You are designing a lamp for the interior of a special executive express elevato

Question

You are designing a lamp for the interior of a special executive express elevator in a new office building. The lamp has two sections that hang one directly below the other. The bottom section is attached to the top one by a thin wire and the upper section is attached to the ceiling by another thin wire. Because the idea is to make each section appear to be floating without support, you want to use the thinnest (and thus weakest) wire possible. You decide to calculate the force each wire exerts on the lamp sections during an emergency stop. The elevator has all the latest safety features and will stop with an acceleration of g/3 in any emergency. Each section of the lamp weighs 7.7 Newtons.

(a) What is the algebraic expression for the maximum tension in the upper wire in terms of the weight of each section of the lamp (W) and the gravitational acceleration (g)?
(b) What is the numerical value of the maximum tension in the upper wire?  N
(c) What is the algebraic expression for the maximum tension in the lower wire in terms of the weight of each section of the lamp (W) and the gravitational acceleration (g)?
(d) What is the numerical value of the maximum tension in the lower wire (make sure to include units and put a space between the number and the units)?  N

Explanation / Answer

Assume wire is mass less since no mass was given for the wire nor were we given a way to calculate the mass of the wire.

Mass each lamp segment

F = MA
7.7 N = M * 9,8 m/s^2
M = 0.785 kg

Lets calculate the maximum acceleration
This follows from the elevator is going down at constant velocity .

acceleration on lamp is g and it is brought to a stop with an acceleration of 0.33 g ==> total acceleration = 1.33 g
The maximum acceleration is 1.33 g.

A = 9.8 m/s^2 * 1.33 = 13.034 m/s^2

First segment - bottom wire

F = MA
F = 0.785kg * 13.034 m/s^2 = 10.231 N ===> Maximum tension in first wire

Second segment is twice the first since it carries the tension in the bottom wire

T = 2 * 10.231= 20.462 N

3rd segment is 3 times the first since it carries the 1st and 2nd wires

T = 10.231+20.462 = 30.69 N

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