Answer to Question #105547 in Physics for Liboma Zenzo

Question

A cube of green jelly rests on a table. The jelly is pushed laterally with a
force of 30 N at it top surface and changes into a parallelogram shape
instead of sliding, with a lateral displacement of 0.125 m. Calculate the
shear strain induced in the cube of jelly.

Answer 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,

$$\tan 30=\frac{0.125}{y}\to y=0.217\ m$$

Thus the area of a face of the cube is,

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

Therefore the shearing stress is,

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

The shearing strain is

$$\theta=30\degree=\frac{\pi}{6}\ rad$$