Minimum Force ( I don't know why this is not correct)

  • #1
3,003
6
An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0300g.

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

[tex]\sum F=m.03g[/tex]
[tex]T=mg(1+.03)[/tex]
41385 N <--this IS correct


What is the minimum force the motor should exert on the supporting cable? I thought this would just be weight?? But that is not correct
N

I am clearly misunderstanding what the question is asking

Casey
 
Last edited:

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
19,606
2,972
What is the minimum force the motor should exert on the supporting cable? I thought this would just be weight??
No. If the motor provided a force equal to the weight, then the elevator would not move. The force of the motor would be balanced by the weight.

When one feels an acceleration it is a differential acceleration, i.e. it is in addition to the acceleration of gravity.

So T = m(g+a) = mg (1 + 0.03) = 4100 kg * 9.8 m/s2 * (1.03) = 41385.4 N
 
  • #3
3,003
6
No. If the motor provided a force equal to the weight, then the elevator would not move. The force of the motor would be balanced by the weight.

When one feels an acceleration it is a differential acceleration, i.e. it is in addition to the acceleration of gravity.

So T = m(g+a) = mg (1 + 0.03) = 4100 kg * 9.8 m/s2 * (1.03) = 41385.4 N

But that is what I did use (exactly^^^) for the maximum force. It is the minimum that I am concerned with.

I got 41385.4 N for MAX force. And it was correct.

Casey
 
  • #4
Astronuc
Staff Emeritus
Science Advisor
19,606
2,972
Are you supposed to get something like 38975 N as a minimum?
 
  • #5
3,003
6
I don't know. It is one of those web assign problems. It is for a friend of mine that I am helping. I never used webassign in school. But I can say that I don't like it. Half the time, I cannot tell what they are asking.

Why? Where are you getting the 38975 from? What is your stream of thought here?

Casey
 
  • #6
Astronuc
Staff Emeritus
Science Advisor
19,606
2,972
I the elevator was dropping with an acceleration, then the tension would bg m (g-a) or mg (1-0.03) = mg (0.97).

I was trying to understand the question you were being asked.


There are two ways to feel the increased weight when standing in an elevator: 1) when it starts upward with some acceleration a from rest, or 2) when it slows down to a stop during a descent. When the elevator is falling (with some acceleration) the tension is less than the weight when it is stopped.

I'm just trying to understand the reference to minimum in this problem.
 
  • #7
3,003
6
I the elevator was dropping with an acceleration, then the tension would bg m (g-a) or mg (1-0.03) = mg (0.97).

I was trying to understand the question you were being asked.

I'm just trying to understand the reference to minimum in this problem.

So am I. It's not as explicit as I would like it to be.

Casey
 
  • #8
3,003
6
Well. That was the correct answer Astronuc!!! I NEVER would have thought that THAT was what was implied by the question. So let me get this straight. The REAL question implied is what is the minimum force that the motor CAN exert and still have the elevator accelerating at .03g?

Does that make sense?

Casey
 

Related Threads on Minimum Force ( I don't know why this is not correct)

Replies
4
Views
585
  • Last Post
Replies
4
Views
997
Replies
6
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
2K
  • Last Post
Replies
5
Views
1K
Replies
4
Views
1K
Top