Answer= The weight of raft=m*g=rho*V*g=550*100*0.1*9.8=53900N Fb(weight of the d

Question

Answer=

The weight of raft=m*g=rho*V*g=550*100*0.1*9.8=53900N
Fb(weight of the displaced water)=rho*h*A*g=1000*h*100*9.8=980000*h
for equilibrium the two forces must be equal
980000*h=53900
h = 53900/980000 = 0.055m
5.5cm of the raft height will be below the water

weight of raft with people is 53900+n*70*9.8=53900+686*n
Fb(raft submerged)=rho*V*g=1000*100*0.1*9.8=98000N
53900+686*n=98000
n=(98000-53900)/686=44100/686=64.3 people

A raft has a surface area of 100m 2 and a height of 10.0cm. If it is constructed of yellow pine and floating in a lake, calculate how much of the raft will be above the waterline Also calculate how many people with an average mass of 70.0kg can be placed on the raft to completely submerge it without sinking it.

Explanation / Answer

buyoant force, B = weight of the displaced water

B = rho_water*Volume of water*g

here, volume of water = volume of raft

B = rho_water*volume of raft*g

B = 1000*100*0.1*9.8

= 98000 N

in equilibrium B = weight of raft + weight of the persons

let n is the no of persons

B = volume of raft*density of raft*g + n*m*g

B = 100*0.1*550*9.8 + n*70*9.8


n = (B - 10*550*9.8)/(70*9.8)

= (98000 - 10*550*9.8)/(70*9.8)

= 64.28

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