40
1 Kinematics and Statics
Fig. 1.28 Centre of mass of
NH 3 molecule
Z CM =
3m H
3m H + m N
× r C N =
3m H
3m H + 14m H
× 0.38 = 0.067 Å
1.50 Take the origin at A at the left end of the boat, Fig. 1.29. Let the boy of mass
m be initially at B, the other end of the boat. The boat of mass M and length
L has its centre of mass at C. Let the centre of mass of the boat + boy system
be located at G, at a distance x from the origin. Obviously AC = 1.5 m:
Fig. 1.29
AG = x =
MAC + mAB
M + m
=
100 × 1.5 + 50 × 3
100 + 50
= 2 m
Thus CG = AG − AC
= 2.0 − 1.5 = 0.5 m
When the boy reaches A, from symmetry the CM of boat + boy system would
have moved to H by a distance of 0.5 m on the left side of C. Now, in the
absence of external forces, the centre of mass should not move, and so to
restore the original position of the CM the boat moves towards right so that the
point H is brought back to the original mark G. Since HG = 0.5 + 0.5 = 1.0,
the boat in the mean time moves through 1.0 m toward right.