Lumberjack Logging

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September 4th, 2009 admin Leave a comment Go to comments

A 76 kg lumberjack stands at one end of a 359 kg floating log?

If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack's speed relative to the shore? Ignore friction between the log and the water.

hmmmmm, fun problem

mometum transfer

m1 = mass of the lumberjack
m2 = mass of the log

if the man walks along the log, then the log starts to move opposite of the direction of the man. since the man is still on the log, then they will in turn move together.

m1*v1 - (m1+m2)*v2 = 0
v2 = m1*v1/(m1+m2)

simply you can say that the velocity of the man relative to the shore = v1 - v2

the long way is ...
fun with relativity....
v1 = v(man/log)

v(man/log) = v(man) - v(log)
v(man/shore) = v(man) - v(shore)
v(log/shore) = v(log) - v(shore)
v(man/log) = v(man/shore) - v(shore) - v(log/shore) + v(shore)
v(man/log) = v(man/shore) - v(log/shore)
v(man/shore) = v(man/log) + v(log/shore)
** v(log/shore) = v2, which you already solved for but it moves opposite of v1
** v(man/log) = v1, given to you in the problem
v(man/shore) = v1 - v2