Alexander

answers: 1

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Answer:

➡ See the attachment free body diagram.

=》T₁ + T₂ = mg

- m is the mass of AB bar = 10 kg

- g is gravitational acceleration = 9.8 m/s²

=》 T₁ + T₂ = 10 × 9.8

=》 T₁ + T₂ = 98 N ------(1)

✅ Consider torque at point O (Centre/mid pt. of AB bar).

=》 (T₁ × r) - (T₂ × r) = 0

- r is the length of OA for T₁ & OB for T₂. But both r are same, because O is the midpoint of AB.

=》 r (T₁ - T₂) = 0

=》 T₁ - T₂ = 0

=》 T₁ = T₂ =

The tension in each wire is 49 N.

[NOTE ➡ If we take g = 10 m/s², then the value of tension will be different.]

- E be the Young's modulus of steel = 2.0 × 10¹¹ Pa

- Force = T₁ = 49 N

- r = radius of steel wire = = 0.30 mm = 0.0003 m

- l = 2.00 m.

The extension of the steel wire is 1.737 × 10⁻³ m.

The energy stored is 42.5 × 10⁻³ J

☆ Extension is same in both steel and brass wire. Because the resultant tension is same in both.

- = 1.0 × 10¹¹ Pa

- Force = T₂ = 49 N

- l = 2.00 m

- = 1.737 × 10⁻³ m

The diameter of the brass wire is 8.45 × 10⁻⁴ m.

197

David

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