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

MECHANICS 195
109. In equilibrium the sum of the forces acting on the picture is zero
(Fig. 312). Therefore, G=F/r+T cos a, and N=T sin (1,. The force of friction
should satisfy the inequality
FIr
Flr~kN or k~N
The equality of the moments relative to point B gives us the equation
~ I sin a=T (l cos a+ Vdl-I I sinl a) sin a
and
Hence,
FIr
N
l cos a+2 V d 2 -1 2 sin 2 a
l sin a
110. Let us first find the direction of the force f with which rod BC acts
on rod CD. Assume that this force has a vertical component directed upward.
Then. according to Newton's third law, rod CD acts on rod BC with a force
whose vertical component is directed downward. This contradicts the symmetry
of the problem, however. Therefore, the vertical component of the force f
should be equal to zero. The force acting on rod CD from rod DE will have
both a horizontal and a vertical components, as shown in Fig. 313a.
Since all the forces acting on CD are equal to zero, F=mg and f=f'.
The equa lity to zero of the moment of the forces with respect to D gi yes us:
or
f sin ~
CD=mg co; ~ CD
mg
tan~=2T
Figure 313b shows the forces acting on rod DE. Since the moment of the
Of,
(a)
/
.0
(b)
Fig. 313
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