Question

2.39. The resultant of two forces acting at an angle f 150°
is 10 N and is perpendicular to one of the forces. The
other force is​

Answer 1

$\underline{\red{\frak{ \: Given:- \: }}}$

• Resultant of two force, R = 10N
• $\theta$ = 150°
• $\sf \vec{R} \perp \vec{A}$

$\underline{\red{\frak{ \: Find:- \: }}}$

• $\sf |\vec{B}|$

$\underline{\red{\frak{ \: Diagram:- \: }}}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(1.02,1.02){\framebox(0.3,0.3)}\put(5.36,1){\vector(1,0){3mm}}\put(1.79,4){\vector(-2,1){3mm}}\put(1,4.5){\vector(0,1){2mm}}\multiput(5.6,1)(0.3,0){9}{\line(1,0){1mm}}\qbezier(4.8,1.5)(6,1.9)(6.5,1)\put(0.5,3){ \bf \overrightarrow{R} }\put(4,2.4){ \bf \overrightarrow{B} }\put(3,0.4){ \bf \overrightarrow{A} }\put(5,1.7){ \bf \theta = {150}^{ \circ} } \end{picture}$

$\underline{\red{\frak{ \: Solution:- \: }}}$

Here, we have

$\to\sf R^2 = A^2 + B^2 + 2AB \cos 159^{\circ}$

$\to\sf 10^2 = A^2 + B^2 + 2AB \bigg( \dfrac{ - \sqrt{3}}{2} \bigg)$

$\to\sf 100 = A^2 + B^2 + AB \bigg( - \sqrt{3} \bigg)$

$\to\sf 100 = A^2 + B^2 - \sqrt{3} AB \quad \bigg\lgroup Equation\:1\bigg\rgroup$

Now,

$\to \sf \tan {90}^{ \circ} = \dfrac{B\sin 150^{\circ}}{A + B\cos 150^{\circ}}$

$\to \sf \tan {90}^{ \circ} = \dfrac{B \times \dfrac{1}{2} }{A + B \bigg( \dfrac{ - \sqrt{3} }{2} \bigg)}$

$\to \sf \tan{90}^{ \circ} =B \times \dfrac{1}{2} \times \dfrac{2}{A + B \big( - \sqrt{3}\big)}$

$\to \sf \tan{90}^{ \circ} =B \times \dfrac{1}{A + B \big( - \sqrt{3}\big)}$

$\to \sf \tan{90}^{ \circ} = \dfrac{B}{2A - \sqrt{3} B}$

$\to \sf \infty = \dfrac{B}{2A - \sqrt{3} B}$

$\to \sf 2A - \sqrt{3} B = 0$

$\to \sf 2A = \sqrt{3} B$

$\to \sf A = \dfrac{\sqrt{3}}{2} B$

Substituting the value of A in eq. 1

$\sf\implies 100 = A^2 + B^2 - \sqrt{3} AB$

$\sf\implies 100 = \bigg( \dfrac{ \sqrt{3} }{2} B\bigg) ^2 + B^2 - \sqrt{3} \bigg( \dfrac{ \sqrt{3} }{2} B\bigg)B$

$\sf\implies 100 = \dfrac{3}{4} B ^2 + B^2 - \sqrt{3} \times \dfrac{ \sqrt{3} }{2} B \times B$

$\sf\implies 100 = \dfrac{1}{4} B ^2$

$\sf\implies 100 \times 4 =B ^2$

$\sf\implies 400 =B ^2$

$\sf\implies \sqrt{400} =B$

$\sf\implies 20N =B$

$\underline{\boxed{\sf\therefore B = 20N}}$