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
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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Q:1 A cube of side 0.4m rests on the floor as shown given that the cube has a mass of 200kg the pressure exerted by the cube on the floor is ( take g = 10 N kg)
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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