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Two vectors have equal magnitudes of 12.7 units. They are oriented as shown in Fig. and their vector sum is . Find (a) the x and)' components of . (b) the magnitude of , and (c) the angle makes with the +x axis

Two vectors   have equal magnitudes of 12.7 units. They are oriented as shown in Fig. and their vector sum is . Find (a) the x and)' components of . (b) the magnitude of , and (c) the angle  makes with the +x axis

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
Assumptions:
Let us assume that the vectors are given as:
236-1420_1.PNG
We assume that the magnitude of vectors above are given as a , b and r respectively.
We also assume that the angle made by vector \overrightarrow{a}from positive x axis, measured anticlockwise is given by\theta _{a} whereas the angle made by vector\overrightarrow{a} from positive x axis, measured counterclockwise is \o.
Given:
236-2360_2.PNG
First we write the vectors\overrightarrow{a} and \overrightarrow{b}in unit vector notations so that we can calculate the vector \overrightarrow{r} using simple addition of their components
The figure below shows the vector diagram, representing vectors \overrightarrow{a}and\overrightarrow{b} respectively.
It can be seen from the figure that the angle (say \theta _{b}) made by vector \overrightarrow{b}from line AB, measured counterclockwise, is:
236-1390_3.PNG

236-478_4.PNG
One should note that the above figure is not subjected to accurate scaling.
Therefore we can write the vector components of vector \overrightarrow{b}as:
236-952_5.PNG
The negative sign in the vector component is due to the fact that the vector componentb_{x}\widehat{i} points in the direction opposite to that with the unit vector\widehat{i} along the positive x axis.
Therefore vector\overrightarrow{b} is:
236-1945_6.PNG
If the angle made by vector\overrightarrow{a} with respect to positive x axis, measured counterclockwise is\theta _{a} , we can write the vector components of the vector as:
236-1988_7.PNG

(a) The vector\overrightarrow{r} can now be given as:
236-854_8.PNG
Substituting the value of components of vectors \overrightarrow{a}and \overrightarrow{b} from equation (1) and (2), we have:
236-976_9.PNG
Now for \theta _{a} = 28.2°,\theta _{b} = 46.8° , a = 12.7 units and b = 12.7 units, we can calculate the vector \overrightarrow{r}using equation (3). Since the magnitude of vectors \overrightarrow{a}and \overrightarrow{b} are equal, we can modify the equation (3) using a = b as:236-1117_10.PNG
One must note that we have just substituted the magnitude of vector\overrightarrow{a} with that of vector\overrightarrow{b} to obtain the equation above.
Now we substitute the given values of,\theta _{a} = 28.2°, b = 12.7 units and calculated value of\theta _{b} = 46.8° in equation (4) to obtain the vector \overrightarrow{r}as:
236-1951_11.PNG
From calculated value of vector\overrightarrow{r} , we have
236-1650_12.PNG
Therefore the horizontal component of vector\overrightarrow{r} is 2.54 units.
The vertical component of vector \overrightarrow{r}is 15.1 units.
(b) The magnitude of vector\overrightarrow{r} is:
236-408_13.PNG
Therefore the magnitude of vector \overrightarrow{r}is 15.3 units
(c) The angle made by vector \overrightarrow{r}with positive x axis, measured counterclockwise can be given as:
236-1808_14.PNG
Therefore the angle subtended by vector \overrightarrow{r}is 80.4°with positive x axis, measured counterclockwise.

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