Chapter 7.5, Problem 9E

To determine

To find: The assembly’s force constant.

Answer to Problem 9E

The answer: $7238\frac{lb}{in}$

Explanation of Solution

Given information:

The force of $21,714\text{lb}$ to compress a coil spring assembly on a New York city transit assembly subway car from its free height $\text{8 in}$ to its fully compressed height of $\text{5 in}$ .

Calculation:

To obtain the force constant, use Hooke's Law for Springs which is given by the formula:

  $F=kx\mathrm{.....}(1)$

where $\text{F}$ is the amount of force needed to compress or extend the spring by $x$ units from its normal length and $\text{k}$ is the spring's force constant.

The natural length $x_0=8\text{ in}$ and

final length $x_f=5\text{ in}$ due to force

  $F=21,714\text{lb}$ use Equation $(1)$ to solve the force constant of the spring:

  $\begin{array}{c}21714=k(8-5)\\ \Rightarrow k=\frac{21714}{3}\\ \Rightarrow k=7238\end{array}$

Therefore, assembly’s force constant is $7238\frac{lb}{in}$ .

$\text{(b)}$

To determine

To find: How much work does it take the compare the assembly the first half inch and second half inch.

Answer to Problem 9E

The answer:

The first half inch is $905\text{ lb-in}$ and

The second half inch is $2714\text{ lb-in}$ .

Explanation of Solution

Given information:

The force of $21,714\text{lb}$ to compress a coil spring assembly on a New York city transit assembly subway car from its free height $\text{8 in}$ to its fully compressed height of $\text{5 in}$ .

Calculation:

To solve for the work done by compress the spring, use the formula:

  

Calculate the work done by compress the spring from its native state to the first half inch using Equation $\text{(2)}$:

  

Therefore, the required first half inch is $905\text{ lb-in}$ .

Calculate the work done by compress the spring from its native state to the second half inch using Equation $\text{(2)}$:

  

Therefore, the required second half inch is $2714\text{ lb-in}$ .