

Subject:
Elevator and Tension
Category: Science > Physics Asked by: alwysnforevr002ga 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 

There is no answer at this time. 

Subject:
Re: Elevator and Tension
From: anonymous3141ga 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: aliangga 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: hedgiega on 06 Aug 2006 00:31 PDT 
To all  who but big numbers into a computer: This comment is about the use of the Enotation: 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 aliangga 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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