newton laws problem?

Question

you are designing a decorative lamp to hang in an express elevator. The lamp has three sections that hang one directly bellow the other, connected by thin wires.The lamp is also attached to the ceiling by a single thin wire. Each section of the lamp weighs 5 N . determine the max tension of each wire if the elevator comes to an emergence stop from a speed of 5m/s over a distance of 3.8 m.

can anyone give me a more detailed explanation where these figures came from

how do u resolve the forces on the lamps

Answer 1

a = 25/7.8 ≈ 3.28947
m = 5/9.80665 ≈ 0.5098581 kg each
F = m(a + g)
F1 = 0.5098581(3.28947 + 9.80665)
F1 = 0.5098581(13.09612)
F1 ≈ 6.677165 N
F2 ≈ 13.35433 N
F3 ≈ 20.03149 N
A safety factor should be included because mass of the wire(s) is not included in the calculations, nor are contingencies.

Answer 2

In response to "additional information" I understand your confusion.
Not having all this sort of stuff memorized, I started at; http://en.wikipedia.org/wiki/Equations_for_a_falling_body#Overview

using the equation for "Instantaneous velocity vi of a falling object that has traveled distance d"
v = sqr( 2gd), and substituting the unknown "a" for "g", I get; 2a = 5m/s^2 / 3.8m, which could also be written a = 25/7.6 (not 7.8, but since I can't type I understand how that happened, but note that he used 7.6 for the calculation as the result 3.28947 is correct)

Now that we know the de-acceleration of the elevator we can calculate weight of each section under the total acceleration (note that gravity remains in effect ) the previous answer calculated the mass of the lamp sections and then calculated the force resulting fro the total acceleration. Another way to do it would, since the given weight (5N) is m * g, is to calculate how much the acceleration of the elevator adds to g and just multiply the given weight by that. Rounding things a bit, I have g = 9.8 and a = 3.29; 3.29/9.8 = 0.336g;
a = 0.336g adding that to g, I get 1.336 ties the static weight.
1.336 * 5N = 6.68N for each section of the lamp.
For the bottom section, the tension in the wire is 6.68N, for the second section it is 6.68N for the second section, plus 6.68N for the bottom section, or 13.36, and for the top section, add another 6.68, for 20.04N (one gets slightly different numbers if one uses more digits in the calculations, but we don;t have many significant digits in the data)

Answer 3

Note to others .. the Emergency Stop is highly unlikely to be a LINEAR de-acceleration !