Physics, asked by juhiiii2771, 10 months ago

A cube of green jelly rests on a table.the jelly is pushed laterally with a force of 30N at its top surface and changes into a parallelogram shap instead of sliding, with a lateral displacement of 0.125m.calculate the shear strain induced in the cube of jelly.

Answers

Answered by poonambhatt213
1

The shearing strain is π/6 or 30°.

Explanation:

=> It is given that,

Lateral displacement = 0.125 m

The angle corresponds to the edge = 30 degrees.

Force = 30 N

=> Thus, tan 30° = 0.125 / y

y = 0.125/0.577 (∵ tan 30° = 0.577)

y = 0.217

=> Now, the area of a face of the cube:

A = y²

A = (0.217 m)²

A = 0.0469 m²

=> Hence, the Shearing stress, σ :

σ = Force/area of a face of the cube

= 30 / 0.0469

≈ 641 Nm⁻²

=> The shearing strain θ is:

θ = 30° = π/6 rad

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Answered by Anonymous
6

\small{\colorbox{cyan}{✓Verified Answer}}

The shearing strain is π/6 or 30°.

Explanation:

=> It is given that,

Lateral displacement = 0.125 m

The angle corresponds to the edge = 30 degrees.

Force = 30 N

=> Thus, tan 30° = 0.125 / y

y = 0.125/0.577 (∵ tan 30° = 0.577)

y = 0.217

=> Now, the area of a face of the cube:

A = y²

A = (0.217 m)²

A = 0.0469 m²

=> Hence, the Shearing stress, σ :

σ = Force/area of a face of the cube

= 30 / 0.0469

≈ 641 Nm⁻²

=> The shearing strain θ is:

θ = 30° = π/6 rad

\huge\colorbox{lime}{sᴀᴋsʜɪ࿐ ❤}

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