# Show that for any object suspended in equilibrium by two wires,the ratio of the tensions in the wires is equal to the ratio of the secants of the horizontal angles formed by the wires. (Note:...

Show that for any object suspended in equilibrium by two wires,

the ratio of the tensions in the wires is equal to the ratio of the secants of the horizontal angles formed by the wires.

(Note: sec x = 1/cos x)

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### 1 Answer

Let the object be suspended by two wires AB and CD. Let the tension in them be T1 and T2. Let the horizontal angle formed by AB be x1 and that formed by CD be x2.

The tension in the wire AB can be divided into a horizontal component given by T1* cos x1 which acts toward the vertical plane on which the point of attachment of the wire is, and a vertical component acting upwards given by T1* sin x1. Similarly the tension in the wire CD can be divided into a horizontal component T2*cos x2 which acts towards the vertical plane on which the point of attachment of the wire is, and a vertical component acting upwards equal to T2* sin x2

As the object is in equilibrium, the sum of the vertical components is equal to the weight of the object and the horizontal components are equal.

=> T1* cos x1 = T2*cos x2

=> T1 / T2 = cos x2 / cos x1

=> T1 / T2 = sec x1 / sec x2

**This proves that the ratio of the tensions in the wires is equal to the ratio of the secants of the horizontal angles formed by the wires**