A flat raft with a \(6 \mathrm{ft} \times 6 \mathrm{ft}\) surface floats in a fresh water lake. Determine the deflection of the raft if a 190 lb man stands at the geometric center. Fig. P1.7

First, we need to calculate the buoyancy force acting on the raft at its initial position. The buoyancy force can be calculated using the following formula: \(F_B = V \times \rho_w \times g\) where \(F_B\) is the buoyancy force, \(V\) is the volume of submerged part of the raft, \(\rho_w\) is the density of fresh water (approximately \(1000 \, \mathrm{kg/m^3}\)), and \(g\) is the acceleration due to gravity (approximately \(9.81 \, \mathrm{m/s^2}\)). We don't know \(V\) directly, so we need to find the height of the submerged part of the raft, which we'll call \(h\). Assume that the initial position of the raft means it is floating without any deflection (all its weight is supported by the buoyancy force). From this assumption, we can calculate \(h\): \(h \times 6 \, \mathrm{ft} \times 6 \, \mathrm{ft} = 190 \, \mathrm{lb}\).

To proceed, we need to convert all the units in SI (i.e., meters and newtons). 1 \(\mathrm{ft} = 0.3048 \, \mathrm{m}\), and 1 \(\mathrm{lb} = 4.44822 \, \mathrm{N}\). So, \(6 \, \mathrm{ft} = 1.8288 \, \mathrm{m}\) and \(190 \, \mathrm{lb} = 846.16 \, \mathrm{N}\).

Now we can solve the equation for \(h\): \(h \times 1.8288 \, \mathrm{m} \times 1.8288 \, \mathrm{m} = 846.16 \, \mathrm{N}\). \(h = \frac{846.16 \, \mathrm{N}}{(1.8288 \, \mathrm{m})^2} \approx 0.253 \, \mathrm{m}\).

When the man stands on the raft, his weight will cause additional force on the raft and hence increases the submerged part of the raft. To calculate the additional submerged part of the raft, we can use the additional force due to man's weight and the formula for buoyancy force. \(\Delta F_B = \Delta V \times \rho_w \times g\) \(\frac{\Delta F_B}{\rho_w \times g} = \Delta V = h_{new} \times 1.8288 \, \mathrm{m} \times 1.8288 \, \mathrm{m} - h \times 1.8288 \, \mathrm{m} \times 1.8288 \, \mathrm{m}\), where \(h_{new}\) is the height of the submerged part of the raft when the man is standing on it. Since the additional force is due to the man's weight, we have \(\Delta F_B = 846.16 \, \mathrm{N}\).

We can now solve the above equation for \(h_{new}\). \(h_{new} \times 1.8288 \, \mathrm{m} \times 1.8288 \, \mathrm{m} - h \times 1.8288 \, \mathrm{m} \times 1.8288 \, \mathrm{m} = \frac{846.16 \, \mathrm{N}}{1000 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2}}\) \(h_{new} \approx 0.253 \, \mathrm{m} + \frac{846.16 \, \mathrm{N}}{1000 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2} \times 1.8288^2 \, \mathrm{m^2}} \approx 0.266 \, \mathrm{m}\).

The deflection of the raft is the difference between the new submerged height of the raft and the initial submerged height of the raft: Deflection \(\Delta h = h_{new} - h \approx 0.266 \, \mathrm{m} - 0.253 \, \mathrm{m} \approx 0.013 \, \mathrm{m}\). Now we convert the deflection back to feet: Deflection \(\Delta h = 0.013 \, \mathrm{m} \times \frac{1 \, \mathrm{ft}}{0.3048 \, \mathrm{m}} \approx 0.043 \, \mathrm{ft}\). So, the deflection of the raft when the man stands at the geometric center is approximately 0.043 ft.