Question

What is the smallest number of whole logs (ρ = 755 kg/m3, radius = 0.0725 m, length = 2.70 m) that can be used to build a raft that will carry four people, each of whom has a mass of 86.0 kg?

Answer 1

is the answer 10?

Answer 2

No

Answer 3

Alright so you want a certain number of logs to provide enough buoyant force to oppose the weight of 4 people.

Lets start off by finding the force of weight exerted on the raft by all 4 people
The weight of one person can be found using:
W = mg = 9.81m/s^2 * 86.0 kg = 844 N
Since there are four people, we have to multiply this by 4.

The total weight is now 3374 N

Now to find the buoyant force of a single log, then we can find out how many logs would be required

Buoyant force is given by the volume of water that is displaced.

One log has a volume of:
V = pi*r^2*L = 3.14*0.0725^2 * 2.70 = .0446 m^3

With the given density of the log, that gives us a mass of
m=pV=.0446*755 = 33.6 kg

The weight of one log would therefore be
W=mg = 33.6*9.81= 330N

The buoyant force of one log would be the logs volume, times the density of water times gravity
B=pVg

The density of water, which was not given, is 997 kg/m^3

this gives each log a buoyant force of:

B=997*.0446*9.81= 436 N

Note that this is a greater force than the weight of the logs themselves. thats how you know they float!

Now to balance our forces and finish up the problem!

For the raft to float with 'n' number of logs, B*n > W_p + n*W_l

with some simple algebra, we can solve for n:
n(B-W_l)=W_p
n=W_p/(B-W_l)
n=3374/(436-330)
n=31.8 therefore you should use 32 logs (assuming my numbers are right!)

Answer 4

Lemme know if you still need help with anything in there! :)

Answer 5

Thank you so much!! Your explanation helped me with 3 other problems!!!

Answer 6

Good to hear!