What do we know?
F = mg, where m is the mass of
the object.
If
r is the radius of the rod then A = πr^{2}.
Length of rod = 4.8 X 10^{-2} m
Radius of Rod = 5.3 X 10^{-2} m
Sheer Force = 1.2 X 10^{3} N
Shear Modulus of Aluminum is 3.0 X
10^{10} Nm^{-2}
Plan to solve:
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Shear Modulus = [^{[Stress]} / _{[Strain]}]
Strain = [^{[ΔX]} / _{[L]}]
So we can get ΔX if we calculate Strain
Thus, the shear stress is: |