Question

A rope of mass M is hanged from two support 'A' & 'B' as shown in figure. Maximum and minimum tension in the rope is –

• A. $\frac { M g \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$ , $\frac { M g \cos \theta _ { 1 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$
• B. Mg, Mg cosq₁
• C. $\frac { M g \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$ , $\frac { M g \cos \theta _ { 1 } \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$
• D. $\frac { M g \cos \theta _ { 2 } } { \cos \left( \theta _ { 2 } - \theta _ { 1 } \right) }$ , $\frac { M g \cos \theta _ { 1 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$

Answer 1

Answer: (C)

$\frac { M g \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$ , $\frac { M g \cos \theta _ { 1 } \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$

Answer 2

Free body diagram of rope AB -

T_A cos q₁ = T_B cos q₂

T_A sin q₁ + T_B sin q₂ = Mg

\ T_A = $\frac { M g \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$

T_B = $\frac { M g \cos \theta _ { 1 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$

Free body diagram of point AC –

Horizontal equilibrium :

T_A cos q₁ = T_C

\ T_C = $\frac { M g \cos \theta _ { 1 } \cos \theta _ { 2 } } { \sin \left( \theta _ { 1 } + \theta _ { 2 } \right) }$

Tension will be maximum at A and minimum at C.