Question

A fisherman in a small fishing boat at rest in a lake hooks a giant log floating in the lake $30$ meters away. The fisherman reels the log in. During this process, the boat moves $12$ meters in the direction of the log. If the mass of the boat and fisherman is $400\,kg$, what is the mass of the log? Assume frictionless.

Answer 1

This question is from the Center of Mass topic and is pretty easy to solve if you recall the mathematical meaning of Center of Mass, which is, the ratio of sum of product of respective distance from Center of Mass and masses with the sum of masses, mathematically represented as:
${x_{CoM}}{\kern 1pt} = {\kern 1pt} \dfrac{{\sum {{x_i}{M_i}} }}{{\sum {{M_i}} }}$

Complete step-by-step solution:
We understand the basic concept that when the log gets into the boat, both, the boat and the log would be exactly where their Center of Mass is. It is so because the Center of Mass can never be moved due to internal forces, and the fisherman pulling the log is only an internal force and there is no other external force acting on the system because the lake is completely at rest and no water is flowing or so.
Hence, we can mathematically, we can write:
${x_{CoM}}\, = \,\dfrac{{{x_{boat}}{M_{boat}}\, + \,{x_{\log }}{M_{\log }}}}{{{M_{boat}}\, + \,{M_{\log }}}}$
Now, we know that the boat and fisherman travelled $12$ meters while pulling the log, this means that the log travelled $18$ meters during this. Let us consider the direction of the boat from the end-point as negative direction and the direction of log from the end-point as positive direction. All this while the position of Center of Mass being considered as the origin point, or it’s position as $0$ .
Now that we have confirmed our sense of direction, the value of position of boat and log before the fisherman started to pull the log becomes:
${x_{CoM}}\, = \,0 \\\ {x_{boat}}{\kern 1pt} = {\kern 1pt} - 12 \\\ {x_{\log }}\, = \,18 \\\$
We already know the mass of boat and fisherman; all we now need to do is to put in the values and find the value of mass of the log.
After putting in the values, we get:
${x_{CoM}}\, = \,\dfrac{{\left( { - 12} \right)\left( {400} \right)\, + \,\left( {18} \right)M}}{{400\, + \,M}}\, = \,0$
After transposing, we get:
$M\, = \,\dfrac{{12\, \times \,400}}{{18}}$
Solving this, we get:
$M\, = \,267\,kg$
Hence, the answer is $M\, = \,267\,kg$ .

Note:- Many students don’t apply the approach that internal forces cannot move Center of Mass, only external forces can. Even after they apply this approach, they forget to take one direction as positive and the opposite as negative. Just this little mistake can waste your time and get you to a wrong answer.