Find Center of Mass for Uniform Log: 2.33m, 94.9kg, 72.7kg, 26.1kg

[Barrynew]

A uniform log of length 2.33 m has a mass of 94.9 kg and is floating in water. Standing on this log is a 72.7-kg man, located 18.9 cm from one end. On the other end is his daughter (m = 26.1 kg), standing 1.15 m from the end. Find center of mass

2. m1r1+m2r2/(m1+m2)=R



3. I used (1.15)*26.1+72.7*0.189/(26.1+72.7+94.9)
the answer is not right...Need some help. Thank you!

 

[gneill]

Hi Barrynew, Welcome to Physics Forums.

First you should establish your coordinate system. Where is the origin from which you want distances to be measured (so when you say something like "the position of the center of mass is X", everyone will be able to agree where X is to be measured from).

It would appear from the equation that you've written that you want the center of the 94.9 kg log to be where your origin is located. Why? Because you've not included an offset for the log's center of gravity in the numerator of your calculations but have included the mass of the log in the denominator...

Next, the problem statement is not clear about where things are measured from! The man is located at a distance of 18.9 cm from one end of the log. Fine. But then it says, "On the other end is his daughter ... standing 1.15 m from the end". Are we to understand that the daughter is standing 1.15 m from the "other" end, or 1.15 m from the same end that the man is standing? My guess would be they intended 1.15 m from the "other" end.

Once you've established your coordinate system, write all distances in terms of that system! You'll want to know how far the man is from the origin, and how far his daughter is from the origin. Those are the distances that go into your center-of-mass formula.