Solving Elevator Problem Homework Statement

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The discussion centers on solving a physics problem involving a woman standing on a scale in a moving elevator. The woman has a mass of 50 kg, and the hoisting cable exerts a force of 9350 N during upward acceleration. Participants clarify the correct approach to calculate the apparent weight, emphasizing the importance of using consistent units and correcting misunderstandings about the mass of the elevator. The final calculation indicates that the scale reads approximately 50 kg, reflecting the woman's weight during the elevator's acceleration. The conversation highlights the need for careful unit management in physics problems.
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Homework Statement


A woman stands on a scale in a moving elevator. Her mass is 50kg and the combined mass of the elevator accelerates upward. During the acceleration , the hoisting cable applies a force of 9350N. What does the scale read during the acceleration ?


Homework Equations


F net= T-MG= Ma
Apparent weight = mg(a/g+1)

The Attempt at a Solution


First we find a and the plug in the a in the apparent weight and we find the scale reading
first attempt: 9530-(50+815) = (815+50)a
and a= 1.22m/s^2
and the apparent weight = 9532N
 
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Anthonyphy2013 said:

Homework Statement


A woman stands on a scale in a moving elevator. Her mass is 50kg and the combined mass of the elevator accelerates upward. During the acceleration , the hoisting cable applies a force of 9350N. What does the scale read during the acceleration ?


Homework Equations


F net= T-MG= Ma
Apparent weight = mg(a/g+1)

The Attempt at a Solution


First we find a and the plug in the a in the apparent weight and we find the scale reading
first attempt: 9530-(50+815) = (815+50)a

You are mixing up Newtons with kg in this equation.
Your problem doesn't state it but I gather the mass of the elevator is 815 kg.
 
Where does the number 815 come from?
Please work with units, it is easier to spot errors that way. For example, you cannot subtract 50kg from 9350N, that does not make sense.

and the apparent weight = 9532N
It would be impossible to stand on a scale in that case (that is the weight of a car), the result cannot be true.
 
815 comes from the combined mass of the elevator and the scale .
9530N-(865*9.8)N=(865)a
a=1.22 m/s2
Apparent weight=N=Mg(a/g+1)
=(865*9.8)(1.22/9.8 + 1)
9532.3N
Is that corrected ?
 
Last edited:
Anthonyphy2013 said:
815 comes from the combined mass of the elevator and the scale .
You mean 815 is the mass of the elevator, I think.
9530N-(865*9.8)N=(865)a
a=1.22 m/s2
That is finally correct.
Apparent weight=N=Mg(a/g+1)
=(865*9.8)(1.22/9.8 + 1)
9532.3N
Is that corrected ?

No. She's not going to weigh much more than at rest. And the scale reads in kg, not N.
So the answer is not much more than 50 kg.
 
Anthonyphy2013 said:
Apparent weight=N=Mg(a/g+1)
=(865*9.8)(1.22/9.8 + 1)
865? Isn't that going to give you the apparent weight of the elevator+passenger?
 
haruspex said:
865? Isn't that going to give you the apparent weight of the elevator+passenger?

so you mean the mass is 50 not 865( 815+50) ?
 
Anthonyphy2013 said:
so you mean the mass is 50 not 865( 815+50) ?

Yes. You're not measuring the weight of the elevator, right?
 
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