Question

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support, as shown in figure 1.0. The upper cables make angles of 41º and 63º with the horizontal. Calculate the tension in each of the three cables.

Answer 1

Given:

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cables make angles of 41º and 63º with the horizontal.

To find:

Tension in the strings

Calculation:

Applying Lami's Theorem ;

$\sf{ \dfrac{T1}{ \sin( {90}^{ \circ} + {63}^{ \circ} ) } = \dfrac{T2}{ \sin( {90}^{ \circ} + {41}^{ \circ} ) } = \dfrac{T3}{ \sin( {180}^{ \circ} - {90}^{ \circ} - {63}^{ \circ} ) }}$

$= > \sf{ \dfrac{T1}{ \cos( {63}^{ \circ} ) } = \dfrac{T2}{ \cos( {41}^{ \circ} ) } = \dfrac{T3}{ \sin( {180}^{ \circ} - {153}^{ \circ} ) }}$

$= > \sf{ \dfrac{T1}{ \cos( {63}^{ \circ} ) } = \dfrac{T2}{ \cos( {41}^{ \circ} ) } = \dfrac{T3}{ \sin( {153}^{ \circ} ) }}$

Putting T3 = 200 N ;

$= > \sf{ \dfrac{T1}{ \cos( {63}^{ \circ} ) } = \dfrac{T2}{ \cos( {41}^{ \circ} ) } = \dfrac{200}{ \sin( {153}^{ \circ} ) }}$

$= > \sf{ \dfrac{T1}{ 0.45} = \dfrac{T2}{ 0.75 } = \dfrac{200}{ \sin( {153}^{ \circ} ) }}$

$= > \sf{ \dfrac{T1}{ 0.45} = \dfrac{T2}{ 0.75 } = \dfrac{200}{ 0.45 }}$

Hence ,

$= > \sf{ \dfrac{T1}{ 0.45} = \dfrac{200}{ 0.45 }}$

$\boxed{ = > \sf{ T1 = 200 \: N }}$

Hence ,

$= > \sf{ \dfrac{T2}{ 0.75 } = \dfrac{200}{ 0.45 }}$

$\boxed{ = > \sf{ T2 = 333.33 \: N }}$

Hope It Helps.