Question 9.9: Floating Down the River GOAL Apply Archimedes’ principle to ...
Floating Down the River
GOAL Apply Archimedes’ principle to a partially submerged object.
PROBLEM A raft is constructed of wood having a density of 6.00 × 10² kg/m³. Its surface area is 5.70 m², and its volume is 0.60 m³. When the raft is placed in fresh water as in Figure 9.25, to what depth h is the bottom of the raft submerged?
STRATEGY There are two forces acting on the raft: the buoyant force of magnitude B, acting upward, and the force of gravity, acting downward. Because the raft is in equilibrium, the sum of these forces is zero. The buoyant force depends on the submerged volume V_{\mathrm{{water}}} = Ah. Set up Newton’s second law and solve for h, the depth reached by the bottom of the raft.
Apply Newton’s second law to the raft, which is in equilibrium:
B-\,m_{\mathrm{raft}}g=0~~\rightarrow~~B=\,m_{\mathrm{raft}}g
The volume of the raft submerged in water is given by V_{\mathrm{{water}}} = Ah. The magnitude of the buoyant force is equal to the weight of this displaced volume of water:
B=\,m_{\mathrm{{water}}}g=\,(\rho_{\mathrm{{water}}}V_{\mathrm{{water}}})g=\,(\rho_{\mathrm{{water}}}A h)g
Now rewrite the gravity force on the raft using the raft’s density and volume:
m_{\mathrm{raft}}g=(\rho_{\mathrm{raft}}V_{\mathrm{raft}})g
Substitute these two expressions into Newton’s second law, B = m_{\mathrm{raft}} g, and solve for h (note that g cancels):
(\rho_{\mathrm{water}}A h)\cancel{g}=(\rho_{\mathrm{raft}}V_{\mathrm{raft}})\cancel{g}
h={\frac{\rho_{\mathrm{raft}}V_{\mathrm{raft}}}{\rho_{\mathrm{water}}A}}
={\frac{(6.00~\times~10^{2}\mathrm{~kg/m^{3}})(0.600~m^{3})}{(1.00~\times~10^{3}\mathrm{~kg/m^{3}})(5.70~m^{2})}}
= 0.063 2 m
REMARKS How low the raft rides in the water depends on the density of the raft. The same is true of the human body: Fat is less dense than muscle and bone, so those with a higher percentage of body fat float better.