Question

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that ofbrass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Answer 1

Given, the length of both the wires is 0.25 cm, the initial length of the steel and the brass wire is 1.5 m and 1.0 m respectively. Also the Young’s modulus of steel is 2.0× 10 11  Pa.

The arrangement of the given system is shown in the figure below,

The total force acting on the steel wire in the loaded condition is,

F 1 =( 4 kg+6 kg )×9.81 m/ s 2 =98.1 N

Let L 1 be the length of the steel wire, d be the diameter of the steel wire, Δ L 1 be the change in the length of the wire and F 1 be the given load on the steel wire.

The Young’s modulus of the steel wire is,

Y 1 = 4 F 1 L 1 π d 2 Δ L 1 Δ L 1 = 4 F 1 L 1 π d 2 Y 1

Substituting the values in the above equation, we get:

Δ L 1 = 4( 98.1 N )( 1.5 m ) π ( 0.25× 10 −2  m ) 2 ( 2.0× 10 11  Pa ) =1.5× 10 −4  m

Hence, the elongation in the steel wire is 1.5× 10 −4  m.

Let L 2 be the length of the brass wire, d be the diameter of the brass wire, Δ L 2 be the change in the length of the wire and F 2 be the given load on brass wire.

The given load on brass wire is,

F 2 =6 kg×9.81 m/ s 2 =58.86 N

The Young’s modulus of the brass wire is,

Y 2 = 4 F 2 L 2 π d 2 Δ L 2 Δ L 2 = 4 F 2 L 2 π d 2 Y 2

Substituting the values in the above equation, we get:

Δ L 2 = 4( 58.86 N )( 1 m ) π ( 0.25× 10 −2  m ) 2 ( 0.91× 10 11  Pa ) =1.3× 10 −4  m

Hence, the elongation in the brass wire is 1.3× 10 −4  m.