Question

A raft is made of 12 logs lashed together. Each is 45 cm in diameter and has a length of 6.5 m. How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of 68 kg? Do not neglect the weight of the logs. Assume the specific gravity of wood is 0.60.

Answer 1

Answer:

324 people

Explanation:

radius r = diameter / 2 = 45cm /2 = 22.5 cm or 0.225 m

We can calculate the total volume of the log by computing the volume of each cylindrical log:

$V = \pi r^2 L = \pi 0.225^2 * 6.5 = 4.6 m^3$

So the total volume of all 12 logs is

$12 * 4.6 = 55.135 m^3$

The weight that the raft can support, is the buoyancy force subtracted by its own weight.

The buoyancy force is basically the weight of the water is displaced, which is water density times volume. The log weight is also log density times its volume

$F = F_b - F_w = g\rho_wV - g\rho_rV = gV(\rho_w - \rho_r)$

where $\rho_w = 1000kg/m^3,\rho_r$ are the density of water and raft, respectively. But since we have the specific gravity of wood is 0.6. That measn

$\rho_r / \rho_w = 0.6$

$\rho_r = 0.6\rho_w$

Therefore $\rho_w - \rho_r = \rho_w - 0.6\rho_w = 0.4\rho_w = 0.4*1000 = 400 kg/m^3$

Let g = 9.8m/s2. We can now calculate the force that the raft can support

$F = gV(\rho_w - \rho_r) = 9.8*55.135*400=216129.2N$

Each person has a mass of 68kg, their weight would be

W = 68*g = 68*9.8 = 666.4N

So the maximum number of people that the raft can hold is

F / W = 216129.2 / 666.4 = 324 people