Question

Consider the 65.0kg ice skater being pushed by two others shown in the figure.

Find the direction and magnitude of ${{F}_{tot}}$ the total force exerted on herbie the others, given the magnitudes of ${{F}_{1}}$ and ${{F}_{2}}$ are 26.4N and 18.6N respectively.

Answer 1

Hint: You could first represent the whole situation in a free body diagram. Then you will find that the resultant of the two vectors representing the applied forces will give the total applied force on the skater. Now, for the direction you could take the tangent of the appropriate angle in the figure.
Formula used:
Vector addition,
$Sum=\sqrt{{{A}^{2}}+{{B}^{2}}+AB\cos \theta }$

Complete answer:
In the question, we have a skater weighing 65kg who is being pushed by two others with forces 26.4N and 18.6N respectively. Also, we have the diagram that shows that the forces being applied are perpendicular to each other. So we could find the resultant force by simply doing vector addition of the two forces.
Let us recall that the sum of the vectors A and B at an angle $\theta $ with each other can be given by,
$Sum=\sqrt{{{A}^{2}}+{{B}^{2}}+AB\cos \theta }$
Here the angle between them is found to be$90{}^\circ $. So, net applied force on the skater of mass 65kg could be given by,
${{F}_{tot}}=\sqrt{{{F}_{1}}^{2}+{{F}_{2}}^{2}+2{{F}_{1}}{{F}_{2}}\cos 90}$
$\Rightarrow {{F}_{tot}}=\sqrt{{{26.4}^{2}}+{{18.6}^{2}}}$
$\therefore {{F}_{tot}}=32.3N$
So we found the magnitude of the total force to be 32.3N on the skater of 65kg.
We are also supposed to find the direction at which the net force is directed. We could find it from the following vector representation.

Here, the tangent of the angle can be given by,
$\tan \theta =\dfrac{{{F}_{2}}}{{{F}_{1}}}$
$\Rightarrow \tan \theta =\dfrac{18.6}{26.4}$
$\Rightarrow \theta ={{\tan }^{-1}}\left( 0.7 \right)$
$\therefore \theta =35.2{}^\circ $
So, we found that the net applied force will be directed at an angle $\theta =35.2{}^\circ $ with ${{F}_{1}}$.

Note:
You may note that throughout the solution, we have stressed the word ‘applied force’. This is because in addition to the applied forces shown in the diagram the skater also has a force due to gravity. So, we should specifically mention that we are discussing only about the applied forces.