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A 5kg mass is hanging from the roof of an elevator accelerating downward at a speed of 1.65 m/s^2 what is the tension of the string (ignore friction, string mass, and air resistance) ? I used this formula to solve the problem $$-T-mg = -ma$$ In this problem, i feel that the tension should be negative since the mass is trying to maintain its constant speed due to inertia and elevator is heading downward an its pulling the string with it. But for some reason tension in this case is positive. While if the same object was hanging from the same string but instead of the elevator, we have a person rotating it vertically with a centripetal acceleration of 1.62 m/s^2, the tension of the string when the object is all the way at the top will be negative. I don't understand why these two almost identical situations end up with a different sign for the tension

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  • $\begingroup$ Have you drawn free body diagrams of these two situations, properly showing the forces acting on the mass, and written down the force balance equations for these two very different cases? That should help you make the comparison more easily. $\endgroup$ – Chet Miller Mar 3 '16 at 0:07
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Thank you for writing out the equation. Please note that the tension in the string can't be negative, because a string can't support negative tension (compression). It's the same as trying to push a wet noodle.

There are two ways of writing the force balance, depending on whether the upward direction is taken as positive or negative. Both ways give the exact same result for the tension.

Method 1: Upward direction is positive. Here, $$T-mg=ma$$where a is the upward acceleration of the mass. Since the acceleration of the mass is the same as the elevator, here, a = -1.65. So, $T=m(g-1.65)$

Method 2: Downward direction is positive. Here, $$mg-T=ma$$where a is the downward acceleration of the mass. Since the acceleration of the mass is the same as the elevator, here, a = + 1.65. So, $T=m(g-1.65)$

Either way you get the same answer.

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  • $\begingroup$ then how come tension can be negative in the rotation problem? $\endgroup$ – user3929076 Mar 3 '16 at 0:43
  • $\begingroup$ I guess my question is, since the elevator is the one causing the mass to go down, and since it's do so via string, shouldn't it thus be negative $\endgroup$ – user3929076 Mar 3 '16 at 0:49
  • $\begingroup$ Regarding your question about rotation, that's a separate problem that can also readily be analyzed. $\endgroup$ – Chet Miller Mar 3 '16 at 0:58
  • $\begingroup$ Regarding your question," since the elevator is the one causing the mass to go down, and since it's do so via string, shouldn't it thus be negative", my response is that the tension actually is less than if the elevator were not accelerating downward. If it were not accelerating downward, the tension would be mg. The downward acceleration reduces this, but not enough to drop the tension to zero. In order for the tension to drop to zero, the downward acceleration of the elevator would have to be g. Then the elevator and the weight would both be in free fall. $\endgroup$ – Chet Miller Mar 3 '16 at 1:02
  • $\begingroup$ and if downward acceleration was faster than g, will the tension now be negative? $\endgroup$ – user3929076 Mar 3 '16 at 1:18
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Weather you define a certain direction as negative positive is really just a matter of personal choice. It will not effect the math or results at all as long as you are consistent.

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  • $\begingroup$ Well my professor said that i did it wrong and marked me off for it $\endgroup$ – user3929076 Mar 3 '16 at 0:00
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In classical mechanics, you cannot simply say whether or not a force is positive or negative. You must define a coordinate axis first. Your tension could be either positive or negative, depending on where you place your coordinates.

For example, if the tension is facing upwards, then it is positive only if you have defined the upward direction to be positive when solving Newton's 2nd Law. However, you may have defined the positive direction to be downward, since this is the direction in which the elevator is moving. In this case, an upward-pointing tension will be negative.

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  • $\begingroup$ I defined up as positive and down as negative. Thus, the tension should be negative since the elevator is pulling the mass down via string right $\endgroup$ – user3929076 Mar 3 '16 at 0:25
  • $\begingroup$ If the box was just hanging from a ceiling, would you class the tension as positive or negative? $\endgroup$ – ODP Mar 3 '16 at 0:25
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    $\begingroup$ i would define it as positive $\endgroup$ – user3929076 Mar 3 '16 at 0:35
  • $\begingroup$ but since the elevator is going down, shouldn't it pull the mass down with it downward? and since that force is being applied via string, shouldn't the tension also be downwards $\endgroup$ – user3929076 Mar 3 '16 at 0:41
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When this type of question is first encountered it can cause problems unless one is consistent with signs.

The question "What is the tension in the string?" can be interpreted as "What is the force on the 5 kg mass due to the string?"

When a string is exerting a force on a mass the string can only "pull" it can never "push".

Whatever way the problem is solved the forces acting on the 5 kg mass are as in Diagram 1.

enter image description here

To solve the problem first choose a positive direction and draw a free body diagram.

Diagrams 2 to 5 show you four possible FBDs.

I know that strings do not push as in Diagram 2 and 4 but I have drawn those diagrams to illustrate the point that whatever directions are guessed for the unknown forces the correct answer will result.

Referring to the chosen positive direction assign signs to the known values, $a$ and $g$ in this example.

Use Newton's second law making sure that the directions which are assigned to the unknown forces, the tension $T$ in this case, are consistent with the direction which was chosen to be positive.

So in Diagram 3 down was chosen to be the positive direction and the direction of the tension was drawn as up so the force due to the tension when applying Newton's second law is $-T$.
The result is that the tension is found to be $+40.80$ and referring to the diagram that is a force of $40.80$ upwards.

In Diagram 4 the tension is found to be $-40.80$.
Referring to the diagram the tension was shown to be downwards but the minus sign shows that the tension force is in the opposite direction to down which is up, so the tension is $40.80$ upwards.

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