Figure represents a light inextensible string A B C D E in which A B=B C=C D=D E and to which are attached masses M, m, and M at the points B, C, and D, respectively. The system hangs freely in equilibrium with ends A and E of the string fixed in the same horizontal line. It is given that tan α=3 / 4 and tan β=12 / 5. Then the tension in the string B C is

Question

Figure represents a light inextensible string A B C D E in which A B=B C=C D=D E and to which are attached masses M, m, and M at the points B, C, and D, respectively. The system hangs freely in equilibrium with ends A and E of the string fixed in the same horizontal line. It is given that tan α=3 / 4 and tan β=12 / 5. Then the tension in the string B C is (A) 2 mg (B) (13/10) m g (C) (3/10) m g (D) (20/11) m g

Tardigrade Admin · Accepted Answer

tan β=(12/5) ∴ cos β=(5/13) T1 cos β+T2 cos β=m g ...(i) T1 sin β=T2 sin β ...(ii) ∴ T1=T2=T ∴ 2 T cos β=m g ⇒ T=(m g/2 cos β) ⇒ T=(13/10) m g

2 mg

(13/10) m g

(3/10) m g

(20/11) m g

tanβ=512​ ∴cosβ=135​ T1​cosβ+T2​cosβ=mg ...(i) T1​sinβ=T2​sinβ ...(ii) ∴T1​=T2​=T ∴2Tcosβ=mg ⇒T=2cosβmg​ ⇒T=1013​mg

$tan β = \frac{12}{5}$
$∴ cos β = \frac{5}{13}$
$T_{1} cos β + T_{2} cos β = m g$ ...(i)
$T_{1} sin β = T_{2} sin β$ ...(ii)
$∴ T_{1} = T_{2} = T$
$∴ 2 T cos β = m g$
$\Rightarrow T = \frac{m g}{2 c o s β}$
$\Rightarrow T = \frac{13}{10} m g$