Two blocks of wood are partially submerged in water. Each block of wood is a cub

Question

Two blocks of wood are partially submerged in water. Each block of wood is a cube, 4.50 cm on each side. The left hand block has a density of 0.75 g/cm3, while the right hand block has a density of 0.85 g/cm3. The density of water is 1.00 g/cm3. How much of each block sticks out of the water? (That is, what is the height d of the top surface above the water surface, in cm?) Strings are now attached to the cubes in part (a) and tension is applied to each string such that each is exactly halfway out of the water (2.25 cm below the surface, 2.25 cm above). What is the tension in each string? The strings in part(b) are now attached to a thin massless bar of length 0.450 m, which is supported by a third string. Where should this third (top) string be attached (what should be the distance x from the left-hand edge), and what should the tension FT top in the third string be, if the two blocks are to remain suspended halfway out of the water?

Explanation / Answer

a)

Vtot = L^3 , given L = 4.5 cm

Vsubmerged = L^2*dsubmerged

density of water* L^2*dsubmerged*g = density of wood * L^3*g

dsubmerged =( density of wood / density of water) *L

d out of water = L - dsubmerged

d out of water = L - ( density of wood / density of water) *L

d out of water = L -* ( 1 -density of wood / density of water)

d out of water left = 4.5 * (1-0.75/1) = 1.125 cm

d out of water right = 4.5 * (1-0.85/1) = 0.675 cm

b)

Ftension = density of wood * Vtotal*g - density of water * Vsubmerged*g

Vtotal = L^3

Vsubmerged = L^3/2

Ftension = g*L^3 * ( density of wood - density of water/2)

Ftension Left = 9.8 * 0.045^3 *( 750-1000/2) = 0.223 N

Ftensionright = 9.8 * 0.045^3 *( 850-1000/2) = 0.312 N

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.