Question

A cube of green jelly rests on a table. You push the jelly laterally with a force F = TOO) at its top surface and find that instead of sliding across the table it takes the shape of a parallelogram (as shown in the diagram) where the lateral displacement is 0.116m. The modulus of rigidity of the jelly is
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Answer 1

Answer:

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,

\tan 30=\frac{0.125}{y}\to y=0.217\ mtan30=

y

0.125

→y=0.217 m

Thus the area of a face of the cube is,

A=y^2=0.0468\ m^2A=y

2

=0.0468 m

2

Therefore the shearing stress is,

\sigma=\frac{30}{0.0468}=641\frac{N}{m^2}σ=

0.0468

30

=641

m

2

N

The shearing strain is

\theta=30\degreeθ=30°