Bukhovtsev-et-al-Problems-in-Elementary-Physics

184 ANSWERS AND SOLUTIONS
Let us assume that the force F is so small that the weight does not
slide over the table. Hence a 1=a2 and F1r=F Gl~G2 ·
By gradually increasing the force F we shall thereby increase the force of
friction F1" If the table and the weight are immovable relative to each other,
however, the force of friction between them cannot exceed the value Ffr.max=kG 2 •
For this reason the weight will begin to slide over the table when
01+01 G2
F F 0
> 'f.max-a--=ka
( I+G2)=10 kgf
. 1 1
In our case F=8 kgf, consequently, the weight will
al=~=Gl~G2 g=:5g~ 314 em/5 2
not slide, and
(2) In this case the equations of motion fo the table with the pulleys and
the weight will have the form:
-F+F,,=G 1 a
g l
F-F,,=G2 a 2
g
The accelerations of the table and the weight are directed oppositely, and the
weight will be sure to slide.
Hence, F1,=k0 2 • .
The acceleration of the table is
-F+kG 2 2
01= g=--g=- 131 em/s
0 2
1 15
and it will move to the left.
72. According to Newton's second law, the change in the momentum of
the system "cannon-ball It during the duration of the shot -r should be equal
to the impulse of the forces acting on the system.
Along a horizontal line
mvocos a-MvI=FIf'"
where FIf" is the impulse of the forces of friction.
Along a vertical line
mvo sin a=N-r-(Mg+mg) '"
where NT. is the impulse of forces of normal pressure (reactions of a horizontal
area), (Mg+mg) -r is the impulse of the forces of gravity, Remembering that
Ftr =kN. we obtain
m m. M+m
[)t=M Vo cos a-k ¥vosln a-k~g-r
or since g'f ~ Vo
VI. ZVo (cos a,-k sin a,)
This solution is suitable for k cot a the cannon will remain