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When a mass of m = 29 kg is hung from the middle of a fixed straight aluminum wire, the wire sags to make an angle of 12∘ with the horizontal as shown in (Figure 1) . The elastic modulus for aluminum is 7.0×1010N/m2. Determine the radius of the wire

When A Mass Of M 29 Kg Is Hung From The Middle Of A Fixed Straight Aluminum Wire The Wire Sags To Make An Angle Of 12 With The Horizontal As Shown In Figure 1 T class=

Sagot :

The solution to the given question is that the radius of the wire will be 0.378 millimetre approximately

At equilibrium,

Lets assume Tension T in the wire

mass of the block is given as 29 Kg

T sin 12° + T sin 12° = mg  

T sin 12° + T sin 12° = 29 × 9.8

2T sin 12° = 29 × 9.8

T = ( 29 × 9.8 )/2 sin 12°

T = 141.1 / sin 12°

T = 141.1 / 0.2

T = 705.5 Newton

Stress = K × Strain

where K is the elastic modulus of the metal

elastic modulus is given as 7.0 × 10¹⁰ N/m²

T/A = 7.0 × 10¹⁰ × (Δl/l)

A = T × (l/Δl) × (1/ (7.0 × 10¹⁰) )

A = π r² = T / ( (7.0 × 10¹⁰) × ( (1/cos 12°) - 1) )

π r² =  705.5 / ( (7.0 × 10¹⁰) × 0.02234 )

r = √( 705.5 / ( π × (7.0 × 10¹⁰) × 0.02234  ) ) metre

r = √( 705.5 / ( 4910.33 × 10⁶   ) ) metre

r = √( 0.1436 × 10⁻⁶ ) metre

r = ( 0.378 × 10⁻³ ) metre

r = 0.378 millimetre

Thus we have find the radius of the wire by applying stress strain relation which came out to be 0.378 millimetre.

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