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Suppose I want to buy a cable that will support & pull a mass of $2\ kg$ upward at an acceleration rate of $2 m/s^2$, I must specify the maximum mass the cable will carry.

I know how to calculate the tension in the cable :

There are 2 forces acting on the mass:

  1. The force of the cable pulling up which is unknown, call it "$F_t$"

  2. The force of the gravity of the earth pulling down call it "$F_w$"

since the mass is moving upward at an acceleration rate of $2m/s^2$, then the resultant force affecting on the mass $ΣF$ must be $F=2kg\times 2m/s^2=4N$

$$ΣF = F_t\; -F_w$$ Add $+F_w$ to both sides: $$F_t = F_w+ΣF$$ $$F_t = mg + ma$$ $$F_t = 2\times9.81 + 2\times2 =23.62N$$

Now how can I express this force in kilogram should I divide it by $9.81m/s^2$ which is acceleration due to gravity or divide it by $2m/s^2$

I can express the same problem horizontally:

A $2kg$ trailer is pulled to the right on a frictionless surface by a cable connected to a train moving at an acceleration rate of $0.5m/s^2$ , what is weight on the cable during acceleration of the trailer?

I know that the tension on the cable is $F=ma=2\times0.5=1N$ . But again if I want to buy that cable I must specify the maximum mass the cable will carry, is it correct to say that this mass $= (1/9.81)=0.102kg$ ?

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  • $\begingroup$ Homework type questions are considered off-topic here. But you’ve got the right idea. Thanks. $\endgroup$
    – joseph h
    Commented Mar 3, 2022 at 1:28
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    $\begingroup$ @josephh sir this isn't my homework, this just an example to help people understand my problem, i didn't write the question and simply asked for an answer , i tried to solve the problem but i've confusion about the issue of the weight of an accelerated mass which i explained in my question $\endgroup$
    – techno tux
    Commented Mar 3, 2022 at 1:35
  • $\begingroup$ Whether it was actually assigned as homework or not is of little relevance. Please read our policy regarding homework-like questions. Also, as stated in my last comment, you have the right idea. To get $m$ you divide $a$ into the force, or $m=\frac Fa$ Cheers. $\endgroup$
    – joseph h
    Commented Mar 3, 2022 at 1:41
  • $\begingroup$ @josephh which acceleration value should i divide by ?? the 9.81 due to gravity or the elevator 2m/$s^2$ , please explain the answer thanks, $\endgroup$
    – techno tux
    Commented Mar 3, 2022 at 1:44
  • $\begingroup$ Are you buying a cable that will lift a maximum of 2kg at the rate of 2m/s^2? $\endgroup$
    – joseph h
    Commented Mar 3, 2022 at 1:54

1 Answer 1

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You would use $9.81$ as your divisor. You correctly computed a tension force of $23.62 N$, and stores would say that a maximum tension of $23.62 N$ would mean that it can carry a maximum mass of $\frac{23.62}{9.81} = 2.4 kg$. It's pretty annoying that "forces" are sometimes measured with kilograms, but as a rule of thumb you should always be using what is more general ($g=9.81 m/s^2$ and it is well known) as opposed to the less general things (you're accleration of $2 m/s^2$ which is specific to your case).

Hope this helps.

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  • $\begingroup$ thanks for your answer sir , but are suggesting that i should use 9.81 just because "it is more general & well known"?? the apparent weight will be much different if i chose wrong?? $\endgroup$
    – techno tux
    Commented Mar 3, 2022 at 2:15
  • $\begingroup$ That's not the reason you should use $9.81$, I'm just pointing out that if you have a future predicament always stick with what is more general if you're measuring quantities or comparing units. The reason you should use $9.81$ is because if a rope is rated for 5kg of "force," we are supposed to interpret that as saying it is rated for $5\cdot 9.81$ Newtons of force. If I say I weigh 50kg, all I'm saying is that I experience a normal force of $50 \cdot 9.81$ newtons when I step on a scale. However, people get lazy and just say kilograms because the $9.81 m/s^2$ is implied. $\endgroup$
    – Eli Yablon
    Commented Mar 3, 2022 at 2:51
  • $\begingroup$ it does seem to me that it's correct to Divide force tension by 9.8 to find an equivalent mass in kilograms, suppose the acceleration is 0, from: FT=ma+mg, Ft=0+mg, so FT=mg , so to get 'm' you divide by g. $\endgroup$
    – techno tux
    Commented Mar 6, 2022 at 22:36

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