Using the \(\widehat i\)- \(\widehat j\)system, it is very easy to add or subtract vectors. Consider the following two vectors:
\[\begin\overrightarrow > = 3\widehat i - 5\widehat j\\\overrightarrow > = - 2\widehat i + \widehat j\end\]
What will be the sum of these two vectors? To calculate the sum, we can simply add the \(\widehat i\)components separately, and the \(\widehat j\)components separately. Thus,
\[\begin\overrightarrow > + \overrightarrow > = \left( \right)> \right)\widehat i + \left( <\left( < - 5>\right) + 1> \right)\widehat j\\\,\,\,\,\,\,\,\,\,\,\,\,\,\quad = \widehat i - 4\widehat j\end\]
The correctness of this answer can be verified through the following figure:
Why does this method of separately adding components work? Think in terms of net effects. Suppose that
\[\begin\overrightarrow > = \widehat i + \widehat j\\\overrightarrow > = \widehat i + \widehat j\end\]
Now, \(\overrightarrow > + \overrightarrow > \) represents the net effect of \(\overrightarrow > \) and \(\overrightarrow > \). How can we calculate this effect?
Starting from any point, say, the origin, \(\overrightarrow > \) will take us \(\) units in the \(\widehat i\)-direction and \(\) units in the \(\widehat j\)-direction:
Now, if we apply a displacement of \(\overrightarrow > \), this will take us another \(\) units in the \(\widehat i\)-direction, and \(\) units in the \(\widehat j\)-direction:
If we now see our net movement in the \(\widehat i\)-direction, it is \(\left( + > \right)\), while our net movement in the \(\widehat j\)-direction is \(\left( + > \right)\). Thus, the net effect of \(\overrightarrow > + \overrightarrow > \) was to move us \(\left( + > \right)\) units in the \(\widehat i\)-direction and \(\left( + > \right)\) units in the \(\widehat j\)-direction. This means that
\[\overrightarrow > + \overrightarrow > = \left( + > \right)\widehat i + \left( + > \right)\widehat j\]
Subtraction works similarly. Suppose that
\[\overrightarrow > - \overrightarrow > = \left( - > \right)\widehat i + \left( - > \right)\widehat j\]
The \(\widehat i\)-components and the \(\widehat j\)-components can be added separately. Verify this graphically using the following two vectors once again:
\[\begin\overrightarrow > = 3\widehat i - 5\widehat j\\\overrightarrow > = - 2\widehat i + \widehat j\end\]
Example 1: Let \(\overrightarrow a \) and \(\overrightarrow b \) be two vectors such that
\[\begin\overrightarrow a = 4\widehat i - 3\widehat j\\\overrightarrow b = - 2\widehat i + 7\widehat j\end\]
Solution: We have:
Example 2: A vector \(\overrightarrow a \) of length 5 units is inclined at an angle of 30 0 to the horizontal. Another vector \(\overrightarrow b \) of length 8 units is inclined at an angle of 120 0 to the horizontal:
Find the following vectors:
Solution: Let us first specify \(\overrightarrow a \) and \(\overrightarrow b \) in the \(\widehat i\)- \(\widehat j\) system. Consider the following figure:
\[\begin\overrightarrow a = \overrightarrow + \overrightarrow = \left( ^0>> \right)\widehat i + \left( ^0>> \right)\widehat j\\ \qquad\qquad\quad \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac>\widehat i + \frac\widehat j\\\overrightarrow b = \overrightarrow + \overrightarrow = - \left( ^0>> \right)\widehat i + \left( ^0>> \right)\widehat j\\ \qquad\qquad\quad \,\,\,\,\,\,\,\,\,\,\,\,\, = - 4\widehat i + 4\sqrt 3 \widehat j\end\]
\[\begin\overrightarrow a + \overrightarrow b = \left( > - 4> \right)\widehat i + \left( + 4\sqrt 3 > \right)\widehat j\\\overrightarrow a - \overrightarrow b = \left( > + 4> \right)\widehat i + \left( - 4\sqrt 3 > \right)\widehat j\\2\overrightarrow a + 3\overrightarrow b = \left( \right) + \left( < - 12\widehat i + 12\sqrt 3 \widehat j>\right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( \right)\widehat i + \left( \right)\widehat j\end\]
Example 3: A boy is pulling a box with a force of 10 N towards the east direction. Another boy is pulling the same box with a force of 8 N at an angle of 60 0 to the east:
If the mass of the box is 4 kg, find its acceleration \(\overrightarrow a \).
Solution: Note that our answer should be a vector, as the acceleration of the box will have a magnitude as well as a direction. Let us establish an \(\widehat i\)- \(\widehat j\) system such that the \(\widehat i\)-direction is the east direction, while the \(\widehat j\)-direction is the north direction:
Now, if the two forces being applied are represented as \(\overrightarrow > \) and \(\overrightarrow > \), we have:
\[\begin\overrightarrow > = 10\widehat i\,\,\,>\\\overrightarrow > = \left( ^0>> \right)\widehat i + \left( ^0>> \right)\widehat j\\\,\,\,\,\,\, = 4\widehat i + 4\sqrt 3 \widehat j\,\,\,>\end\]
\[\overrightarrow > + \overrightarrow > = 14\widehat i + 4\sqrt 3 \widehat j\,\,\,>\]
Clearly, the magnitude of the resultant force is
Let \(\theta \) be the inclination of this force with the \(\widehat i\)-direction:
Using the component form of the resultant vector, we can directly write:
The acceleration can be written as
The magnitude of the acceleration is
The direction of the acceleration is the same as the direction of the net force.
Example 4: Three forces act on a box, as shown in the following figure:
Find the resultant force.
Solution: We take the direction of the first force as the \(\widehat i\)-direction, and the direction perpendicular to it as the \(\widehat j\,\)-direction:
Now, we specify the three forces in this \(\widehat i\, - \,\widehat j\,\)\(\) system (the units are Newtons):
\[\begin\overrightarrow > = 6\widehat i\\\overrightarrow > = \left( ^0>> \right)\widehat i + \left( ^0>> \right)\widehat j\,\\\,\,\,\,\,\, \;= - 2\widehat i + 2\sqrt 3 \widehat j\,\\\overrightarrow > = \left( ^0>> \right)> \right)\widehat i + \left( ^0>> \right)> \right)\widehat j\,\\\,\,\,\,\,\, \;= - 4\sqrt 3 \widehat i - 4\widehat j\,\end\]
The net force will be
The magnitude of the net force is
If \(\theta \) is the angle of inclination of the net force with the \(\widehat i\)-direction, we have:
Make sure you understand this last step carefully by drawing an accurate diagram for the net force.