(^134) Aptitude Test Problems in Physics
equal to T 2 ) to the minimum tension (mg) is ob-
viously the same as in the former case: T 2 /(mg) =
1/(1 — k)= mg/Ti. Hence we obtain
() 2
T2—
mg
=90 N.
T 1
1.32. Let us see what happens when the driver turns
the front wheels of a stationary car (we shall con-
sider only one tyre). At the initial moment, the
wheel is undeformed (from the point of view of
torsion), and the area of the tyre region in contact
with the ground is S. By turning the steering
wheel, the driver deforms the stationary tyre until
the moment of force dis applied to the wheel and
tending to turn it becomes larger than the max-
imum possible moment of static friction acting on
the tyre of contact area S. In this case, the forces
of friction are perpendicular to the contact plane
between the tyre and the ground.
Let now the motorcar move. Static frictional
forces are applied to the same region of the tyre of
area S. They almost attain the maximum values
and lie in the plane of the tyre. A small moment of
force e7/ applied to the wheel is sufficient to turn
the wheel since it is now counteracted by the total
moment of "oblique" forces of static friction which
is considerably smaller than for the stationary
car. In fact, in the case of the moving car, the
component of the static friction responsible for the
torque preventing the wheel from being turned is
similar to liquid friction since stagnation is not
observed for turning wheels of a' moving car. Thus,
a small torque can easily turn a moving wheel, and
the higher the velocity (the closer the static friction
to the limiting value), the more easily can the
wheel be turned.
1.33. Let us choose the reference frame as shown
in Fig. 144. Suppose that vector OA is the vector
of the initial velocity v. Then vector AB is the
change in velocity during the time interval At.
Since the force acting on the body is constant,
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