A shearing force of 50 N is applied to an aluminum rod with a length of 10 m, a cross-sectional area of 1.0 x 10 ^x ⁵ m, and shear modulus of 2.5 x ¹⁰N/m².As a result the rod is sheared through a distance of: A)zero B)2.0 mm C)2.0 cm D)20 cm E)2.0 m

Question

Quizplus · Accepted Answer

The shear strain (𝛾) can be calculated using the formula:

𝛾 = (F/A)/G

where F is the applied shearing force, A is the cross-sectional area, and G is the shear modulus of the material.

Plugging in the given values:

𝛾 = (50 N / 1.0 × 10^-4 m^2) / 2.5 × 10^10 N/m^2

𝛾 = 2.0 × 10^-3

This is the shear strain.

The distance through which the rod is sheared (𝛿x) can be calculated using the formula:

𝛿x = 𝛾 × L

where L is the length of the rod.

Plugging in the given values:

𝛿x = 2.0 × 10^-3 × 10 m

𝛿x = 2.0 × 10^-2 m

𝛿x = 2.0 mm

Therefore, the correct choice is B.

A Shearing Force of 50 N Is Applied to an Aluminum