Calculating the apparent load on a suspension point during deceleration of an object in motion

Question

Ok, so I work in the entertainment industry and deal with hanging scenic elements/lighting/video walls pretty regularly. For reference I do the installation part not the planning so don't get too scared by me asking this question.

What is the Force?/Apparent Weight?(in rigging we often use the term "felt weight") during deceleration of an object in motion.

Often times electric chain hoists are used in suspension systems and the most common speed is 4m/min.

So what formula can I use to assess the felt weight on the suspension point when a hoist lowering a 500kg mass goes from 4m/min to stopped in approx 0.5seconds?

Is it something like:

$v_i = \frac{meters/minute}{60} = \frac{4}{60}$

$v_i = 0.067$

$a = \frac{v_f - v_i}{t} = \frac{0 - 0.067}{0.5}$

$a = 0.134$

$F = ma = 100 \cdot 0.134$

$F = 13.4\;\;\; \text{#I am unsure of the units at this point...}$

At the end of that does the felt weight on the suspension point during deceleration equal the static load of 100kg + the 13.4(kg?)?

I appreciate your insights!

Answer 1

You are pretty much entirely correct. What follows would be the computation that includes the units and their conversion. Taking upwards as $+$ve,

$$ \begin{align} \tag1v_i&=-\left(\frac{4\,\text m}{\text{min}}\right)\left(\frac{\text{min}}{60\,\text s}\right) =-\frac 4{60}\,\text m/\text s\approx-0.067\,\text m/\text s\\ \tag2a&=\frac{v_f-v_i}t =\frac{0-(-\frac4{60}\,\text m/\text s)}{\frac12\,\text s}=+\frac8{60}\,\text m/\text s^2\approx+0.133\,\text m/\text s^2\\ \tag3\text{N2L}\ :\qquad\sum\vec F&=m\vec a\\ \tag4T-W&=ma\\ \tag5T&=W+ma\\ \tag6&=mg+ma\\ \tag7&=m(g+a)\\ \tag8&=\left(m+m\frac ag\right)g\\ \tag9&=\left[m\left(1+\frac ag\right)\right]g \end {align}$$ Where $T$ is the upwards tension in the chain, $W=mg$ is the weight of the object, and we now see that $m(1+a/g)$ is the new apparent equivalent mass, if you so wish. If we substitute, the values, we will see $$\tag{10}m\left(1+\frac ag\right)\approx100\,\text{kg}\left(1+\frac{0.133}{9.81}\right)\approx101.359\,\text{kg}$$ Note that while you are about a factor of 10 away, it is coming from you not knowing that you have to compare the above value with the weight, as computed from the acceleration due to gravity $g=9.81\,\text m/\text s^2=9.81\,\text N/\text{kg}\approx10\,\text m/\text s^2$. In particular, you computed $$\tag{11}ma=100\,\text{kg}\cdot0.133\,\text N/\text{kg}=13.3\,\text N$$ that you should be comparing with the weight $$\tag{12}W=mg=100\,\text{kg}\cdot9.81\,\text N/\text{kg}=981\,\text N\approx1000\,\text N$$ and then you will get the same answers.