Question

For forces acting on an object, given by A=43.3 N East, B=46.3 N north, C=70.7 N west, D=66.7 N south. B. What is the direction of the force? (enter your answer in degrees counterclockwise from the X axis)

Answer 1

Given:

The forces;

A=43.3 N east

B=46.3 N north

C=70.7 N west

D=66.7 N south

To find:

The direction of the net force.

Explanation:

Let us assume that the east is the positive x-axis and the north is the positive y-axis.

The forces in the vector forms can be written as,

$\begin{gathered} \vec{A}=43.3\hat{i} \\ \vec{B}=46.3\hat{j} \\ \vec{C}=70.7(-\hat{i}) \\ \vec{D}=66.7(-\hat{j}) \end{gathered}$

Where i and j are the unit vectors along the x and y axes respectively.

The vector sum of the forces is,

$\vec{R}=\vec{A}+\vec{B}+\vec{C}+\vec{D}$

On substituting the known values,

$\begin{gathered} \vec{R}=43.3\hat{i}+46.3\hat{j}-70.7\hat{i}-66.7\hat{j} \\ =-27.4\hat{i}-20.4\hat{j} \end{gathered}$

The direction of a vector is,

$\begin{gathered} \theta=\tan^{-1}(\frac{-20.4}{-27.4}) \\ =36.7\degree \end{gathered}$

Final answer:

Thus the direction of the resultant force is 36.7° in counter-clockwise from the x-axis.