Question

two weights w1 and w2 are connected by a light thread which passes over a light smooth pulley if the pulley is raised upwards with an acceleration equal to g , then the tension in the thread will be

@Kainui @nincompoop @iambatman @ParthKohli @skullpatrol

Answer 1

Answer is \(\huge \sf \frac{4W_1W_2}{W_1+W_2}\)
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Answer 2

I was not aware of this...

Answer 3

It's ok

Answer 4

Drawing that goes with it?

Answer 5

yes

Answer 6

Oh it's not loading for me

Answer 7

one min

Answer 8

Can u explain this ?

Answer 9

I'm not sure what the problem is here, it's just using newtons second law, pretty simple :P

Answer 10

yeah :p

Answer 11

Specific part? Or just not sure what's going on in general?

Answer 12

Specific part

Answer 13

\(\sf m_2g-T=m_2(g-a)\)

Answer 14

Not able to understand the sign conventions !

Answer 15

Look at the acceleration and gravity

Answer 16

@No.name

Answer 17

Wait reading the question!

Answer 18

ok

Answer 19

Is it quite urgent , because in 2 hours my exam is going to start , i am studying for it

Answer 20

It's ok .....

Answer 21

Yeah i am getting the same answer

Answer 22

How ?

Answer 23

@kropot72

Answer 24

do you still need help?

Answer 25

try breaking it down into parts

Answer 26

@nincompoop yes i still need help...u there ?

Answer 27

@compphysgeek @RaphaelFilgueiras

Answer 28

I have this solution given, can u guys xplain it ?

Answer 29

m2.g-T=m2(g-a), but g=a then T=m2.g i don't get this line

Answer 30

the pseudo g, you can feel this effect when the elevator is going up dont tou feel your feet pressing the floor ?it's that effect

Answer 31

\(\color{blue}{\text{Originally Posted by}}\) @RaphaelFilgueiras
m2.g-T=m2(g-a), but g=a then T=m2.g i don't get this line
\(\color{blue}{\text{End of Quote}}\)


Same here !

Answer 32

I don't understand the sign convention they have used.

Answer 33

is this the saraeva book?

Answer 34

Idk...found it on internet......
I hv onemore solution from other source.......would u like to see it ?

Answer 35

what part of the solution you don't understand?

Answer 36

How they deduced eq 1 and eq 2

Answer 37

they actaully thought that m2 mass is greater than m1 so for m1 tension in the string is greater than the weight so they took T-m1 and for m2 ,its weight is greater than tension,so they took m2-T.. I THINK SOO BUT NOT SURE

Answer 38

are you familiar with the newton's laws?

Answer 39

yes i am..

Answer 40

can you break the illustration into components according to what you know? forget the picture you were given, I want to know how you will make your own diagram.

Answer 41

yes,
ok let's suppose the lift is not moving and it is stationary....then it should be

T-m1g = -m1a [if m1 > m2]

T-m2g = m2a

Answer 42

correct ?

Answer 43

@nincompoop ?

Answer 44

when stationary, we use the first law
T +(-w1) = 0
T = w1

Answer 45

No i didn't said that the pulley is stationary.....i said the lift is stationary...!

Answer 46

There are two simultaneous acceleration working according to the question...

Answer 47

One is accn due to normal pulley thing and the other is due to the accelerated motion of lift inside which the pulley is kept.

Answer 48

See this diagram, it represents the question

Answer 49

The pulley is raised upwards with an acceleration=g

Answer 50

i see it now

Answer 51

It's like raising a beam balance with two different weights on it

Answer 52

I understand that now. look at your equation when the pulley is not being raised

Answer 53

No, w1 = m1g

Answer 54

m1a is the accelaration by which the block will move downwards.