Question

Forces 2N and 1FN act at 45 degrees to give resultant root 10 then F=?? ​

Answer 1

hope this solution helps you.

Answer 2

ANSWER :–

$\begin{lgathered}{ \bold{f = - \sqrt{2} \: \: and \: \: {f} = 3 \sqrt{2} }} \\\end{lgathered}$

EXPLANATION :–

GIVEN : —

▪︎ Two forces are (2)N and (1F)N act at 45⁰ .

▪︎ Resultant of vectors are √10 N.

TO FIND :–

Value of 'F'.

SOLUTION :–

☞ We know that when two forces a and b act at θ degree angle then resultant is –

$\begin{lgathered}\\ \implies{ \boxed{ \bold{R = \sqrt{ {a}^{2} + {b}^{2} - 2ab \cos( \theta) } }}} \\\end{lgathered}$

▪︎ Now put the values –

$\begin{lgathered}\\ \implies{ \bold{ \sqrt{10} = \sqrt{ {(2)}^{2} + {(1f)}^{2} - 2(2)(1f) \cos( {45}^{0} ) }}} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{ \sqrt{10} = \sqrt{4 + {f}^{2} - 4f( \frac{1}{ \sqrt{2} }) }}} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{ \sqrt{10} = \sqrt{4 + {f}^{2} - 2 \sqrt{2} f }}} \\\end{lgathered}$

• Now square both side –

$\begin{lgathered}\\ \implies{ \bold{10 = 4 + {f}^{2} - 2 \sqrt{2} f }} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{{f}^{2} - 2 \sqrt{2} f - 6 = 0}} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{{f}^{2} - 3 \sqrt{2} f + \sqrt{2} f - 6 = 0}} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{f({f} - 3 \sqrt{2} ) + \sqrt{2}( f - 3 \sqrt{2} )= 0}} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{(f + \sqrt{2}) ({f} - 3 \sqrt{2} ) = 0}} \\\end{lgathered}$

$\begin{lgathered}\\ \implies{ \bold{f = - \sqrt{2} \: \: , \: \: {f} = 3 \sqrt{2} }} \\\end{lgathered}$