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3 years ago

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Your friend has been hired to design the interior of a special executive express elevator for a new office building. This elevat

or has all the latest safety features and will stop with an acceleration of g/4 in the case of an emergency. The management would like a decorative lamp hanging from the unusually high ceiling of the elevator. He designs a lamp which has three sections which hang one directly below the other. Each section is attached to the previous one by a single thin wire, which also carries the electric current. The lamp is also attached to the ceiling by a single wire. Each section of the lamp weighs 10.0 N. Because the idea is to make each section appear that it is floating on air without support, he wants to use the thinnest wire possible. Unfortunately the thinner the wire, the weaker it is.
Required:
Calculate the force on each wire in case of an emergency stop.

1 answer:

Xelga [282]3 years ago

4 0

Answer:

The force on each wire is

$T_1 = 12.5 \ N$

$T_2 = 25 \ N$

$T_3 = 50 \ N$

Explanation:

From the question we are told that

The acceleration at which the elevator will stop is $a = \frac{g}{4}$

The weight of each section of the wire is $W = \ 10 \ N$

Generally $W_1 = W_2 = W_3$ here $W_1 , W_2 , W_3$ are weight at each section

Generally considering the first section, the force acting along the y-axis is mathematically represented as

$\sum F_y_1 = T_1 - W_1 = m * a$

Here $T_ 1$ represents the tension on the wire at the first section while $W_1$ represents the weight of the lamp at the first section

So

$T_1 - 10 = m * \frac{g}{4}$

=> $T_1 - 10 = \frac{W_1}{4}$

=> $T_1 - 10 = \frac{10}{4}$

=> $T_1 = 12.5 \ N$

Generally considering the second section, the force acting along the y-axis is mathematically represented as

$\sum F_y_2 = T_2 -T_1- W_2 = m * a$

=> $T_2 - T_1- 10 = m * \frac{g}{4}$

=> $T_2 - 12.5- 10 = \frac{W_2}{4}$

=> $T_2- 12.5- 10 = \frac{10}{4}$

=> $T_2 = 25 \ N$

Generally considering the third section, the force acting along the y-axis is mathematically represented as

$\sum F_y_3 = T_3- T_2 -T_1- W_3 = m * a$

$T_3 - T_2 - T_1- 10 = m * \frac{g}{4}$

$T_3 - 25 - 12.5- 10 = \frac{W_2}{4}$

$T_3 - 25 - 12.5- 10 = \frac{10}{4}$

$T_3 = 50 \ N$

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2 years ago

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Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

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$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

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Now, each speed is the magnitude of respective velocity vector:

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$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

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$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

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