Question

The magnitude of the resultant of two forces of magnitude f and 2½f is f. Then the angle between them is

Answer 1

Answer:

Given:

Magnitude of forces are f and 2½f

Resultant comes as f

To find:

Angle between the vectors.

Calculation:

Let the vectors be A and B

|A| = f ; |B| = 2½f = (5/2)f

As per Addition of Vector using Parallelogram Law :

$\vec r = \vec A + \vec B$

$=> | \vec r | = \sqrt{ {f}^{2} + { ( \dfrac{5f}{2}) }^{2} + \{2 \times f \times \dfrac{5f}{2} \cos( \theta) \}}$

$=> f = \sqrt{ {f}^{2} + {( \dfrac{5f}{2}) }^{2} + \{2 \times f \times \dfrac{5f}{2} \cos( \theta) \} }$

Squaring on both sides :

$=> {f}^{2} = {f}^{2} + {( \dfrac{5f}{2}) }^{2} + \{2 \times f \times \dfrac{5f}{2} \cos( \theta) \}$

$=>5 {f}^{2} \cos( \theta) = -\dfrac{25 {f}^{2} }{4}$

$=>\cos( \theta) = -\dfrac{5}{4}$

$=>\theta = {cos}^{ - 1} \bigg( -\dfrac{5}{4} \bigg)$

The value of cosine can never be less than (-1) . Hence this type of situation ( this resultant) is not possible.

Answer 2

$\underline{\boxed{ \huge\purple{ \rm{Answer}}}}$

Given :

The magnitude of the resultant of two forces of magnitude F and 2(1/2)F is F.

To Find :

Angle between them...

Formula :

Resultant of two Force vectors is given by...

$\: \: \dag \: \underline{ \boxed{ \rm{ \bold{ \pink{\rm| \vec{F}| = \sqrt{ { | \vec{F_1}| }^{2} + { | \vec{F_2}| }^{2} + 2 | \vec{F_1}| | \vec{F_2}| cos \theta } }}}}} \: \dag$

Calculation :

$\implies \rm \: F = \sqrt{ {F}^{2} + { (\frac{5F}{2}) }^{2} + 2(F)( \frac{5F}{2} )cos \theta} \\ \\ \implies \rm \: {F}^{2} = {F}^{2} + \frac{25{F}^{2}}{4} + 5{F}^2cos \theta \\ \\ \implies \rm \: - \frac{25}{4} = 5cos \theta \\ \\ \implies \rm \: cos \theta = - \frac{5}{4} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{ \theta = {cos}^{ - 1}( \frac{5}{4}) }}}}} \: \red{ \clubsuit}$

I think there is some mistake in question because maximum value of cosine is -1 .