Question

A particle is acted upon by four forces simultaneously, $30N$ due east $20N$ due north $15N$ due west $40N$ due south find the resultant force of the particle?

Answer 1

Step 1: Given data

Four forces act on a particle as shown in the figure.

Step 2: To find

The resultant force on the particle.

Step 3: Calculation for magnitude of resultant force

We know, resultant of the forces acting in opposite direction can be found by taking their difference.

Here, force in the north-south direction,

$F_{NS}=40-20=20N$ towards south.

Force in the east-west direction,

$F_{EW}=30-15=15N$ towards east.

Resultant force,

$$\begin{gathered} F=\sqrt{F_{NS}^2+F_{EW}^2+2F_{NS}{F}_{EW}\cos \theta } \\ F=\sqrt{{20}^2+{15}^2+2\times 20\times 15\times \cos 90°}\left(\theta =90°\because theforcesare\perp r\right) \\ F=\sqrt{400+225+\left(600\times 0\right)} \\ F=\sqrt{625} \\ F=25N \end{gathered}$$

Step 4: Calculation for angle made by the resultant force

$$\begin{gathered} \tan \left(\theta \right)=\frac{20}{15} \\ \tan \left(\theta \right)={53}^0 \end{gathered}$$

Hence, the resultant force of the particle is $25\mathrm{N}$ and it will make ${53}^{°}$with the west.