[mattmravunac]

Suppose two friends, with masses m2 = 1.3 m1, are on a perfectly smooth, frictionless, frozen lake.
They are both holding the end of a rope of length Lo .

A) Find the position of the center of mass, in terms
of Lo , from the smaller person.

B) If the two pull half of the rope in such that the
final length of the rope is L = Lo/2, find the new
position of the center of mass from the smaller
person.

C) Find the distance each person moves from their original positions.

[ebaines]

Please show us what you have done to try and solve these problems, and then we can help you wheer you get stuck. Keep in mind that the center of mass is that point on the rope where the distance to M1 times the mass M1 equals the distance to M2 times the mass M2.

[alpal24]

I need help with this one to. I got A and B but I am not really sure how to do C? Could you maybe help me set it up?

[mattmravunac]

I know that
X2=Lo and m2=1.3m1 and have the equation
Xcm=(m2X2) / (m1+m2)
I am just know sure how to go about it from this step

[hrab]

Xcm=(1.3m1*Lo)/(2.3m1)

[hrab]

So... its 0.57Lo=Xcm

[mattmravunac]

Thank you!
For part 2 would I just need to take the .057Lo/2? Which would give me .29 (?) or would I need to go about it like a completely new problem?

[hrab]

I think you need to set x1=Lo/4 and x2=3/4(Lo). And plug everything in and solve

[hrab]

So your cm will be half as far towards the heavier person. . 57/2=.285 .285+.25=Xcm if Lo/2

[mattmravunac]

why would I need to set x1 to Lo/4 (?) if half the rope is gone then wouldn't that just be equal to Lo/2 and then X2 would be set to (1.3m1Lo)/2 (?)