A raft 4.6 m long, 1.8 m wide, and 32 cm thick, made from solid balsa wood, is floating in a freshwater lake. How many 63 kg people can it support? Round down your final answer to the nearest integer.

Answer 1

Using Archimedes' Principle, the buoyancy force F_B equals density of the fluid (water, ρ_w = 1000 kg/m³) times volume of fluid displaced V_w times acceleration of gravity g. In equilibrium, F_B = weight m_t g where total mass m_t = mass of raft m_r plus mass of N people, 63 N. From definition of density, m_r = ρ_r V_r.

Did the instructions or your reference material specify the density of balsa wood? If so, substitute that in the following. For now I'm using a rough average of 140 kg/m³ gleaned from web searches that gave values from 110 to over 200.

So F_B = ρ_w V_w g = (ρ_r V_r + 63 N) g. When barely floating, V_w = V_r = (4.6)(1.8)(0.32) = 2.65 m³. Divide both sides by the common factor g:

(1000 kg/m³) (2.65 m³) = (140 kg/m³) (2.65 m³) + (63 kg) N. Then 63 N = 2650 - 371, and N (rounded down) = 36 people with this value for the density of balsa wood.

Answer 2

Hello Nilufar

^{A raft 4.6 m long, 1.8 m wide, and 32 cm thick, made from solid balsa wood, is floating in a freshwater lake. How many 63 kg people can it support? Round down your final answer to the nearest integer.}

Maximum up thrust = Volume of raft *density of water* accel. due to gravity 4.6*1.8*0.32*1000*10 ( Taking g= 10m/s²

Total down ward force = n*M*g where n = Max number of persons , M = 63 kg, g= 10m/s²

Solve for n , The number of persons

n*M*g=n*63*10= 4.6*1.8*0.32*1000*10