Question

Calculate the smaller force if the bigger force is $40\mathrm{N}$ and the resultant force is perpendicular to the smaller force. The two forces are acting at an angle of ${120}^{\circ }$

Answer 1

Step 1:Given data & Formula used:

Let the angle $\mathrm{\theta }={120}^0$

The smaller force is $20\mathrm{N}$.

$\mathrm{R}.\mathrm{F}=\sqrt[2]{{{\mathrm{F}}_1}^2+{{\mathrm{F}}^2}_2+2{\mathrm{F}}_1{\mathrm{F}}_2\mathrm{cos\theta }}$

Let A be the bigger force and B be the smaller force.

Where

R.F=Resulting force

${\mathrm{F}}_1,{\mathrm{F}}_2$=Two known forces

$\mathrm{\theta }$=angle between two forces

Step 2:Calculating the smaller force:

Given that,

Let $\underset{\mathrm{A}}{\to }$ and $\underset{\mathrm{B}}{\to }$ be two vectors.

$\mathrm{A}=40\mathrm{N}$

We know that $\mathrm{\theta }={120}^0$

Thus, ${\left|\mathrm{RF}\right|}^2={\left|\mathrm{A}\right|}^2+{\left|\mathrm{B}\right|}^2+2\mathrm{ABcos\theta }$

we know that $\mathrm{\theta }={120}^{\circ }$substituting in the above formula,

Value for cos $\cos {120}^{\circ }=\frac{1}{2}$

${\mathrm{RF}}^2={\mathrm{A}}^2+{\mathrm{B}}^2-\mathrm{AB}$

Given that the resultant makes a right angle with the smaller force

So $\mathrm{B}=-\mathrm{A}\mathrm{cos\theta }$ (1)

Substituting the known values that are,

$\mathrm{A}=40\mathrm{N}$

$\mathrm{\theta }={120}^{\circ }$

Substituting in equation 1 we get,

$\mathrm{B}=40\times -\frac{1}{2}$

$\mathrm{B}=20\mathrm{N}$

Thus the smaller force is $20\mathrm{N}$.