Why Are My Elevator Tension Calculations Incorrect?

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Homework Help Overview

The discussion revolves around calculating the tensions in cables connected to two masses inside a downward-accelerating elevator. The masses involved are m1 = 2.00 kg and m2 = 4.00 kg, with the elevator accelerating at g/2. Participants are analyzing free body diagrams and the forces acting on each mass.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of free body diagrams for both masses and question the assumptions regarding the forces in the y-direction. There are inquiries about the correct application of Newton's second law and the interpretation of gravitational force for each mass.

Discussion Status

Some participants have identified potential mistakes in the original calculations and are actively checking algebraic steps. There is a mix of agreement on certain values, but also uncertainty regarding the final answers, particularly for the tension in cable B.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can provide or the methods they can use. The instructor's feedback has also introduced additional uncertainty regarding the correctness of the calculated tensions.

ur5pointos2sl
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My answers for this problem aren't correct. Can anyone please help me spot my mistake. Thanks.


Inside the elevator, masses m1 = 2.00kg and m2 = 4.00 kg are conncted by a cable A. Mass m1 is connected to the roof of the elevator by a cable B. The elevator has a downward acceleration of g/2. Calculate the tensions in cables A and B.

m1 = 2.00 kg
m2 = 4.00 kg
a = 9.8/2

I set up two free body diagrams.. One for m1 and one for m2

So for m1 FBD

SUMFy= 0 = Fb - 2(9.8/2) - Fa
Fb = 9.8 N

for m2 FBD

SUMFy = 0 = Fa - mg
Fa = 4(9.8/2) = 19.6 N NOW PLUGGING BACK IN
 
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There are no answers to check. When you post them, we will check them. Nevertheless, I can see mistakes already. You say

SUMFy = 0 = Fa - mg

Why is the sum of all the forces in the y-direction zero? Are the masses not accelerating?

Also, when you say "mg" which mass are you talking about, 1 or 2?
 
kuruman said:
There are no answers to check. When you post them, we will check them. Nevertheless, I can see mistakes already. You say

SUMFy = 0 = Fa - mg

Why is the sum of all the forces in the y-direction zero? Are the masses not accelerating?

Also, when you say "mg" which mass are you talking about, 1 or 2?

m1 = 2.00kg m2 = 4.00 kg

ok let me try this over..

FBD of m1...so here the mg will be 2kg (9.8) or would it be 2(9.8/2)?

so SUMFy = Fb - Fa - mg = ma
= Fb - Fa -(2)(9.8)=2(-9.8/2)
Fb = -9.8 N which couldn't be possible since its in tension

FBD of m2...so here mg will be 4kg (-9.8/2)?

so SUMFy= Fa - 4(9.8)= 4(-9.8/2)
Fa = 19.6 N

again I am getting the same answers as before..
 
ur5pointos2sl said:
so SUMFy = Fb - Fa - mg = ma
= Fb - Fa -(2)(9.8)=2(-9.8/2)
Fb = -9.8 N which couldn't be possible since its in tension

How did you get -9.8 N? I don't get that when I put in the numbers. Check your algebra.

so SUMFy= Fa - 4(9.8)= 4(-9.8/2)
Fa = 19.6 N
This looks correct. Do you think it is not?
 
kuruman said:
How did you get -9.8 N? I don't get that when I put in the numbers. Check your algebra.


This looks correct. Do you think it is not?


The instructor said 19.6 N was not the correct answer. I am not really sure of the actual correct answer.

I checked my algebra again for Fb and ended up getting 49N. I really am not sure what to think now.
 
The Fa = 19.6 N is correct. I agree with your instructor. Now show me exactly what you do after you write down

Fb - Fa -(2)(9.8)=2(-9.8/2)
 
kuruman said:
The Fa = 19.6 N is correct. I agree with your instructor. Now show me exactly what you do after you write down

Fb - Fa -(2)(9.8)=2(-9.8/2)

Fb - 19.6 -(2)(9.8)=2(-9.8/2)

Fb = 19.6 + 98/5 - 49/5
= 29.4 N
 
Looks good. I think you are done.
 

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