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Q: Elevator and Tension ( No Answer,   3 Comments )
Question  
Subject: Elevator and Tension
Category: Science > Physics
Asked by: alwysnforevr002-ga
List Price: $2.00
Posted: 22 Jul 2006 17:47 PDT
Expires: 21 Aug 2006 17:47 PDT
Question ID: 748632
An elevator weighing 2.00 x 105 N is supported by a steel cable. What
is the tension in the cable when the elevator is accelerated upward at
a rate of 3.00 m/s2? (g = 9.81 m/s2)
 
for this equation i think i may have found the answer but i am really
not sure. I do not know if there is an equation for tension. Here is
what i thought:
 
Weight = mass x acceleration
mass = weight in N divided by gravity
2 x 10(to the fifth) / 9.81 m/s
= 20387.4 x 3
=61162.1N
Answer  
There is no answer at this time.

Comments  
Subject: Re: Elevator and Tension
From: anonymous3141-ga on 22 Jul 2006 20:55 PDT
 
Tension in the cable = mass * acceleration.
You calculated the mass already.
Multiply by mass (9.81+3)
Subject: Re: Elevator and Tension
From: aliang-ga on 26 Jul 2006 05:45 PDT
 
The sum of all forces must accelerate the elevator upwards. We know
the gravitational force on the elevator already (20000 Newtons) from
the question. We also know that the only two forces acting on the
elevator are a downwards force from gravity and an upwards force from
the tension in the cable. Assume upwards is positive.

Therefore:

SUM OF ALL FORCES = mass * actual acceleration
Tension ? force of gravity = mass * actual acceleration

Therefore:
Tension = mass * actual acceleration + force of gravity

We know everything there except the mass, but we know the
gravitational force on the elevator which is proportional to mass
(F=ma), therefore m=F/a=force of gravity/(9.81)

Tension = actual acceleration * force of gravity / 9.81 + force of gravity

Tension = 3.00 * 20000 / 9.81 + 20000 Newtons = 26116 Newtons
Subject: Re: Elevator and Tension
From: hedgie-ga on 06 Aug 2006 00:31 PDT
 
To all - who but big numbers into a computer:

This comment is about the use of the E-notation:
In the question Alwys  says: 
2.00 x 105 N 
2 x 10(to the fifth)   ...

 Notation developed particularly for computer and scientific calculators
 would represent this number   as

2.0E5

For details, see 

http://www.nyu.edu/pages/mathmol/textbook/scinot.html
http://members.aol.com/profchm/sci_not.html


BTW
Calculation by  aliang-ga  looks OK.  If this would be a real life case,
not just a school test, I would worry about that number a=3 m/s *s :

 Elevators accelerate only during first second; once they start moving,
 a is zero, and tension is weight + friction.

 Acceleration during that first second is not a constant; Time profile of that
 peaks depends on the motor, and is not easy to mesure (without a
scope, or a computer .. ).
   So, if this would really be  about required dimnension of the cable, I would
   suggest not to depend on this simplistic calculation.

Hedgie

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