The given situation is shown in the following figure
Effective mass of the system of n masses placed on the table, M=m+
m
2
+
m
6
+...+
m
n!
=m(1+
1
2
+
1
6
+...+
1
n!
) =m(
1
1!
+
1
2!
+
1
3!
+...+
1
n!
)
=m(1+
1
1!
+
1
2!
+
1
3!
+...+
1
n!
−1)
M=m(e−1)
[∵e=1+
1
1!
+
1
2!
+
1
3!
+...+
1
n!
]
The given system of masses can be redrawn as
If a be the acceleration of system of masses, then according to Newton's second law of motion, T=Ma⇒T=m(e−1)a....(i) and mg−T=ma⇒T=mg−ma T=m(g−a)...(ii) From Eq. (i) and Eq. (ii), we get m(e−1)a=m(g−a)⇒a=g∕e
