5 Marks Questions

Question 15 Marks

Refer to figure. Find

The magnitude.
x and y components
The angle with the X-axis of the resultant of $\overrightarrow{\text{OA}},\overrightarrow{\text{BC}}$ and $\overrightarrow{\text{DE}}.$

Answer

x component of $\overrightarrow{\text{OA}}=2\cos30^{\circ}=\sqrt{3}$

x component of $\overrightarrow{\text{BC}}=1.5\cos120^{\circ}=-0.75$

x component of $\overrightarrow{\text{DE}}=1\cos270^{\circ}=0$

y component of $\overrightarrow{\text{OA}}=2\sin30^{\circ}=1$

y component of $\overrightarrow{\text{BC}}=1.5\sin120^{\circ}=1.3$

y component of $\overrightarrow{\text{DE}}=1\sin270^{\circ}=-1$

R_x = x component of resultant $=\sqrt{3}-0.75+0=0.98\text{m}$

R_y = resultant y component $=1+1.3-1=1.3\text{m}$

$\therefore\ \text{Resultant, R}=\sqrt{(\text{R}_\text{x})^2+(\text{R}_\text{y})^2}$

$=\sqrt{(0.98)^2+(1.3)^2}$

$=\sqrt{0.96+1.69}$

$=\sqrt{2.65}$

$=1.6\text{m}$

So, R = Resultant $=1.6\text{m}$

If it makes and angle $\alpha$ with positive x-axis

$\tan\alpha=\frac{\text{y component}}{\text{x component}}$

$\Rightarrow\tan\alpha=\frac{1.3}{0.98}=1.332$

$\Rightarrow\alpha=\tan^{-1}1.32$

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Question 25 Marks

Add vectors $\overrightarrow{\text{A}},\overrightarrow{\text{B}}$ and $\overrightarrow{\text{C}}$ each having magnitude of 100 unit and inclined to the X-axis at angles 45°, 135° and 315° respectively.

Answer

x component of $\overrightarrow{\text{A}}=100\cos45^{\circ}=\frac{100}{\sqrt{2}}\text{ unit}$

x component of $\overrightarrow{\text{B}}=100\cos135^{\circ}=\frac{100}{\sqrt{2}}$

x component of $\overrightarrow{\text{C}}=100\cos315^{\circ}=\frac{100}{\sqrt{2}}$

Resultant x component $=\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$

y component of $\overrightarrow{\text{A}}=100\sin45^{\circ}=\frac{100}{\sqrt{2}}\text{unit}$

y component of $\overrightarrow{\text{B}}=100\sin135^{\circ}=\frac{100}{\sqrt{2}}$

y component of $\overrightarrow{\text{C}}=100\sin315^{\circ}=\frac{100}{\sqrt{2}}$

Resultant y component $=\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$

Resultant $=100$

$\tan\alpha=\frac{\text{y component}}{\text{x component}}=1$

$\Rightarrow\alpha=\tan^{-1}(1)=45^{\circ}$

The resultant is 100 unit at 45° with x-axis.

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