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Suppose we have a system as shown in the diagram above, inside an elevator, where object $Q$ has mass $M$, object $P$ has mass $m$, and $M>m$.

When this elevator accelerates upward with acceleration $a$, is it correct to say that in the frame of reference of the elevator, the forces $p$ and $q$ are $m(a+g)$ and $M(a+g)$ respectively?

If so, and if I assume that $Q$ moves downward with acceleration $f$ as observed from inside the elevator, can I say $M(a+g)-T=Mf$?

Or should I just say $Mg-T=Mf$, as they do in this video? Why? How is it that the upward acceleration of the elevator has got no effect in this equation?

  • $\begingroup$ Your reasonning looks fine to me. Applying this to both masses leads to $f=\frac{M-m}{M+m}(a+g)$ and $T=\frac{Mm}{M+m}(a+g)$, which seems to behave correctly. Didnt watch the video though. $\endgroup$
    – Nicolas
    Mar 19 '20 at 7:05

Here is a still from the video and in this case the term absolute in the context of the acceleration means magnitude relative to the ground.

enter image description here

Applying Newton's second law in the ground frame of reference for each of the masses with up as positive:

$T-6g = 6(b+\frac g8)$ and $T - 7g = 7(\frac g8 -b)$

and with two equations you can solve for the two unknowns $T$ and $b$.

Notice that in the video up is positive fro the left hand mass and down is positive for the right hand mass which I prefer not to do.
The two equations are essentially the same because mutiplying my second equation by $-1$ produces the lecturer's second equation.

Now let me rearrange my two equations by putting the term which contains $\frac 98$ on the left hand side.

$T-6g-6\frac g8 = T-6(g+\frac g8) = 6b$ and $T-7g-7\frac g8 = T-7(g+\frac g8) = 7(-b)$

Now these two equations are Newton's second law applied in the accelerating frame of the lift with two fictitious/pseudo force equal to $6\frac g8$ and $7\frac g8$ added to the masses.

Another way of looking at this is to say that the (local = relative to the lift) acceleration of free fall is $g+\frac g8$ downwards.

If the lift was accelerating downwards at $\frac g8$ then all that would happen is that there would be a change of sign for the fictitious/pseudo force.

$T-6g+6\frac g8 = T-6(g-\frac g8) = 6b$ and $T-7g+7\frac g8 = T-7(g-\frac g8) = 7(-b)$

This is perhaps what you might expect in that if the lift is accelerating downwards at $g$ then you find that both the tension $T$ is zero and the acceleration of the masses relative to the lift is also zero.


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