Solution
a. Force applied F = (A) (stress)
π(5.00 ×10
−3
m)
2
(4.00×10
8
N/m
2
)=3.14×10
4
N
b. The area over which the shear stress occurs is equal to the circumference of the hole times its thickness.
Thus, A = (2 π r) t = 2π (5.00 ×10
−3
m) (5.00 ×10
−3
m)
= 1.57 ×10
−4
m
2
So, F = (A) (stress)
= (1.57 ×10
−4
m
2
) (4.00 ×10
8
N/m
2
) = 6.28 ×10
4
N