A 61.0 kg survivor of a cruise line disaster rests atop a block of styrofoam ins

Question

A 61.0 kg survivor of a cruise line disaster rests atop a block of styrofoam insulation using it as a raft. The styrofoam has dimensions of 2.00m X 2.00m X 0.0850m . The bottom 0.021m of raft is submerged.

A) write Newtons second law for the system in one dimension, using B for buoyancy, w for weight, and wr for weight of the raft. (set a=0, solve for fy, let up be positive y direction.

B) Calculate numeric value for Buoyancy.

C) Calculate wr of styrofoam using values of weight and buyancy.

D) What is density of the Styrofoam?

E) What is maximum Buoyant force corresponding to the raft being submerged up to its top surface?

D) What total mass of survivors can raft support?

Explanation / Answer

a)w = weight of survivor = mg = 597.8 N

wr = 2.00m X 2.00m X 0.0850m*density of styrofoam =

Newtons second law for the system =

Fy= w+wr -B=ma =0

so, B=w+wr

so,B=w+wr

b)volume submerged = 0.021*2*2= 0.084 m^3

water displaced = 1000*0.084 = 84 kg

so, B= 84*9.8 = 823.2 N

c)so, wr = B-w = 823.2-597.8 = 225.4 N

d)mass of styrofoam = 225.4/9.8 = 23 kg

so, density = mass/volume = 23/(2*2*0.085) = 67.647 Kg/m^3

e)If the total raft is submerged,

then

volume submerged = 0.085*2*2= 0.34 m^3

water displaced = 1000*0.34 = 340 kg

so, Bmax = 340*9.8 = 3332N

f)w max = Bmax- wr = 3332- 225.4 = 3106.6 N

then the no.of persons = 3106.6/(60*9.8) = 5.283 persons, so around 5 persons

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