Question

A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Given the magnitude and direction of the acceleration of the body.

Answer 1

We can make use of vector law of addition to solve this question. The sum of two forces is given as $\sqrt{{{F}_{1}}^{2}+F_{2}^{2}+2{{F}_{1}}{{F}_{2}}\cos \theta }$, here $\theta$ is the angle between the two forces ${{F}_{1}}\And {{F}_{2}}$. A special case is when the Angle between the two forces is $\theta =90{}^\circ$ then the resultant force is given as $\sqrt{{{F}_{1}}^{2}+F_{2}^{2}}$, since $\cos 90{}^\circ =0$. The angle made by the resultant force can be given as $\alpha ={{\tan }^{-1}}\left( \dfrac{{{F}_{y}}}{{{F}_{x}}} \right)$.

Complete step by step answer:
We know that force is a vector quantity and therefore can be added vectorially. Sum of the two forces can be given as $\sqrt{{{F}_{1}}^{2}+F_{2}^{2}+2{{F}_{1}}{{F}_{2}}\cos \theta }$, where θ is the angle between the two forces. Here the angle between the two forces is ${{90}^{\circ }}$.Therefore,
$\sqrt{{{F}_{1}}^{2}+F_{2}^{2}}$
Substituting the values, we get,
$F=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( 8 \right)}^{2}}}$
$\Rightarrow F =\sqrt{36+64}$
$\Rightarrow F =\sqrt{100}\\\ \Rightarrow F =10N$
According to Newton's second law of motion force is defined as a product of mass and acceleration. And therefore the resultant acceleration is,
$a=\dfrac{F}{m}\\\ \Rightarrow a =\dfrac{10}{5}\\\ \Rightarrow a =2m/{{s}^{2}}$
The angle made by the resultant force is given as,
$\alpha ={{\tan }^{-1}}\left( \dfrac{{{F}_{y}}}{{{F}_{x}}} \right)$.
$\therefore\alpha ={{\tan }^{-1}}\left( \dfrac{-6}{8} \right)=-36.87{}^\circ$
The negative sign indicated an angle made in clockwise direction from the x-axis. The direction of the resultant force is the same as the direction of the resultant acceleration and is therefore of magnitude of $2m/{{s}^{2}}$ and directed 36.87° in clockwise direction from the force of 8 N.

Note: One should know the general formula for addition of two vectors to solve such type of problem. The angle made by the resultant force$\left( \alpha \right)$ with the x axis can be given as $\alpha ={{\tan }^{-1}}\left( \dfrac{{{F}_{y}}}{{{F}_{x}}} \right)$. According to Newton's second law of motion force is defined as a product of mass and acceleration. This acceleration takes place in the direction of net force applied.