Answer to Question #105531 in Molecular Physics | Thermodynamics for sangulukani

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

In the question the angle between the perpendicular y-axis and the top displaced edge of the cube is not mentioned So I am assuming it to be 30^o

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,

$\tan 30^o = \dfrac{0.125} y \\y=0.217 m$

Thus the area of a face of the cube is,

$A=y^2=(0.217)^2=0.0468m^2$

The applied force is F=30 N. Therefore the shearing stress is,

$S=\dfrac{F}{A}=\dfrac{30}{0.0468}=641N/m^2$

Shearing strain is $\theta=\pi/6$