Question

The force of 5N south acts concurrently with force of 5N 30 degrees South of East

A. Draw the diagram - TAIL to TAIL

B. Label both vectors (see page one for example)

C. Draw and LABEL the resultant and equilibrant

D.Find the magnitude and direction of the resultant and equilibrant


(scale: 1cm = 1N)

Answer 1

Answer:

The magnitude and direction of the resultant is 8.66 N and $60^{\circ}$ south of East.

The magnitude and direction the equilibrant is 8.66 N and $60^{\circ}$ North of West respectively.

Explanation:

Let

$P=5\ \text{N}\ \text{south}$

$Q=5\ \text{N}\ 30^{\circ}\ \text{south of east}$

$\theta$ = Angle between P and Q = $60^{\circ}$

Magnitude of resultant

$R=\sqrt{P^2+Q^2+2PQ\cos\theta}\\\Rightarrow R=\sqrt{5^2+5^2+2\times 5\times 5\cos60^{\circ}}\\\Rightarrow R=8.66\ \text{N}$

Direction is given by

$\theta=\tan^{-1}\dfrac{Q\sin\theta}{P+Q\sin\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{5\sin60^{\circ}}{5+5\sin60^{\circ}}\\\Rightarrow \theta=30^{\circ}$

The magnitude and direction of the resultant is 8.66 N and $30^{\circ}+30^{\circ}=60^{\circ}$ south of East.

The magnitude and direction the equilibrant is 8.66 N and $60^{\circ}$ North of West.