Question

The little lion is playing in a pool, in the coast of the city of Antofagasta, with a (closed) cube of edge s and density
homogeneous ρ c. If you fully submerge the block, as the figure shows, such that your
center of mass is at a depth H. Find:

a) The mass of the block.
b) The net force on the upper face and on the face bottom of the block.
c) The thrust that the block experiences.
d) The apparent weight of the block for the child.
e) If the child drops the block, with what acceleration falls to the bottom of the pool?

Answer 1

TH Given side of the cube(a) = 5 Density of cube le Distance b/w free surface of water in the pool and centre of mass of Cube = H find h mass of the block we know to 71312 Density x volume M = 4.53 ③ Net force acting on upper and bottom face ok cube mass volume mos Density 1 Fluid is at Rest, so shear stress on top and bottom surface is zero, Only Hydrostatic force will act, whic is given as - Fa . . A where I es density of fluid A is area of surface h = distance b/w free surface of fluid and c. G of surface. Here ħ for top Surface - H-1 and for bottom surface th) = H + 5 / Area = ² fluid is water = fw = 1000 kg/m3 Hydrostatic force on top surface Igah 1000 x 9,81 xs? (H) 9810 210 (13-) Hydrostatic force on botom surface (FH) bottom = ggah 103 * 9.81 Xs”(H+) 2310(1843 Net force = {fu ) bottom - ( foreStop 9810/4s? + ) - 9810(+32 ) So

9810(H8*+-+*+ IFnet Free 981053 as Thrust on block - Thrust (Upward) acting on block is called buoyancy force which is given Fe = wt. of fluid displace - fra vor of fluid xg - Spa x vol of body immesse x xg FB = lo lovo X9.81.5 FB 981053 This is d) The apparent weight of block - The child will feel that wt of block is slightly less than its original wt. because of buoyancy force acting Glock. apparent weight - original wx - Buoyanty ferie Sxgx vol of block S x 9.8185 981053 $379.814 – 9810) weight FB 3 apparent