Respuesta :
Answer:
n = 9 logs
Explanation:
given,
dimension of whole logs
ρ = 715 kg/m³
radius = 0.0815 m
length = 2.76 m
Weight of 4 people
W = 4 m g
W = 4 x 84 x 9.8
W = 3292.8 N..................(1)
Weight of logs
L = n x ρ V g
L = n x 715 x π x 0.0815²x 9.8 x 2.76
L = 403.56 n............................(2)
now, equating both the equation (1) and (2)
3292.8 = 403.56 n
n = 8.159
n = 9 logs
so, 9 logs will be used to carry 4 People of 84 Kg.
The smallest number of whole logs that can be used to build a raft that will carry four people with each mass of 84.0 kg is 23
The volume of the log is = πr²l
=π × (0.0815 m) ² × 2.76m
= 0.058 m³
The mass of the log;
[tex]\mathbf{m_{log} = \rho_{log} \times volume _{log}}[/tex]
[tex]\mathbf{m_{log} =715 \ kg/m^3\times 0.0508 \ m^3}[/tex]
[tex]\mathbf{m_{log} =36.322 \ kg}[/tex]
However, the water displaced by the log is calculated as:
[tex]\mathbf{m_{water} = \rho_{water} \times volume _{log}}[/tex]
[tex]\mathbf{m_{water} =1000 \ kg. m^{-3} \times 0.0508 \ m^{3}}[/tex]
[tex]\mathbf{m_{water} =50.8 \ kg}[/tex]
Now, the extra lifting thrust is estimated by determining the difference in the mass of the log and the water displaced by the log;
i.e.
ΔF = (50.8 - 36.322)g
ΔF = 14.478 × 9.8 m/s²
ΔF = 141.88 m/s²
Hence, the force required to carry 4 people is:
F = 84.0 kg × 4 × 9.8 m/s²
F = 3292.8 m/s²
∴
Number of logs (N) required = [tex]\mathbf{\dfrac{F}{\Delta F} }[/tex]
= [tex]\mathbf{\dfrac{3292.8}{141.88}}[/tex]
= 23.2
≅ 23
Therefore, the smallest number of whole logs that can be used to build a raft that will carry four people with each mass of 84.0 kg is 23.
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