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

Question

Quizplus · Accepted Answer

Shear strain is defined as the deformation of the material per unit length. In this case, the deformation is $2.00 \times 10^{-6} \mathrm{~m}$ and the length over which this deformation occurs is $2.0 \mathrm{~cm} = 0.02 \mathrm{~m}$. Thus, shear strain = deformation/length = $(2.00 \times 10^{-6} \mathrm{~m})/(0.02 \mathrm{~m}) = 1.00 \times 10^{-4}$.

A Rod with a Diameter Of 3.00 mm3.00 \mathrm{~mm}3.00 mm , Projects 2.0 cm2.0 \mathrm{~cm}2.0 cm

A Rod with a Diameter Of $3.00 \mathrm{~mm}$ , Projects $2.0 \mathrm{~cm}$