Question-217506

Tinku Tara March 15, 2025 Others 0 Comments

Facebook Tweet Pin

Question Number 217506 by Tawa11 last updated on 15/Mar/25

Commented by Tawa11 last updated on 15/Mar/25

2.5m/s^2 ?

2 . 5 m / s^{2} ?

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

yes, if taken g=10 m/s^2 . what are the accelerations of the pulleys?

y e s, i f t a k e n g = 10 m / s^{2} .

w h a t a r e t h e a c c e l e r a t i o n s o f t h e

p u l l e y s ?

Commented by Tawa11 last updated on 15/Mar/25

Sir, I only solve acceleration of the block. Show me the acceleration of pulleys sir.

Sir, I only solve acceleration of the block .

Show me the acceleration of pulleys sir .

Commented by Tawa11 last updated on 15/Mar/25

sir. Q217504, am I correct?

sir .

Q 217504, am I correct ?

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

i don′t understand the question.

i {d o n}^{'} t u n d e r s t a n d t h e q u e s t i o n .

Commented by Tawa11 last updated on 15/Mar/25

Hahahahahaha. I think the question is missing a parameter.

Hahahahahaha .

I think the question is missing a

parameter .

Commented by Tawa11 last updated on 15/Mar/25

Thanks sir.

Thanks sir .

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

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

2F_1 =F=100 ⇒F_1 =50 N 2F_2 =F_1 =50 ⇒F_2 =25 N with g=10 m/s^2 F_1 −m_1 g=m_1 a_1 a_1 =((F_1 −m_1 g)/m_1 ) a_1 =((50−1×10)/1)=40 m/s^2 (↑) similarly a_2 =((25−2×10)/2)=2.5 m/s^2 (↑) a_3 =((25−3×10)/3)=−(5/3) m/s^2 (↓) A_2 −a_2 =a_3 −A_2 ⇒A_2 =((a_2 +a_3 )/2)=((2.5−(5/3))/2)=(5/(12)) m/s^2 (↑) A_1 −a_1 =A_2 −A_1 ⇒A_1 =((a_1 +A_2 )/2)=((40+(5/(12)))/2)=20(5/(24)) m/s^2 (↑)

2 F_{1} = F = 100 \Rightarrow F_{1} = 50 N

2 F_{2} = F_{1} = 50 \Rightarrow F_{2} = 25 N

w i t h g = 10 m / s^{2}

F_{1} - m_{1} g = m_{1} a_{1}

a_{1} = \frac{F_{1} - m_{1} g}{m_{1}}

a_{1} = \frac{50 - 1 \times 10}{1} = 40 m / s^{2} (↑)

s i m i l a r l y

a_{2} = \frac{25 - 2 \times 10}{2} = 2.5 m / s^{2} (↑)

a_{3} = \frac{25 - 3 \times 10}{3} = - \frac{5}{3} m / s^{2} (↓)

A_{2} - a_{2} = a_{3} - A_{2}

\Rightarrow A_{2} = \frac{a_{2} + a_{3}}{2} = \frac{2.5 - \frac{5}{3}}{2} = \frac{5}{12} m / s^{2} (↑)

A_{1} - a_{1} = A_{2} - A_{1}

\Rightarrow A_{1} = \frac{a_{1} + A_{2}}{2} = \frac{40 + \frac{5}{12}}{2} = 20 \frac{5}{24} m / s^{2} (↑)

Commented by Tawa11 last updated on 15/Mar/25

God bless you sir. I really appreciate you sir.

God bless you sir .

I really appreciate you sir .

Commented by Tawa11 last updated on 15/Mar/25

For mass 3kg = 5ms^(−1) For mas 2kg = 2.5 ms^(−1) for mass 1kg = 3.75ms^(−1)

For mass 3 kg = {5 ms}^{- 1}

For mas 2 kg = 2.5 {ms}^{- 1}

for mass 1 kg = 3 . {75 ms}^{- 1}

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

wrong!

w r o n g!

Commented by Tawa11 last updated on 15/Mar/25

I will try more

I will try more

Commented by Tawa11 last updated on 16/Mar/25

1.8 ms^(− 2) is this correct sir. for the 3kg, 2kg and 1kg pulleys

1.8 {ms}^{- 2}

is this correct sir .

for the 3 kg, 2 kg and 1 kg pulleys

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

no! how can they all have the same acceleration? they are not fixed with each other.

n o! h o w c a n t h e y a l l h a v e t h e s a m e

a c c e l e r a t i o n ? t h e y a r e n o t f i x e d

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

Commented by Tawa11 last updated on 16/Mar/25

noted sir.

noted sir .

Commented by Tawa11 last updated on 16/Mar/25

Please solve sir.

Please solve sir .

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

i have given an example above. if you have understood this, i think you should be able to solve it by yourself. but i have opened a new thread. you can follow the question and solution there.

i h a v e g i v e n a n e x a m p l e a b o v e . i f

y o u h a v e u n d e r s t o o d t h i s, i t h i n k

y o u s h o u l d b e a b l e t o s o l v e i t b y

y o u r s e l f . b u t i h a v e o p e n e d a n e w

t h r e a d . y o u c a n f o l l o w t h e q u e s t i o n

a n d s o l u t i o n t h e r e .

Commented by Tawa11 last updated on 16/Mar/25

you have not solve Q217565 sir

you have not solve Q 217565 sir

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

i know. but it is solved! you can compare with your own solution.

i k n o w . b u t i t i s s o l v e d! y o u c a n

c o m p a r e w i t h y o u r o w n s o l u t i o n .

Commented by Tawa11 last updated on 16/Mar/25

Commented by Tawa11 last updated on 16/Mar/25

Commented by Tawa11 last updated on 16/Mar/25

Is this correct now sir?

Is this correct now sir ?

Commented by mr W last updated on 17/Mar/25

since you take F_1 =F/2, F_2 =F_1 /2, the rest must be wrong. these are such basic things! even when the objects are in rest, we have that F=2F_1 +M_1 g > 2F_1 . how can you say F_1 =F/2?

s i n c e y o u t a k e F_{1} = F / 2, F_{2} = F_{1} / 2,

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

s u c h b a s i c t h i n g s! e v e n w h e n t h e

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

F = 2 F_{1} + M_{1} g > 2 F_{1} . h o w c a n y o u

s a y F_{1} = F / 2 ?

Commented by mr W last updated on 17/Mar/25

it is assumed that there is no friction between the strings and the pulleys. in this case the pulley don′t rotate. otherwise you can not assume that the string on both sides of a pulley has the same tension, but have assumed F_(1, left) =F_(1, right) =F_1 , therefore no friction and no rotation of pulleys!

i t i s a s s u m e d t h a t t h e r e i s n o

f r i c t i o n b e t w e e n t h e s t r i n g s a n d

t h e p u l l e y s . i n t h i s c a s e t h e p u l l e y

{d o n}^{'} t r o t a t e . o t h e r w i s e y o u c a n n o t

a s s u m e t h a t t h e s t r i n g o n b o t h s i d e s

o f a p u l l e y h a s t h e s a m e t e n s i o n, b u t

h a v e a s s u m e d F_{1, l e f t} = F_{1, r i g h t} = F_{1},

t h e r e f o r e n o f r i c t i o n a n d n o

r o t a t i o n o f p u l l e y s!

Commented by Tawa11 last updated on 17/Mar/25

Noted sir.

Noted sir .

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Facebook Tweet Pin

Leave a Reply Cancel reply