Find the buoyant force on a cube of iron 0.05m on each side that is immersed in water
Answers
Explanation:
a) The pressure (including the contribution from the atmosphere) at a depth of h
top
=L/2 (corresponding to the top of the block) is
p
top
=p
atm
+ρgh
top
=1.01×10
5
Pa+(1030kg/m
3
)(9.8m/s
2
)(0.300m)=1.04×10
5
Pa where the unit Pa(pascal) is equivalent to N/m
2
. The force on the top surface (of area A = L^2=0.36\hspace{0.05cm}m^2is F_{top}=p_{top}A=3.75\times 10^4\hspace{0.05cm}N$$.
(b)The pressure at a depth of h
top
=3L/2(thatofthebottomoftheblock)isp_{top}=p_{atm}+\rho gh_{bot}=1.01\times 10^5\hspace{0.05cm}Pa +(1030\hspace{0.05cm}kg/m^3)(9.8\hspace{0.05cm}m/s^2)(.900\hspace{0.05cm}m) =1.10\times 10^5\hspace{0.05cm}PawherewerecallthattheunitPa(pascal)isequivalenttoN/m^2.TheforceonthebottomsurfaceisF_{bot}=p_{bot}A=3.96\times 10^4\hspace{0.05cm}N$$.
(c) Taking the difference F
bot
−F
top
cancels the contribution from the atmosphere (including any numerical uncertainties associated with that value) and leads to F
bot
−F
top
=ρgh(h
bot
−h
top
)A=ρgL
3
=2.18×10
3
N which is to be expected on the basis of Archimedes' principle. Two other forces act on the block: an upward tension T and a downward pull of gravity mg. To remain stationary, the tension must be T=mg−(F
bot
−F
top
)=(450kg)(9.80m/s
2
)−2.18×10
3
N=2.23×10
3
N.
(d) This has already noted in the previous part: F
b
=2.18×1
3
N, and T+F
b
=mg.