Question

Two forces f and x acting at a point have a resultant of√3f . And is perpendicular to f. Then value of x is answer

Answer 1

Given:

Two forces f and x acting at a point have a resultant of√3f. x is perpendicular to f .

To find:

Value of x ;

Calculation:

Angle between f and x = 90° ;

$\therefore \: r = \sqrt{ {f}^{2} + {x}^{2} + 2fx \cos( \theta) }$

$= > \: r = \sqrt{ {f}^{2} + {x}^{2} + 2fx \cos( {90}^{ \circ} ) }$

$= > \: r = \sqrt{ {f}^{2} + {x}^{2} + 2fx(0) }$

$= > \: r = \sqrt{ {f}^{2} + {x}^{2} }$

Now , putting available values;

$= > \: \sqrt{3} f= \sqrt{ {f}^{2} + {x}^{2} }$

Squaring on both sides;

$= > \: 3 {f}^{2} = {f}^{2} + {x}^{2}$

$= > {x}^{2} = 3 {f}^{2} - {f}^{2}$

$= > {x}^{2} = 2 {f}^{2}$

$= > x = \sqrt{2 {f}^{2} }$

$= > x = \sqrt{2 } f$

So, final answer is:

$\boxed{ \bold{ \large{ x = \sqrt{2 } f}}}$

Answer 2

Answer:

But there is no calculation

Explanation:

Given:

Two forces f and x acting at a point have a resultant of√3f. x is perpendicular to f .

To find:

Value of x ;

Calculation:

Angle between f and x = 90° ;

\therefore \: r = \sqrt{ {f}^{2} + {x}^{2} + 2fx \cos( \theta) }∴r=f2+x2+2fxcos(θ)

= > \: r = \sqrt{ {f}^{2} + {x}^{2} + 2fx \cos( {90}^{ \circ} ) }=>r=f2+x2+2fxcos(90∘)

= > \: r = \sqrt{ {f}^{2} + {x}^{2} + 2fx(0) }=>r=f2+x2+2fx(0)

= > \: r = \sqrt{ {f}^{2} + {x}^{2} }=>r=f2+x2

Now , putting available values;

= > \: \sqrt{3} f= \sqrt{ {f}^{2} + {x}^{2} }=>3f=f2+x2

Squaring on both sides;

= > \: 3 {f}^{2} = {f}^{2} + {x}^{2}=>3f2=f2+x2

= > {x}^{2} = 3 {f}^{2} - {f}^{2}=>x2=3f2−f2

= > {x}^{2} = 2 {f}^{2}=>x2=2f2

= > x = \sqrt{2 {f}^{2} }=>x=2f2

= > x = \sqrt{2 } f=>x=2f

So, final answer is:

\boxed{ \bold{ \large{ x = \sqrt{2 } f}}}x=2f