w r o n g . j u s t t h i n k!

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.

Seen sir .

They {don}^{'} t need to resolve it again .

It should be 10 \times 1.41 only .

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