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

236
o
ANSWERS AND SOLUTIONS
a
A
H
Fig. 859
0'
Fig. 360
0'
Therefore.
F =YT2+N2=m y g2+ro4 l 2 sin 2 cp
197. The forces acting on the bead are shown in Fig. 360: f is the
of friction, mg the weight and N the force of the normal reaction.
force
Newton's equations for the projection of the forces on a horizontal
vertical directions will have the form
and a
f sin q> 1= N cos q>=mro2 l sin q>
f cos q> ± N sin q>-mg=O
The upper sign refers to the case shown in Fig. 360 and the lower one to
the case when the force N acts in the opposite direction. We find from these
equations that
f=mro 2 l sin 2 cp+mg cos q>
N= ± (mg sin q> - mro 2 l sin cp cos q»
In equilibrium f ~ kN or
k sin cp-cos cp g
1
(k
cos cP+.) sin q> • 2 when k ~ cot cp
(J)
and
k sin g
I ~. (k .) • 2 when k ~
sin q> cos cp-sln
tan q>
q> co
198. Figure 361 shows the forces acting on the weights. Here T 1 and T 2
are the tensions of the string. Let us write Newton's equations for the proj
ectlons onto a horizontal and a vertical directions.