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State and explain Hookes Law according to Class 11 Physics Chapter 8 Mechanical Properties of Solids. What is its significance?

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1 Answer

Answered by

Pallavi Pathak | Contributor-Level 10

6 months ago

Hooke's Law states that within the elastic limit, the stress applied to a material is directly proportional to the strain produced. Mathematically,

E stands for the modulus of elasticity also called Young's modulus in case of linear deformation.
The Hooke's Law defines the elastic behaviour of materials, determines its elastic limit and helps in material selection for construction and manufacturing. It is used to design springs and elastic components in machines. Hook's law is applicable for the solids within their elastic limit.
Hooke's law for a spring and other elastic material is shown as:

F is the force applied on the mat

...more

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4.22 $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? What are the components of a vector A= 2 $\hat{i}$ + 3 $\hat{j}$ along the directions of $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? [You may use graphical method]

Vishal Baghel

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$

...more

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made by vector $\overset{?}{Q}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(Q x / Q y) θ$ = ${- \tan}^{- 1} ? (1 / 1)$ , $θ$ = - 45 $°$ with the x axis

It is given that $\overset{?}{A}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ . We can write Ax $\overset{?}{i}$ + Ay $\overset{?}{j}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$

Comparing the components on both sides, we get

Ax =2 and Ay = 3

$|\overset{?}{A}|$ = $\sqrt{{(2}^{2}} + 3^{2}$ ) = $\sqrt 13$

Let $\overset{?}{A_{x}}$ make an angle $θ$ with the x axis, then

$\tan ? θ$ = (Ax / Ay), $θ$ = $\tan^{- 1} (? 3 / 2)$ = 56.31 $°$

So the angle between (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ + $\overset{?}{j}$ )

$θ'$ = 56.31 – 45 = 11.31 $°$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{P}$ , making an angle $θ$

= ( A cos $θ'$ ) $\overset{?}{P}$

= ( A cos 11.31) $\frac{(\overset{?}{i} + \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ $\times \frac{0.9806}{\sqrt{2}}$ ( $\overset{?}{i}$ + $\overset{?}{j}$ )

= 2.5 $\times$ $\sqrt{2}$

= $\frac{5}{\sqrt{2}}$

Let $θ'$ be the angle between the vectors (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ - $\overset{?}{j})$

$θ " = 45 + 56.31 = 101.31 °$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{Q}$ , making an angle $θ$

= (A cos $θ "$ ) $\frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ cos ( $101.31 °) \frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= -0.5( $\overset{?}{i}$ - $\overset{?}{j})$

= $- \frac{1}{\sqrt{2}}$

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