Question

A log of wood of length l and
mass M is floating on the surface of a river perpendicular to the banks. One end of the log touches the banks. A man of mass m standing at the other end walks towards the bank. Calculate the displacement of the log when he reaches the nearer end of the log. ​

Answer 1

Simple question

Solution:- Here from the given question we can say,that only internal forces(if we are taking log and man as a system) are actinng on it

so,in this case

Mx1=mx2

where x1 and x2 is the displacement of the log and man with respect to the ground respectively..

here =mass of log+=mass of man

since we know that,if the man will move on block ,then he will apply a force on backward,which make the block to pull backward...

so,

if the log will move x backward,then displacement of man with respect to ground=L-x(see the attachment for understanding)

so, using this we can say

M×x=m(l-x)

M×x=ml-mx

Mx+mx=ml

x(m+M)=ml

x=ml/(m+M)

Second method(Using the exact formula of centre of mass):-

As we know that their is no any external forces

so

intial position of centre of mass=final position of centre of mass

initial Xcm=final Xcm

intial Xcm={(m×l +M×l/2)}/(m+M)={ml+Ml/2}/{(m+M)}

if log will move to x,,then man will move l-x Distance

final Xcm={(M×(l/2+x)+m×x}/(m+M)=

making final and initial equal we get

{ml+Ml/2}/{(m+M)}=(Ml/2 +mx+Mx)/(m+M)

(ml+Ml/2)=Ml/2+mx+Mx

ml=mx+Mx

ml=x(M+m)

x=ml/(m+M)

Hence from both the method we are getting that displacement of the block when the man reach to the other end=ml/(m+M)

Answer 2

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(20,-20){2}{\multiput(0,0)(0,3){2}{\line(1,0){50}}\put(24,0.6){\scriptsize\textsf{M}}}\put(0,3){\line(-1,0){10}}\put(0,3){\line(0,-1){23}}\multiput(0,0)(20,-20){2}{\multiput(0,0)(50,0){2}{\line(0,1){3}}}\multiput(49,0)(-27.8,-20){2}{\put(0,10){\circle{2}}\put(-0.05,9){\line(0,-1){6}}\put(-0.7,10.2){\circle*{0.1}}\put(-1.5,13){\sf{m}}}\put(50,-3){\vector(-1,0){30}}\put(50,1.5){\vector(1,0){20}}\put(33,-6){ \sf{-x_m} }\put(60,-1.5){ \sf{x_l} }\put(20,-23){\vector(1,0){50}}\put(45,-28){ \sf{l} }\multiput(0,3)(-2,0){5}{\line(-1,-1){3.56}}\multiput(0,1)(0,-2){11}{\line(-1,-1){3.56}}\end{picture}$

Here the center of mass of the system of log of wood and man remains unchanged since there's no horizontal force acting on the system.

Let the displacement of the log be $\sf{x_l}$ and that of man be $\sf{-x_m}$ since he moves opposite to the log.

From the fig., we get that the sum of magnitudes of displacements of man and log is equal to length of log, i.e.,

$\longrightarrow\sf{x_m+x_l=l}$

$\longrightarrow\sf{x_m=l-x_l\quad\quad\dots(1)}$

Since the displacement of center of mass is zero,

$\longrightarrow\sf{\dfrac{M(x_l)+m(-x_m)}{M+m}=0}$

$\longrightarrow\sf{Mx_l-mx_m=0}$

$\longrightarrow\sf{mx_m=Mx_l}$

$\longrightarrow\sf{x_m=\dfrac{Mx_l}{m}}$

From (1),

$\longrightarrow\sf{l-x_l=\dfrac{Mx_l}{m}}$

$\longrightarrow\sf{l=x_l\left(\dfrac{M}{m}+1\right)}$

$\longrightarrow\sf{l=x_l\left(\dfrac{M+m}{m}\right)}$

$\longrightarrow\sf{\underline{\underline{x_l=\dfrac{ml}{m+M}}}}$