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My Physics Teacher is Obsessed with Elevators

  • Thread starter Medgirl314
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Homework Statement



A 1200 kg elevator is supported by a cable. If the elevator is accelerating down at 1.1 m/s2, what is the tension in the cable?

Homework Equations


F=ma

The Attempt at a Solution



W=mg
W=1200 kg(1.1 m/s2 )
W= 1320 kg

1.1= 1320-F cable/ 350 kg
Tension=1705

Does this look correct? It seems plausible.

Thank you!
 

Answers and Replies

  • #2
Simon Bridge
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The way to check your working is to go through the reasoning behind it.
The trick is to think of the math as a language for describing relationships - seeing if it makes sense then is a matter of checking the grammar.

For this problem, you'd draw a free-body diagram - you'll notice that physicists are also obsessed with drawing pictures.

The tension points up and weight down - acceleration is also down.
Pick a direction (up or down) to be positive.
Write ##\sum F = ma## and solve the equation for the unknown.
Do not put numbers into the equations until the end.

Try it that way and compare.
 
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  • #3
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The symbol means "change in force", correct? So would that be 1320-F cable? The hard part here seems to be when I got to the equation. I couldn't figure out how to solve it because of where the subtraction sign is. Was I headed in the right direction before that part?

Thank you!
 
  • #4
PeroK
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If you're tired of elevators, try this problem instead:

A 1200 kg elephant is supported by a cable. If the elephant is accelerating down at 1.1 m/s2, what is the tension in the cable?
 
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  • #5
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Hahah! Nice sense of humor. ;-) I actually don't mind the elevators. It's so interesting to think about how the weight on a scale would change if you bring it on an elevator. Do you have an idea if I got the answer right, or if my equation was similar to SimonBridge's? Thank you!
 
  • #6
PeroK
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Here's how I would look at it.

If the elephant was just hanging there, then the tension would be 1200g.

If the elephant was in freefall (at 9.8m/s2) then the tension would be 0.

So, depending on how gently you lower the poor creature, the tension will go from 0 to 1200g.

So, if it's only accelerating at 1.1m/s2, then the tension will be much nearer to 1200g. Much more than 1700N. More like 10,000N at a guess.

Simon's idea to look at the nett force is the best way, I think. You have gravity pulling it down, the tension holding it up, giving a nett force F = ma.
 
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  • #7
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Okay, thanks! I want to try Simon's equation, that's what I'm trying to figure out. Would the sigma part be 1320-F cable? Thanks again!
 
  • #8
PeroK
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No, 1320N is the nett force (downwards).
 
  • #9
Simon Bridge
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The symbol means "change in force", correct? So would that be 1320-F cable? The hard part here seems to be when I got to the equation. I couldn't figure out how to solve it because of where the subtraction sign is. Was I headed in the right direction before that part?

Thank you!
The cap sigma means "sum of" so ##\sum \vec F## is the sum of the force vectors.

It means to pick a direction and call it "positive" ... write down all the forces pointing in that direction with plus signs, followed by all the forces in the opposite direction with minus signs, then write an "=" sign, then "ma".

A good direction to pick for positive is the direction of the acceleration: which means "down" is positive. But it usually doesn't matter as long as you remember that acceleration is also a vector.

So on the LHS you have weight mg in the + direction and tension T in the - direction ... so you write it down:

##mg-T=ma##

Solve for T.
 
  • #10
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mg−T=ma

For some reason, I get confused on whether to divide or subtract the appropriate variables in situations like these. Would it be -T=mg/ma ?

Thank you!!!
 
  • #11
BvU
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Poor girl, poor Simon. His post #2 is so good. Did you make the drawing? You have the earth pulling down, the cable pulling up. The net result is a downward acceleration. So who is the bigger of the two?

And then again in #9. He hands it to you on a plate! If you get confuseed, try something like "how would I solve 100 - T = 90 ?" because mg and ma are known and T is not. Would you really try -T = 100/90 ? I don't think so.

I shouldn't have interfered, but can't bear to see this going on.
There is also a sincere interest in finding out what makes this so confusing for you. You are definitely not the only one to whom this happens, so if I (we) can learn about causes we might be able to give better explanations or provide better examples to others liek you.
Keep it up!
 
  • #12
Simon Bridge
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Not too bad ...
Here's the process - do the addition and subtraction first.

mg - T = ma

The way you read this is: "+mg + (-T) = ma"

on the LHS you have a term "mg" that you don't want.
there is no minus sign in front of it so the operation you are looking for is subtraction.

whatever you do to one side you have to do to the other side - picture: moving weights around on a see-saw while keeping it balanced ... if you remove a weight from one side, you also have to remove the same weight from the other.

subtract mg from both sides:

(mg-T)-mg = (ma)-mg

gives you:

-T = ma-mg

... there is nothing more added or subtracted here.
... you want a +T on the LHS and what you have is a -T ... which is (-1)xT ... so there is a -1 you don't want.
... since the -1 is multiplied in, the operation you want is devision: dividing both sides by -1 (is the same as multiplying by -1 but hey ho):

-(-T)=-ma+mg

gives you (tidying up):

T = m(g-a)

Now you can finally put the numbers in - by waiting you get an easier calculation.
With more complicated situations, the payoff is even bigger.

You may like to peruse:
http://www.scribd.com/doc/149266846/Math-Bits
... sadly it's unfinished and has glaring typos ... but the concepts usually help.
At some point I'll get around to finishing and polishing.

[edit] @BvU: I believe she got it quite early on but is attempting to learn a different, more systematic, way of thinking about the problem. Otherwise I wouldn't have handed it over like this :)
There is more than one way to skin a cat - and we benefit from knowing several ways.
 
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  • #13
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Thank you, Simon. You're correct, I got the concept early on, and solved a similar problem(which you saw) correctly, which is separate from trying to go about it in a more systematic way and refreshing skills gone unused for some time. :smile: I see exactly what you did, what a helpful post! It looks like I did most everything right until I tried dividing instead of subtracting. Is that correct?

T=1200(9.8-1.1)=1200(10.78)=12936 N.

I haven't worked with Newtons(or elevators) at all until this week, so I can't really tell if the number is too large. I don't see any mistakes, do you?

Thanks again for a polite and helpful answer!
 
  • #15
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Apparently anything can happen late at night after my brain refuses to work.:wink: No idea what happened there, I multiplied but don't remember why.

T = m(g-a)=1200(9.8 - 1.1)=1200(8.7)=10440 N.

Does that look better? :smile:
 
  • #16
BvU
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It does and you know it does, don't you. Well done!
 
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  • #17
Simon Bridge
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there you go - well done.
now it's just a matter of practise.
 
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