Question

Pb.7. A weight Q is suspended from a small ring, C, supported by two cords, AC and BC.
The cord AC is fastened at A while the cord BC passes over a smooth frictionless pulley at B
and carries a weight P as shown in Fig.28. If P = Q and a = 50°, find the angle B
​

Answer 1

$\huge\mathbb{\color{coral}QUESTION}$

♕ A weight Q is suspended from a small ring C supported by two cords AC and BC. The cord AC is fastened at A while cord BC passes over a frictionless pulley at B and carries a weight P. If P = Q and $\rm{\alpha}$ = 50°, find the value of $\rm{\beta}$.

$\huge\mathbb{\color{olive}SOLUTION}$

$\Large\bf{\color{indigo}GiVeN,}$

• Mass of Q = Mass of P

• $\rm{\alpha}$ = 50°

➡ See the attachment free body diagram.

☆ S is an unknown symbol as shown in the diagram.

$\bf\red{Resolving\:horizontally,}$

$\longmapsto\:\:\bf\blue{S\:\sin{50}\:=\:Q\:\sin{\beta}\:}$

$\longmapsto\:\:\bf{S\:=\:Q\:\dfrac{\sin{\beta}}{\sin{50}}\:}--(1)$

$\bf\red{Resolving\:Vertically,}$

$\longmapsto\:\:\bf\purple{S\:\cos{50}\:+\:Q\:\cos{\beta}\:=\:Q\:}$

$\longmapsto\:\:\bf{S\:\cos{50}\:=\:Q\:-\:Q\:\cos{\beta}\:}$

$\longmapsto\:\:\bf{S\:\cos{50}\:=\:Q\:\Big(1\:-\:\cos{\beta}\Big)\:}$

✔ Putting the value of S from equation(1), we get

$\longmapsto\:\:\bf{Q\:\dfrac{\sin{\beta}}{\sin{50}}\times{\cos{50}}\:=\:Q\:\Big(1\:-\:\cos{\beta}\Big)\:}$

$\longmapsto\:\:\bf{\dfrac{\cos{50}}{\sin{50}}\:=\:\dfrac{(1\:-\:\cos{\beta})}{\sin{\beta}}\:}$

$\longmapsto\:\:\bf{\cot{50}\:=\:\dfrac{(1\:-\:\cos{\beta})}{\sin{\beta}}\:}$

$\longmapsto\:\:\bf{0.839\:=\:\dfrac{(1\:-\:\cos{\beta})}{\sin{\beta}}\:}$

$\longmapsto\:\:\bf{0.839\:{\sin{\beta}}\:=\:1\:-\:\cos{\beta}\:}$

✒ Squaring both sides,

$\longmapsto\:\:\bf{(0.839\:{\sin{\beta}})^2\:=\:(1\:-\:\cos{\beta})^2\:}$

$\longmapsto\:\:\bf{0.703\:{\sin^2{\beta}}\:=\:1\:+\:\cos^2{\beta}\:-\:2\cos{\beta}\:}$

$\longmapsto\:\:\bf{0.703\:(1\:-\:\cos^2{\beta})\:=\:1\:+\:\cos^2{\beta}\:-\:2\cos{\beta}\:}$

$\longmapsto\:\:\bf{0.703\:-\:0.703\:\cos^2{\beta}\:=\:1\:+\:\cos^2{\beta}\:-\:2\cos{\beta}\:}$

$\longmapsto\:\:\bf{1.703\:\cos^2{\beta}\:-\:2\cos{\beta}\:+\:0.297\:=\:0\:}$

$\longmapsto\:\:\bf{\cos^2{\beta}\:-\:1.174\:\cos{\beta}\:+\:0.174\:=\:0\:}$

☆ Solving the above equation, we get

$\longmapsto\:\:\bf\green{\beta\:=\:63.13°}$