Question

Two force 5N and 10N inclined at an angle of 30 degree are acting on a particle.Find the resultant force and its direction

Answer 1

Explanation: R^2=P^2 +Q^2 +2PQ×COS theta

R^2=5^2+10^2+2×5×10×cos30°

R^2=25+100+100×cos30

R^2=125+100×cos30

R^2=125+100×(3^1/2)/2

R^2=125+50×(3^1/2)

R=[125+50×(3^1/2)]^1/2

Answer 2

Given:

• 5N & 10N forces are inclined at 30°.

To find:

Resultant vector and direction of the resultant?

Calculation:

As per Triangle Law of Vector Addition, the resultant value will be :

$| \vec{r}| = \sqrt{ {5}^{2} + {10}^{2} + 2.5.10. \cos( {30}^{ \circ} ) }$

$\implies | \vec{r}| = \sqrt{ 25+ 100 + 100 \cos( {30}^{ \circ} ) }$

$\implies | \vec{r}| = \sqrt{ 25+ 100 + 50 \sqrt{3} }$

$\implies | \vec{r}| = \sqrt{ 125+ 50 \sqrt{3} }$

$\implies | \vec{r}| = \sqrt{ 125+ 86.6}$

$\implies | \vec{r}| = \sqrt{ 211.6}$

$\implies | \vec{r}| = 14.54 \: N$

So, value of resultant is 14.54 N.

For direction of vector, let angle made by r with 5N force be $\phi$.

Then , we can say:

$\tan(\phi) = \dfrac{10 \sin( {30}^{ \circ} ) }{5 + 10 \cos( {30}^{ \circ} ) }$

$\implies \tan(\phi) = \dfrac{5 }{5 + 5 \sqrt{3} }$

$\implies \tan(\phi) = 0.36$

$\implies \phi = {19.7}^{ \circ}$

Hope It Helps.