Answer:
average shear strain in the rubber is 0.25rad
Explanation:
The base has a dimension of 20mm and 50mm.
frictional force is 50 N
using the expressed shear stress,
we will determine τ = [tex]F_t/A[/tex]
τ = 50 / 1000
= 0.05MPa
shear stress τ = γ .G
τ = shear stress
γ = shear strain
G = 0.20MPa
γ = τ / G
= 0.05 / 0.20
= 0.25rad
average shear strain in the rubber is 0.25rad