A rod with a diameter of 3.00 mm, projects 2.0 cm horizontally outward from a wall. A 50.0 N weight is hung from the end of the rod and the rod deflects downward by 2.00 × 10^-6 m. The shear strain of the rod is A) 1.00 × 10^-4. B) 1.50 × 10^-4. C) 2.00 × 10^-4. D) 2.50 × 10^-4. E) 3.00 × 10^-4.

Question

Quizplus · Accepted Answer

Shear strain is calculated as the change in length (deflection) over the original length. In this case, the deflection is $2.00 \times 10^{-6}$ m and the original length (height in this context, since we're considering the vertical deflection as the change in length) is the diameter of the rod, which is $3.00 \times 10^{-3}$ m. Thus, shear strain = $\frac{2.00 \times 10^{-6} \, \text{m}}{3.00 \times 10^{-3} \, \text{m}} = \frac{2}{3} \times 10^{-3} = 0.6667 \times 10^{-3}$, which rounds to $6.67 \times 10^{-4}$, but given the options, it seems there might be a misunderstanding in the interpretation of the question or a mistake in calculation. The correct approach to find shear strain in this context would involve understanding the deformation in terms of shear, but given the options and typical understanding, none directly match the calculation. Re-evaluating the calculation for a direct match with the provided options under typical assumptions of shear strain might not directly apply here due to a misunderstanding in the application of the concept to the given problem's specifics. The correct answer based on the typical calculation of shear strain does not match due to a conceptual misunderstanding in applying shear strain to this scenario.

A Rod with a Diameter of 3