Grade 11Mechanics

A particle undergoes three successive displacements in a plane, as follows: 4.13 m southwest, 5.26 m east, and 5.94 m in a direction 64.0° north of east. Choose the x axis pointing east and the)' axis pointing north and find (a) the components of each displacement, (b) the components of the resultant displacement, (c) the magnitude and direction of the resultant displacement, and (d) the displacement that would be required to bring the particle back to the starting point.

Simran Bhatia

11 Years agoGrade 11

Aditi Chauhan

11 Years ago

Assumption:

Let us assume that the three successive displacement of the particle is defined using vectors , $\overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c}$ respectively. Also the magnitude of these vectors are represented by a, b and c.
It is also assumed that the displacement vector $\overrightarrow{r}$ of the particle from its starting point is given by vector such that the components of vector is r_x and r_y respectively.
The position vector $\overrightarrow{r}$ defines the position of the particle relative to its starting point, when the particle has moved southwest (defined by vector $\overrightarrow{a}$ ) and eastward (defined by vector ).
Given:
a = 4.13 m
b = 5.26 m
c = 5.94 m
The vector diagram of the motion of the particle is shown below:

It is important to note that the above diagram is not subjected to exact scaling.
(a) Vector $\overrightarrow{a}$ defines the southwest displacement of the particle thereby subtending an angle of 45° with the negative x axis, measured counterclockwise.
Therefore, vector $\overrightarrow{a}$ is given as:

Where a_x and a_y are the components along the x axis and y axis respectively.
Since the particle has going southwest, the horizontal and the vertical component of vector $\overrightarrow{a}$ is negative and is opposite to the direction of unit vector $\widehat{i}\ and\ \widehat{j}$ respectively.
If $\theta _{1}$ is the angle subtended by vector $\overrightarrow{a}$ with negative x axis, which in this case is 45°, we can write vector $\overrightarrow{a}$ as:

Therefore the horizontal vector component of vector $\overrightarrow{a} is -2.92\ m\ \widehat{i}$ whereas the vertical vector component of vector $\overrightarrow{a} is -2.92\ m\ \widehat{j}$ .
Vector $\overrightarrow{b}$ defines the eastward motion of the particle, and runs parallel to the x axis. Therefore the vector has horizontal component, along the unit vector $\widehat{i}$ and no vertical component.
Thus vector $\overrightarrow{b}$ can be written as:

Since the particle has gone eastward, the magnitude of its displacement is equal to the horizontal vector component $b_{x}\ that\ is\ b_{x}\ = b\ while\ b_{y} = 0$ . Substitute these conditions in the vector above to have

Therefore the horizontal vector component of vector $\overrightarrow{b}\ is 52.6\ m\ \widehat{i}$ whereas the vertical vector component is 0 m $\widehat{j}$ .
Vector $\overrightarrow{c}$ defines the motion of particle directed at 64.0° north of east. Therefore the angle subtended by vector from positive x axis is 64.0°, measured counterclockwise (refer vector diagram above).
Therefore the vector can be resolved into horizontal and vector components as:

The positive sign in both the components indicates that the components points in the direction of unit vector $\widehat{i}\ and\ \widehat{j}$ respectively.
Therefore the horizontal vector component of vector $\overrightarrow{c}\ is\ 2.60\ m\ \widehat{i}$ and the vertical vector component of vector $\overrightarrow{c}\ is\ 5.33 \ m \widehat{j}$ .
(b) To get the resultant displacement vector $\overrightarrow{r}$ , first obtain the position vector $\overrightarrow{r_{1}}$ (refer figure above) as:
$\overrightarrow{r_{1}} = \overrightarrow{a}+\overrightarrow{b}$
The net resultant displacement vector $\overrightarrow{r}$ can now be calculated from the position vector $\overrightarrow{r_{1}}$ and vector $\overrightarrow{c}$ as:

Substitute the value of vectors $\overrightarrow{a} , \overrightarrow{b} and \overrightarrow{c}$ , and to obtain vector in unit vector $\overrightarrow{r}$ notation as:

Therefore the horizontal vector component of displacement vector is 0.26 m $\widehat{j}$ whereas the vertical vector component is $8.25\ m\ \widehat{j}$
(c) As assumed, the vector is given as:
$\overrightarrow{r} = r_{x}\widehat{i} + r_{y}\widehat{j}$
On comparing the above vector with the equation (1), we have
r_x = 0.26 m
r_y = 8.25 m
The magnitude of the displacement vector $\overrightarrow{r}$ is:
$|\overrightarrow{r}| = \sqrt{(r_{x})^{2}+ (r_{y})^{2}}$

Substitute the values of r_x and r_y from above

Therefore the magnitude of resultant displacement vector is 8.25 m .
If the resultant vector displacement $\overrightarrow{r}$ makes an angle $\o$ with the positive x axis, the angle can be calculated as:

Therefore the resultant vector makes an angle of 1.85° with positive x axis, measured counterclockwise.
(d) To bring the particle back to its starting position, the particle must move along the vector $-\overrightarrow{r}$ , that is opposite to the direction of the resultant displacement vector.
Therefore the displacement vector (say $\overrightarrow{s}$ ) that will bring the particle back to its starting position is:

Therefore the displacement vector to bring the particle back to its starting point is .
-0.26 m $\widehat{i}$ -8.25 m $\widehat{j}$

Simran Bhatia

1 Answer

Aditi Chauhan

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