Maximum / Minimum Force

  • Thread starter Hotsuma
  • Start date
  • #1
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Homework Statement



An elevator mass 4950 kg is to be designed so that the maximum acceleration is 6.10×10-2g.

What is the maximum force the motor should exert on the supporting cable?

What is the minimum force the motor should exert on the supporting cable?

Homework Equations



[tex]\Sigma[\tex]F = ma

The Attempt at a Solution



I know this should be so easy, but Mastering Physics is not liking my answers. I have tried multiplying the combined forces of acceleration and gravity multiplied over the mass to get a large force, that answer was incorrect. I noticed that the units of acceleration is in 'g', which caused me to believe they were referencing that amount multiplied by gravity instead of added to it, which resulted in another answer MP did not like. My results:

F = 48800 N
F = 2960 N

Three significant figures, neither of those answers were correct (this is for Fmax). I am afraid I am missing something really critical, and wanted to get another opinion. Maybe I'm approaching this problem in the wrong way?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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"combined forces of acceleration and gravity" sounds really good! But doesn't lead to the numbers you got. It would be interesting to see the details of your calculation.
F = ma + mg = ???
 
  • #3
41
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Well mg + ma does equal 48800 N rounded to the number of significant figures the problem asks for.

The calculation I referred to is: 4950*(g+a), which equals 48900... Hmm...

Well, I only have one guess left, and I know that 48900 is not correct.
 
  • #4
Delphi51
Homework Helper
3,407
11
Ah, you are using a=0.061! It should be a = 0.061*9.81.
 
  • #5
41
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So it should be calculated as such:

F = mga + mg = 51500N. Is this right?
 
  • #6
41
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What about finding the minimum force?
 

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