Solutions (^147)
where al is the projection of the acceleration of the
cylinder on the xi axis.
The wedge is acted upon by the force of gravity
Ina, the normal reaction N 2 of the right inclined
plane, and the normal reaction of the cylinder,
which, according to Newton's third law, is equal
to —N 3. We shall write the equation of motion of
the wedge in terms of projections on the zraxis
directed along the right inclined plane:
m 2 a 2 = —m 2 g sin a + N3 cos a. (2)
During its motion, the wedge is in contact
with the cylinder. Therefore, if the displacement
of the wedge along the xraxis is Ax, the centre of
the cylinder (together with the vertical face of the
wedge) will be displaced along the horizontal by
Ax cos a. The centre of the cylinder will be thereby
displaced along the left inclined plane (x 1 -axis) by
Ax. This means that in the process of motion of
the wedge and the cylinder, the relation
a (^2) ( 3 )
is satisfied.
Solving Eqs. (1)-(3) simultaneously, we deter-
mine the force of normal pressure N =N 3 exerted by
the wedge on the cylinder:
2minz 2
N3= tan a.
m1+ m2
1.47. As long as the load touches the body, the ve-
locity of the latter is equal to the horizontal com-
ponent of the velocity of the load, and the accelera-
tion of the body is equal to the horizontal compo-
nent of the acceleration of the load.
Let a be the total acceleration of the load.
Then we can write a ----- at ac, where ac is the
centripetal acceleration of the load moving in the
circle of radius 1, i.e. ac = v 2 /1, where v is the
velocity of the load (Fig. 154). The horizontal
component of the acceleration is
v2
ah=...- at sin a— — 1 cos a.
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