Answer to Question #105355 in Physics for Jake

Question

1. 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.

2. A metal cube of side 2 cm and weight of 24 g in air. Find the apparent weight when submerged in a. Water, b. Alcohol of density 0.8 g/cm3

Answer 1

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

$$\theta=30\degree$$

2) a)

$$W_w=mg-\rho_wga^3$$

$$W_w=9.8(0.024-1000(0.02)^3)=0.157\ N$$

b)

$$W_a=mg-\rho_aga^3$$

$$W_w=9.8(0.024-800(0.02)^3)=0.172\ N$$