Chapter 6, Problem 69P

(a)

To determine

To sketch: A graph of the applied force versus the extension of the spring.

Introduction: According to Hook’s law, there is a linear relation between the force applied to the spring and the extension of the spring.

Explanation of Solution

Given Information:

The data that represents the different loads on a spring and the different length of the stretches is shown below:

Force $F$ (N)	Length $L$ (mm)
2.0	15
4.0	32
6.0	49
8.0	64
10	79
12	98
14	112
16	126
18	149
20	175
22	190

Assuming there is no extension in the spring when there is no force applied on the spring, at $x=0$ then $F=0$. So, include the origin point also to draw the graph.

Redraw the table for the given data,

Force $F$ (N)	Length $L$ (mm)	Length $L$ (m)
0.0	0.0	0.0
2.0	15	0.015
4.0	32	0.032
6.0	49	0.049
8.0	64	0.064
10	79	0.079
12	98	0.098
14	112	0.112
16	126	0.126
18	149	0.149
20	175	0.175
22	190	0.190

Plotting the graph of applied force versus the extension of the spring from the following data given.

Figure I

Conclusion:

Therefore, the graph of the applied force versus the extension of the spring is shown in figure I.

(b)

To determine

The straight line that best fits the data by least square fitting.

Answer to Problem 69P

Solution: The straight line that best fits the data by least square fitting is $F=\left(116.2\text{N}/\text{m}\right)L+0.45\text{N}$.

Explanation of Solution

Using least square method, estimate the slope of the best fit straight line. Drawing the table for the force and the length of the spring,

$F\left(\text{N}\right)$	$L\left(\text{m}\right)$	$F\cdot L \left(\text{N}\cdot \text{m}\right)$	$L^2\left({\text{m}}^{\text{2}}\right)$
0.0	0.0	0	0
2.0	0.015	0.03	0.000225
4.0	0.032	0.128	0.001024
6.0	0.049	0.294	0.002401
8.0	0.064	0.512	0.004096
10	0.079	0.79	0.006241
12	0.098	1.176	0.009604
14	0.112	1.568	0.012544
16	0.126	2.016	0.015876
18	0.149	2.682	0.022201
20	0.175	3.5	0.030625
22	0.190	4.18	0.0361
${\displaystyle \sum \text{F}}=132$	${\displaystyle \sum \text{L}=1.089}$	${\displaystyle \sum \text{F}\cdot \text{L}}=16.876$	${\displaystyle \sum {\text{L}}^{\text{2}}=0.1409}$

Formula to calculate the slope of the best fit straight line is,

$m=\frac{{\displaystyle \sum F\cdot L-\frac{\left({\displaystyle \sum L}\right)\cdot \left({\displaystyle \sum F}\right)}{n}}}{{\displaystyle \sum L^2-\frac{{\left({\displaystyle \sum L}\right)}^2}{n}}}$

• $m$ is the slope of the best fit line.

Substitute $132$ for ${\displaystyle \sum \text{F}}$, $1.089$ for ${\displaystyle \sum L}$, $16.876$ for ${\displaystyle \sum \text{F}\cdot \text{L}}$, $0.1409$ for ${\displaystyle \sum {\text{L}}^{\text{2}}}$ and $12$ for $n$.

$\begin{array}{c}m=\frac{16.876-\frac{132\times 1.089}{12}}{0.1409-\frac{{\left(1.089\right)}^2}{12}}\\ =116.2\end{array}$

Formula to calculate the intercept on force axis is,

$b=\left(\frac{{\displaystyle \sum F}}{n}\right)-m\left(\frac{{\displaystyle \sum L}}{n}\right)$

• $b$ is the intercept on the force axis.

Substitute $132$ for ${\displaystyle \sum \text{F}}$, $1.089$ for ${\displaystyle \sum L}$ $116.2$ for $m$ and $12$ for $n$.

$\begin{array}{c}b=\left(\frac{132}{12}\right)-116.2\left(\frac{1.089}{12}\right)\\ =0.45\end{array}$

Hence the best fit line using the least square method is $F=\left(116.2\text{N}/\text{m}\right)L+0.45\text{N}$.

Conclusion:

Therefore, the straight line that best fits the data by least square fitting is $F=\left(116.2\text{N}/\text{m}\right)L+0.45\text{N}$.

(c)

To determine

Whether to use best fit line using least square method, all the given points must be take or some points can be ignored.

Introduction: The least square method is use to fit the best straight line for a given data. It gives the best fit line.

Explanation of Solution

Since the graph that is plot in part (b) is made from the best fit line method using least square method. All the given points are used to find the straight line. If some points are ignored then the value of the best fit line can be changes.

So, all the given point must be taken, no points can ignore while plotting the best fit line using the least square method.

Conclusion:

Therefore, to plot the best fit straight line using least square method all the given points must include.

(d)

To determine

The value of spring constant from the slope of the best fit line.

Answer to Problem 69P

Solution: The value of spring constant from the slope of the best fit line is $116.2\text{N}/\text{m}$.

Explanation of Solution

Given Information:

The best fit straight line equation is $F=\left(116.2\text{N}/\text{m}\right)L+0.45\text{N}$.

Formula to calculate the straight line equation is,

$y=mx+c$

• $m$ is the slope of the line.
• $c$ is the intercept of the line on y-axis.

Compare the best fit straight line equation with the standard straight line equation.

$m=116.2\text{N}/\text{m}$

Formula to calculate the spring force in the spring is,

$F=kx+c$

• $F$ is the applied force on the spring.
• $k$ is the spring constant.
• $x$ is the extension in the spring due to force.
• $c$ is any constant.

The slope of the force equation is $k$. So, the value of spring constant from the slope of the best fit line is,

$k=m$

Substitute $116\text{N}/\text{m}$ for $m$ to find $k$,

$\begin{array}{c}k=116.2\text{N}/\text{m}\\ \approx 116\text{N}/\text{m}\end{array}$

Conclusion:

Therefore, value of spring constant from the slope of the best fit line is $116\text{N}/\text{m}$.

(e)

To determine

The force exerted on the suspended object by the spring.

Answer to Problem 69P

Solution: The force exerted on the suspended object by the spring is $\text{12}\text{.65}\text{N}$.

Explanation of Solution

Given Information:

The best fit straight line equation is,

$F=\left(116.2\text{N}/\text{m}\right)L+0.45\text{N}$.

Formula to calculate the force on the object by the spring is,

$F=\left(116.2\text{N}/\text{m}\right)L+0.45$

• $L$ is the extension in the spring.

Substitute $105\text{mm}$ for $L$ to find $F$.

$\begin{array}{c}F=\left(116.2\text{N}/\text{m}\right)\left(105\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)+0.45\text{N}\\ =12.65\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on the suspended object by the spring is $\text{12}\text{.65}\text{N}$