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A 61 kg survivor of a cruise line disaster rests atop a block of styrofoam insul

ID: 1374109 • Letter: A

Question

A 61 kg survivor of a cruise line disaster rests atop a block of styrofoam insulation, using it as a raft. The styrofoam has dimensions 2.00m x 2.00m x 0.0880m. The bottom .025m of the raft is submerged. (1) Write Newton's second law for the system in one dimension, using B for buoyancy, w for the weight of the survivor, and wr for the weight of the raft. (Set a = 0. Solve for Fy, the y- component of the net force. Let upward be positive y direction. (2) Calculate the numeric value for buoyancy. (seawater density = 1025 kg/m3.) (3) Using the value from buoyancy and the weight of the survivor, calculate the wight of the styrofoam. (4) What is the density of the styrofoam? (5) What is the maxium buoyant force, corresponding to the raft being submerged up to its top surface? (6) What total mass of survivors can the raft support? Thank you for the help in advance!

Explanation / Answer

Volume of the raft V = 2 *2* 0.08m
= 0.32m^3
Volume of the raft submerged in the water is V_s = 2*2*0.025m
= 0.1m^3

a)The buoyancy force B = w + w_r

b)) Buoyancy force is B = ? V g
= 1025 kg / m^3 * 0.1m^3 * 9.8 m/s^2
= 1004.5 N

c)The weight of the styroform is
w_r = B - w
= 1004.5 N - 61 kg * 9.8 m/s^2
= 406.7 N

d)density of the styroform is
m = w_r / g

= 406.7 N / 9.8 m/s^2
m= 41.5 kg

? = m / V
= 41.5 kg / 0.32 m^3
p= 129.6 m^3

e)Maximum buoyance force
B = ? V g
= 1025 kg / m^3 * 0.36 m^3 * 9.8 m/s^2
B= 3616.2 N

f)Total mass is w = B - w_r
mg = B - w_r
m = 3616.2 N - 406.7N / 9.8 m/s^2
m= 327.5 kg

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