Tension in the cable of the lift

[jatin1990]

Homework Statement

The tension in the cable supporting a lift moving upward is twice the tension when the lift moves downward. What is the acceleration of the lift?

Homework Equations

The Attempt at a Solution

i think its only conceptual but still please show me the way how to think about it.
Thank you.

 

[tiny-tim]

welcome to pf!

hi jatin19901! welcome to pf!

apply good ol' Newton's second law to find the relationship between a g and T for the two cases …

what do you get? ​

 

[jatin1990]

hi jatin19901! welcome to pf!

apply good ol' Newton's second law to find the relationship between a g and T for the two cases …

what do you get? ​

i am doing it like this , when lift is moving downward then , Tension(T) = Mass(m) * Net Acceleration ( Acc. due to gravity(g)+ Lift acceleration(a))
And when lift is moving upward then : 2T = M * (g-a) , and finding out the relation between a and g , i am getting , a=-g/3 , i have a option of g/3 , and i am getting -g/3 .
please correct me where i am wrong.

 

[tiny-tim]

Tension(T) = Mass(m) * Net Acceleration ( Acc. due to gravity(g)+ Lift acceleration(a))

no no no no noooo …

never do that, there is only one acceleration for one body

in other words: you can put as many forces as you like on the LHS of F = ma, but you can only put one acceleration on the RHS​

(of course, you can subtract accelerations of different bodies, to get the relative acceleration of the two bodies, but that's not this case)

g is not an acceleration (on the RHS), mg is a force (on the LHS)

try again! ​

 

[jatin1990]

no no no no noooo …

never do that, there is only one acceleration for one body

in other words: you can put as many forces as you like on the LHS of F = ma, but you can only put one acceleration on the RHS​

(of course, you can subtract accelerations of different bodies, to get the relative acceleration of the two bodies, but that's not this case)

g is not an acceleration (on the RHS), mg is a force (on the LHS)

try again! ​

I do not got your last line.

 

[tiny-tim]

you mean? …

g is not an acceleration (on the RHS), mg is a force (on the LHS)

in F = ma, the only a on the RHS is the actual acceleration (what you, for some reason, are calling the "net acceleration" )

on the LHS, you put all the forces, and that includes the weight, mg

 

[jatin1990]

you mean? …


in F = ma, the only a on the RHS is the actual acceleration (what you, for some reason, are calling the "net acceleration" )

on the LHS, you put all the forces, and that includes the weight, mg

oh ok , yes got the answer as A = g/3 , but i don't know whether it is correct or not , can u please check it. One more thing i have some questions to ask , can you please answer them here or do i need to start a another thread?
Thank you.

 

[tiny-tim]

oh ok , yes got the answer as A = g/3 , but i don't know whether it is correct or not , can u please check it.

if you want us to check your calculations, you need to show them

One more thing i have some questions to ask , can you please answer them here or do i need to start a another thread?

always start another thread

 

[jatin1990]

when lift is moving upward i got , 2T+mg=-ma and when lift is moving downward , T+mg=ma , and then equate the T from both equations , is it right?

 

[tiny-tim]

when lift is moving upward i got , 2T+mg=-ma and when lift is moving downward , T+mg=ma

uhhh?

how can the tension be in the same direction as the weight??​

 

[jatin1990]

uhhh?

how can the tension be in the same direction as the weight??​

Sorry for my stupidity , now i got it. As i am touching physics after two years , so this is the result. I hope i will get my command back upon physics with the help of this forum.
Thank you Tiny Tim for answering my questions very quickly every time.