Question

A 63.0-kg survivor of a cruise line disaster rests atop a block of Styrofoam insulation, using it as a raft. The Styrofoam has dimensions 2.00 m ? 2.00 m ? 0.0820 m. The bottom 0.025 m of the raft is submerged.

(a) Draw a force diagram of the system consisting of the survivor and raft. (Submit a file with a maximum size of 1 MB.)

This answer has not been graded yet.

(b) Write Newton's second law for the system in one dimension, using B for buoyancy, w for the weight of the survivor, and w_r for the weight of the raft. (Set a = 0. Solve for F_y, the y-component of the net force. Let upward be the positive y-direction.)
F_y =

= 0

(c) Calculate the numeric value for the buoyancy, B. (Seawater has density 1025 kg/m³. Enter answer to at least the ones digit.)
N

(d) Using the value of B and the weight w of the survivor, calculate the weight w_r of the Styrofoam.
N

(e) What is the density of the Styrofoam?
kg/m³

(f) What is the maximum buoyant force, corresponding to the raft being submerged up to its top surface?
N

(g) What total mass of survivors can the raft support?
kg

Answer 1

a)

b) Force equation is

F_b - W - W_raft ( where F_b is the buoyant force)

for given symbols in question , it would be

B - w - wr = 0

c) B = 1025*9.8*2*2*0.025

B = 1004.5 N

d) weight of raft wr = 1004.5 - 63*9.8 = 387.1 N

e) density = mass / volume = 39.5 / 0.328 = 120.426 kg/m³   

Note - we need mass of raft, we have its weight calculated in (d) just divide that by 9.8 to get mass .

f) Max buoyant force = 1025*9.8*2*2*0.082

max buoyant force = 3294.76 N

g) Using the equation from part (b)

3294.76 - m(9.8) - 387.1 = 0

2907.66 = 9.8m

m = 296.7 kg