I Apparent Weight problem (falling beam)

[Jackolantern]

Each weight: 2 Kg
Hello All, I'm trying to understand an "apparent weight" problem and check my answer. Please use the picture attached.

A weightless beam is at first resting over my palm, it has one 2 kg weight on each end of it. It is suspended to the ceiling by a rubber band. I drop it and it falls for 1 second. At this instant it is has a velocity of 9.8 m/s and the rubber band is exerting an elastic force on it of 106 Newtons. Now, the task is to estimate the "apparent" weight of the 2 kg weights on the beam in this instant as felt by the beam.

-First, I calculate the vertical acceleration induced by the force of the rubber band on the beam:

With a free body diagram of the beam, the only forces acting on it are the weights and the force from the rubber band

Each weight has a force of 19.62 N ( 2 kg * 9.81).
(F)sum = m * a

106 - 2*(19.62) = 4 kg * a
a = 16.69 m/s^2

The apparent weight of an object is its mass multiplied by the vector difference between the gravitational acceleration and the acceleration of the object.

F = m * a

F = 2 kg * (9.81 + 16.669) = 53 N

So, my question is, at the 2nd instant illustrated by my photo above, the apparent weight of each of the weights on the end of the beam is 53 N?
This is just half of the rubber band force...for every action there is an equal and opposite reaction?

 

[Orodruin]

Each mass has apparent weight 53 N so the total apparent weight of the beam with the masses is 106 N as expected.

Note that some things will often be clearer if you introduce symbols to represent physical quantities and only insert measured values at the very end. In your case, you could introduce the following:

F = 106 N
m = 2 kg
g = 9.1 m/s^2
a = acceleration of the masses
W = apparent weight of one mass

Your algebra would have become:

Force relation to acceleration:
$$
F - 2mg = 2ma \quad \Longrightarrow \quad a = \frac{F}{2m} - g
$$

Apparent weight of one mass:
$$
W = m(a+g) = \frac{F}{2}.
$$

It is then clear that the apparent weight of each mass is exactly half of the tension in the rubber band and that the masses and the acceleration and gravitational acceleration do not really matter for the result.

 

[Jackolantern]

Each mass has apparent weight 53 N so the total apparent weight of the beam with the masses is 106 N as expected.

Note that some things will often be clearer if you introduce symbols to represent physical quantities and only insert measured values at the very end. In your case, you could introduce the following:

F = 106 N
m = 2 kg
g = 9.1 m/s^2
a = acceleration of the masses
W = apparent weight of one mass

Your algebra would have become:

Force relation to acceleration:
$$
F - 2mg = 2ma \quad \Longrightarrow \quad a = \frac{F}{2m} - g
$$

Apparent weight of one mass:
$$
W = m(a+g) = \frac{F}{2}.
$$

It is then clear that the apparent weight of each mass is exactly half of the tension in the rubber band and that the masses and the acceleration and gravitational acceleration do not really matter for the result.

That's much more clear to me now, thank you Orodruin.