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erica [24]
3 years ago
14

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.
Physics
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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