Huckleberry Finn says hi

In its original state, the raft is in equilibrium with the weight of the raft balanced by the upthrust. $mg = \rho Vg m = \rho V$ . Given that the Mississippi river is primarily fresh water, \rho = 1000~kg/m^3/). We can thus work out the mass of the raft. \(m = 1000 \times 5 \times 5 \times (0.1-0.07) = 750~kg . When Huckleberry and Jim stand on the raft, a new equilibrium is established, and the upthrust exerted by the water is higher to support the additional downward force. Thus, letting x be the unknown distance, $(m + 130) = \rho (V_1) 750 + 130 = 1000 \times 5 \times 5 \times (0.1 - x) \rightarrow x = 0.0648~m$ .

According to Archimedes' principle, the buoyant force on an object lowered into a fluid is $F=\rho g V$ , where $V$ is the volume of the displaced fluid. Note that since the raft is in equilibrium, the magnityde of this force must be equal to the rafts weight, $mg$ , so $F=\rho g V=mg$ , and $\rho V=m$ We know the volume of the displaced fluid. Its area is $5^2=25 [m^2]$ , and its depth is $0.1-0.07=0.03 [m]$ (note that the square brackets denote units). The density of water is very close to $\rho=10^3 [kg/m^3]$ . So $m=\rho V=25\times 0.03\times 10^3=750 [kg]$

Now, we add $130 [kg]$ , so the new mass is $750+130=880 [kg]$ . Let $x$ be the new depth (the area is still $25 [m^2]$ such that. $V=25x$ . So, $880=25x\times 10^3$ .

$x=\frac{880}{25\times 10^3}=0.0352 [m]$ . This is the depth, so how far above the water the raft is, $h$ , is just the total height minus $x$ .

$h=0.1-0.0352=0.0648 [m]$

Archimedes' Principle state that mg= \rho Vg where \rho is the density of water, V is the volume of the object immersed in water, m is the mass of the object and g is the gravitational acceleration.

The mass of the raft: m= \rho V, noted that the g has been cancel out.

m=1000kg m^{-3} \times (5m \times 5m \times 0.03m)

m=750kg

130kg(mass of two person) is then added and mass of raft and two person is 880kg.

Use the formula again, but with V=Area \times height of raft in water. 880kg=1000kg m^{-3} \times 25m^2 \times h

h=0.0352m

Therefore height of raft above the water surface is 0.1m-0.0352m=0.0648m

The mass of the raft is equal to the mass of the water displaced by the raft by itself, which is 5m * 5m * 0.03m * 1000kg/m^3 = 750kg. The total mass of the raft and the two boys is 880kg. Hence the volume of the water displaced by the raft with two boys is 880kg / 1000kg/m^3 = 0.88m^3.

Hence 0.88m^3 / 5m / 5m = 0.0352m of the raft is underwater. So the top of the raft is 0.1m - 0.0352m = 0.0648m above water.

initially 0.07m of raft is above the surface of water for extra weight,

equating weight and buoyant force V d g = 130*g 25 * x * 1000 = 130,
where x is the height of raft to be immersed and density of water = 1000kg/m^3

x= 0.0052

so new height = 0.07- 0.0052 = 0.0648

The raft initially sinks 0.03 m into the water before it reaches equilibrium. Therefore we can find the mass of the raft using Newton's laws, the bouyant force, and the force of gravity:

$F_{Net}=F_b+F_g=\rho V_{d} g - m_{raft}g=0$ ,

where $\rho$ is the density of fresh water ( $1000 kg/m^3$ ) and $V_d$ is the displaced volume of water. We can therefore solve for

$m_{raft}=\rho V_d=1000 \times 0.03 \times 5 \times 5=750~kg$ .

Once Jim and Huck step on we can do another force balance,

$\rho W_d g - (m_{raft}+130)g=0$ ,

and solve for $W_d$ , the displaced volume of the raft, as $W_d=880/1000=0.88~m^3$ . Using $W_d=25 h$ gives that the depth of the raft in the water is 0.0352 m which means the top is $0.1-0.0352=0.0648$ m above the surface of the Mississippi.

Assuming a density of water 1000kg / m3 . The square raft will provide 250 kg per cm of immersion.

That figure come from flotation Area 25 m2 times 0.01m which amount to 250kg/cm Since the combined load is 130, the raft will sink 130/250 cm, therefore the free board will be reduced in such amount and we get 0.0648 m as new free board.