1) Here it is given that the lateral displacement of the tip of the vertical edge is 0.125 m. This corresponds to the edge turning through 30 degrees. So we have,
tan30=y0.125→y=0.217 m
Thus the area of a face of the cube is,
A=y2=0.0468 m2 Therefore the shearing stress is,
σ=0.046830=641m2N The shearing strain is
θ=30°
2) a)
Ww=mg−ρwga3
Ww=9.8(0.024−1000(0.02)3)=0.157 N b)
Wa=mg−ρaga3
Ww=9.8(0.024−800(0.02)3)=0.172 N
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