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Vector components Quiz 2

physics

 If the sine and cosine of an angle made by the vector of length $6\text{ cm}$ is given as $0.5\text{ }$ and $0.86$ respectively. Find it’s $x$ and $y$ components. 

 From the trigonometric relations relating sine and cosine, calculate the vector components. 

  From trigonometric relations, $ \\ x=6\cos \theta\\ $  But, $\cos \theta =0.86$ and $r=6$ $ \\ $ $\Rightarrow x=6\left( 0.86 \right) \\ \therefore x=5.16 \\ y=6\sin \theta $ $ \\ $ But, $\sin \theta =0.5 \\ \Rightarrow y=6\left( 0.5 \right) \\ \therefore y=3$

Calculate the length of the vector, if the angle made by the vector with respect to x-axis is $30{}^\circ $ and projection along its x-axis is $4\text{ cm}$. 

The vector is resolved into two components and is written from the trigonometric relations. 

  From the trigonometric relations, we know that, $ \\ \cos \theta =\dfrac{x}{r} \\ $  So, $\Rightarrow r=\dfrac{x}{\cos \theta }$ $ \\ $ Where, $\theta =30{}^\circ \\ \Rightarrow r=\dfrac{4}{0.866} \\ \therefore r=4.6$

A bicycle moves with an initial velocity of $30\text{ }m{{s}^{-1}}$ at an angle of $60{}^\circ $ with the vertical axis. Calculate the components of the velocity. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3b5dc11a-23a5-4804-90f5-8a049c97245f4441020548826547727.png"/> 

The angle that it makes with x-axis is, $90{}^\circ -60{}^\circ =30{}^\circ $. The components of the vector are obtained from the trigonometric relations.

 The angle that the velocity vector makes with the x-axis will be $90{}^\circ -60{}^\circ =30{}^\circ $.Now, the resultant velocity vector is given by,  $ \\ v={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j} \\ {{v}_{x}}=r\cos \theta \\ $ But, $r=30$ and $\theta =30{}^\circ \\ \Rightarrow {{v}_{x}}=30\left( \cos 30{}^\circ  \right) \\ \Rightarrow {{v}_{x}}=25.9\text{ m}{{\text{s}}^{-1}} \\ $  Similarly, y component is given as, $ \\ {{v}_{y}}=r\sin \theta \\ \Rightarrow {{v}_{y}}=30\left( \sin 30{}^\circ  \right) \\ \therefore {{v}_{y}}=15\text{ m}{{\text{s}}^{-1}} \\ $  The resultant vector is written as, $ \\ \overrightarrow{V}=25.9\hat{i}+15\hat{j} \\ $ The x- component of the velocity vector is $25.9$ and the y-component of the velocity vector is $15$.

 If vector $\overrightarrow{{{F}_{1}}}=\hat{i}+6\hat{j}+5\hat{k}$   represents the force exerted by one object on a table and $\overrightarrow{{{F}_{2}}}=2\overrightarrow{{{F}_{1}}}$ is the force exerted by the second object on the table. Calculate the magnitude of the net force exerted by both the objects on the table and the unit vector along the resultant vector? 

 The magnitude of the vector is given as the square root of the sum of the squares of its vector components. The direction of the resultant vector is obtained by finding the unit vector along the resultant vector. 

 The combined force will be, $ \\ \overrightarrow{F}=\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}} \\ \Rightarrow \overrightarrow{F}=\hat{i}+6\hat{j}+5\hat{k}+2\left( \hat{i}+6\hat{j}+5\hat{k} \right) \\ \Rightarrow \overrightarrow{F}=3\hat{i}+18\hat{j}+15\hat{k} \\ \left| \overrightarrow{F} \right|=\sqrt{9+324+225} \\ \Rightarrow \left| \overrightarrow{F} \right|=23.62 \\ $  This is the magnitude of the resultant vector. Now, the unit vector along the resultant vector is,  $ \\ \hat{n}=\dfrac{\overrightarrow{A}}{\left| \overrightarrow{A} \right|} \\ \therefore\hat{n}=\dfrac{3\hat{i}+18\hat{j}+15\hat{k}}{\sqrt{558}}$ 

If the vector $'\overrightarrow{A'}$  of length $5$  makes an angle of $40{}^\circ $ represents the force exerted on a block and if $\vec{B}=2\vec{A}$ is another force exerted on the block. Calculate the net force exerted on the block? Also find the direction of the net force. 

The resultant force or the net force is obtained by adding both the force. The unit vector is obtained by dividing the vector by its magnitude. 

 The angle made by the vector $\vec{A}$ with the horizontal axis is $40{}^\circ $  The length of the vector is given to be equal to 5 $ \\ $ So, the resultant vector is given by,$ \\ A=r\cos \theta \hat{x}+r\sin \theta \hat{y} \\ {{A}_{x}}=r\cos \theta \\ \Rightarrow {{A}_{x}}=5\left( \cos 40{}^\circ  \right) \\ \Rightarrow {{A}_{x}}=3.8 \\ $ Now, the y-component is given by, $ \\ {{A}_{y}}=r\sin \theta \\ \Rightarrow {{A}_{y}}=5\left( \sin 40{}^\circ  \right) \\ \Rightarrow {{A}_{y}}=3.21 \\ $ Now the resultant vector is given by, $\vec{A}=3.8\hat{x}+3.21\hat{y}$ $ \\ $ Now, given that, $\vec{B}=2\vec{A}$ $ \\ $ So, $\Rightarrow \vec{B}=7.6\hat{x}+6.42\hat{y}$ $ \\ $ The resultant force vector is given by, $\vec{A}+\vec{B}=3.8\hat{x}+3.21\hat{y}+7.6\hat{x}+6.42\hat{y}$ $ \\ $ So, $\Rightarrow \vec{A}+\vec{B}=11.4\hat{x}+9.63\hat{y}$ $ \\ $ The unit vector along the resultant force vector is, $ \\ \dfrac{11.4\hat{x}+9.63\hat{y}}{\sqrt{{{\left( 11.4 \right)}^{2}}+{{\left( 9.63 \right)}^{2}}}} \\ \therefore \left| \vec{A}+\vec{B} \right|=\dfrac{11.4\hat{x}+9.63\hat{y}}{14.9}$ 