Effective mass seen from a force/acceleration perspective

Dec 2019
2
0
Denmark
Hi

I've had this discussion with several people, but never found anyone, who agree with me. So now I would like to hear your opinion.

The setting is as follows:

M: A mass hanging in a wire from a drone.
Md: Mass of the drone.
Ft: Thrust of the drone (constant).
Fw: Force transferred by the wire.

The system is descending at a constant negative velocity relative to earth, meaning that

Ft = a * (M+Md)

where "a" is the acceleration of the system, in this case the gravitational acceleration.
An external upward force Fe is then asserted M (the mass hanging from the drone).

The air resistance in this scenario is neglected and the Force transfered by the wire Fw > 0 N.

The question is then. Will the acceleration be Fe/M or Fe/(M+Md)??

From my point of view, it is obvious that it is the latter Fe/(M+Md), but what do you think?
 
Oct 2017
577
297
Glasgow
It depends on the magnitude of \(\displaystyle F_e\). If \(\displaystyle F_e < M_d g\), then the string will remain taut and the force balance is

\(\displaystyle F = F_t + F_e - (M+M_d)g \)

But because \(\displaystyle F_t\) is set to be the same as weight upwards, then

\(\displaystyle F = F_e\)

Effectively the external force offsets the weight of the mass carried by the drone slightly. Since the drone's thrust is constant, there will be an acceleration equal to:

\(\displaystyle (M + M_d) a = F_e\)

So

\(\displaystyle a = \frac{F_e}{(M + M_d)}\)

As the magnitude of \(\displaystyle F_e\) is increased, the tension in the wire decreases. If the force is increased such that \(\displaystyle F_e >= M_d g\) the wire will become slack. In this case, the whole weight is offset by the external force. The drone no longer has to pull the weight and the drone no longer needs to consider it as a load; it effectively acts as an independent object at that point. Consequently, the drone and the mass will accelerate independently of each other.

For the drone:

\(\displaystyle F = F_t - Mg\)
\(\displaystyle = M_d g\)

Its acceleration will be:
\(\displaystyle a_t = \frac{M_d g}{M}\) upwards

For the mass:
\(\displaystyle F = F_e - M_d g > 0\)

Its acceleration will be:
\(\displaystyle a_d = \frac{F_e}{M_d} - g\) upwards


When \(\displaystyle F_e = M_d g\), the acceleration of the drone can be expressed as
\(\displaystyle a_t = \frac{F_e}{M}\) upwards

and the acceleration of the mass is zero.

Note that this is the limiting case, not the general formula. The acceleration of the drone is not dependent on the force applied to the mass \(\displaystyle M_d\) when the wire is slack.

Final point: all of this assumes that the system is descending with constant negative velocity, so these equations ring true for that scenario only; as the drone + package accelerate upwards, eventually the velocity of the drone will be upwards and there will be a time later on when the wire will be taut again and pull the package upwards. The drone + package then need to be treated as a linked system again.
 
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Dec 2019
2
0
Denmark
Thank you very much for your answer. It is very elaborate and I really appreciate it.