2012: Problem 20

Topic: Fluids
Concepts: Buoyant force, Newton’s laws, Scale reading


Solution:

Note that the scales determine mass by dividing their measured force by gg. Thus, in the beginning, the first scale measures the normal force on itself as F1=M1gF_1=M_1g. The second scale measures its tension as F2=M2gF_2=M_2g.

Once the block is submerged in the water, the buoyant force FBF_B acts upward on the block reducing the tension to T=F2FBT=F_2-F_B. By Newton’s 3rd law, the block also exerts a force FBF_B down on the water so the normal force on the bottom increases to N=F1+FBN=F_1+F_B.

By Archimedes’ principle, we have

FB=ρwater(V2)gF_B=\rho_{\text{water}}\left(\frac{V}{2}\right)g

where VV is the volume of the block. Since the block has mass M2M_2,

M2=ρwoodVM_2=\rho_{\text{wood}}V
V=M2ρwoodV=\frac{M_2}{\rho_{\text{wood}}}

Substituting this back into FBF_B,

FB=ρwaterρwoodM2g2F_B=\frac{\rho_{\text{water}}}{\rho_{\text{wood}}}\frac{M_2g}{2}

Then the mass measured by the first scale is

M1=Ng=F1+FBg=M1+ρwaterρwoodM22=45kg+10kg=55kgM_1’=\frac{N}{g}=\frac{F_1+F_B}{g}=M_1+\frac{\rho_{\text{water}}}{\rho_{\text{wood}}}\frac{M_2}{2}=45\,\mathrm{kg}+10\,\mathrm{kg}=55\,\mathrm{kg}

The mass measured by the second scale is

M2=Tg=F2FBg=M2ρwaterρwoodM22=12kg10kg=2kgM_2’=\frac{T}{g}=\frac{F_2-F_B}{g}=M_2-\frac{\rho_{\text{water}}}{\rho_{\text{wood}}}\frac{M_2}{2}=12\,\mathrm{kg}-10\mathrm{kg}=2\,\mathrm{kg}

so the answer is E.

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