physics qns?

Question

TOTAL 3 qns:
A child slides across a floor in a pair of rubber-soled shoes. The friction force acting on each foot is 22 N, the footprint area of each foot is 16 cm2, and the thickness of the soles is 5.0 mm. Find the horizontal distance traveled by the sheared face of the sole. The shear modulus of the rubber is 3.0 106 Pa.

Another qn:
9.16 A 10.0 kg block of metal is suspended from a scale and immersed in water as in Figure P9.30. The dimensions of the block are 12.0 cm 10.5 cm 10.5 cm. The 12.0 cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water

a) What are the forces exerted by the water on the top and bottom of the block? (Do not ignore the effect of the air above the water. Take P0 = 1.0130 105 N/m2.) Ftop N
Fbottom N
(b) What is the reading of the spring scale?
N
(c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

Answer 1

The shear modulus G is (shear stress)/(shear strain). Shear stress is the lateral force divided by the area, or in your case S = 22/16 N/cm^2. Shear strain is ∆x/h, where ∆x is the horizontal displacement, and h=thickness. Therefore ∆x/h = S / G, or

∆x = h*S/G.

The force on the top of the block is rho*ht*g*A + Pa*A, where rho = density of water, ht is distance from the top of the block to the water surface, g = accel of gravity, A = area of the top of the block. Pa is the air pressure. The same applies to the bottom of the block, only hb is distance from surface to bottom of the block.

Bouyant force = volume of the block * rho