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A rod with a diameter of 2.00 mm, projects 20.0 mm horizontally outward from a wall. A 10.0 N weight is hung from the end of the rod and the rod deflects downward by 4.00 × 10-6 m. The shear strain of the rod is

Question

A rod with a diameter of 2.00 mm, projects 20.0 mm horizontally outward from a wall. A 10.0 N weight is hung from the end of the rod and the rod deflects downward by 4.00 × 10-6 m. The shear strain of the rod is

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Solution

The shear strain (γ) is defined as the change in length (ΔL) divided by the original length (L). In this case, the change in length is the deflection of the rod, which is 4.00 × 10^-6 m, and the original length is the length of the rod, which is 20.0 mm or 0.020 m.

So, the shear strain can be calculated as follows:

γ = ΔL / L γ = 4.00 × 10^-6 m / 0.020 m γ = 2.00 × 10^-4

So, the shear strain of the rod is 2.00 × 10^-4.

This problem has been solved

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