The discussion centers on calculating the acceleration of a lift based on the tension in its cable during upward and downward motion. It is established that the tension in the cable when the lift moves upward is twice that when it moves downward. By applying Newton's second law, the correct relationship between the lift's acceleration (A) and gravitational acceleration (g) is derived, resulting in A = g/3. Participants emphasize the importance of correctly applying forces and accelerations in the equations.
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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))tiny-tim said: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 said:
…tiny-tim said: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!
)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?tiny-tim said: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
jatin1990 said:
jatin1990 said:
tiny-tim said:uhhh?
how can the tension be in the same direction as the weight??