Question

the resultant of two forces 2P and √2P is √10P. The angle between the forces is......................................................................

Answer 1

Answer:

HOPE IT HELPS YOU.....

Answer 2

ANGLE BETWEEN THE FORCES IS 45°.

Given:

• Resultant = (√10)P
• The forces are 2P and (√2)P

To find:

• Angle between the vectors ?

Calculation:

Let the angle between the vectors be $\theta$.

$| \vec{r}| = \sqrt{ {(2P)}^{2} + {( P\sqrt{2}) }^{2} + 2.(2P).(P \sqrt{2} ) \cos( \theta) }$

$\implies P \sqrt{10} = \sqrt{ {(2P)}^{2} + {( P\sqrt{2}) }^{2} + 2.(2P).(P \sqrt{2} ) \cos( \theta) }$

$\implies P \sqrt{10} = \sqrt{4 {P}^{2} + 2{P }^{2} + 4 \sqrt{2} {P}^{2} \cos( \theta) }$

• Squaring both sides:

$\implies 10 {P}^{2} = 4 {P}^{2} + 2{P }^{2} + 4 \sqrt{2} {P}^{2} \cos( \theta)$

$\implies 10 {P}^{2} = 6 {P}^{2} + 4 \sqrt{2} {P}^{2} \cos( \theta)$

$\implies 4 = 4\sqrt{2} \cos( \theta)$

$\implies \cos( \theta) = \dfrac{1}{ \sqrt{2} }$

$\implies \theta = {45}^{ \circ}$

So, the angle between the vectors is 45°.