Answer
The children will need to use at least 5 logs.
Work Step by Step
We can find the volume of one log:
$V = (\pi)(0.15~m)^2(1.80~m) = 0.127~m^3$
The buoyant force is equal to the weight of water that is displaced.
We can find the buoyant force when one log is completely submerged:
$F_b = m_{water}~g$
$F_b = \rho~V~g$
$F_b = (1000~kg/m^3)(0.127~m^3)(9.8~m/s^2)$
$F_b = 1245~N$
We can find the weight of one log:
$weight = m_{log}~g$
$weight = \rho~V~g$
$weight = (800~kg/m^3)(0.127~m^3)(9.8~m/s^2)$
$weight = 995.7~N$
We can find the additional weight that can float on one log:
$1245~N-995.7~N = 249.3~N$
We can find the number of required logs:
$\frac{(3)(356~N)}{249.3~N} = 4.3$
Therefore, the children will need to use at least 5 logs.
