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$\tan\alpha=\frac{\text{y component}}{\text{x component}}$
$\Rightarrow\tan\alpha=\frac{1.3}{0.98}=1.332$
$\Rightarrow\alpha=\tan^{-1}1.32$

View full question & answer→

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.

View full question & answer→

Physics 5 Marks Questions

Take a timed test