108  Prof.  J.  Perry  on 
We  assume  that  the  rope  with  its  attached  cage  is  moving 
downwards  with  a  velocity  V  when  the  upper  end  of  it  is 
suddenly  stopped  and  held  fixed. 
Let   ^w  be  the  displacement  downwards  of  a  point  in  the 
rope  distant  z  from  its  lower  end, 
a    the  area  of  cross  section  of  the  rope, 
M  the  mass  of  cage, 
p    the  density  of  the  rope, 
I    the  length  of  the  rope, 
E  Young's  Modulus. 
Then  equating  the  mass-acceleration  of  a  small  element  of 
the  rope  to  the  difference  of  the  pulls  on  its  two  ends  we  get 
or  putting 
of  which  the  solution  is 
a=f(at-z)+F(at  +  z) (1) 
Again,  equating  the  mass-acceleration  of  the  cage  to  the 
pull  in  the  rope  at  its  lower  end,  we  obtain 
m1S=E«|^,     when     ,=0, 
or  using  (1) 
f"(at)  +  F"(at)='^{F'(at)-f(at)}, 
or  putting 
E«        J^         ,  _  Mass  of  Cage 
Ma*-mZ'  re     m~  Mass  of  Pope' 
fXat)  +  r'(at)  =  ±l{F>(at)-f>(at)}.     .     .     (2) 
The  conditions  of  the  problem  show  that  for  all  values 
of  t  <  0  we  may  put 
B2o> 
By© 
=  p-bz*> 
E_ 
9  ' 
=  a2, 
B2*> 
B*2  " 
and  F{at  +  z)=-iv(t+^\ 
0<z<l 
