Solutions 133using the formula x (t) = x 1 sin coot. For this
purpose, we determine t 2 from the formulaxo = x1 sin 0 02.Since xl = 175x 0 , co 0 t 2 = n/4. Consequently, t 2 =
:0(404) = t 1 /2. Since t 3 = t 1 t 2 , where t 3 is the
time of motion of the sledge after the jerk, we ob-
tain the required ratio of the braking times
t 3 1
ti — 2 •1.31. The force of gravity mg = 60 N acting on the
load is considerably stronger than the force with
which the rope should be pulled to keep the load.
This is due to considerable friction of the rope
against the log.
At first, the friction prevents the load from
slipping under the action of the force of gravity. The
complete analysis of the distribution of friction
acting on the rope is rather complex since the
tension of the rope at points of its contact with the
log varies from F 1 to mg. In turn, the force of pres-
sure exerted by the rope on the log also varies,
being proportional at each point to the corre-
sponding local tension of the rope. Accordingly, the
friction acting on the rope is determined just by
the force of pressure mentioned above. In order to
solve the problem, it should be noted, however,
that the total friction Fir (whose components are
proportional to the reaction of the log at each
point) will be proportional (with the corresponding
proportionality factors) to the tensions of the rope
at the ends. In particular, for a certain coefficient
k, it is equal to the maximum tension: Fi r = kmg.
This means that the ratio of the maximum tension
to the minimum tension is constant for a given
arrangement of the rope and the log: mg/ T 1 =
1/(1 — k) since T 1 = mg — kmg.
When we want to lift the load, the ends of the
rope as if change places. The friction is now direct-
ed against the force T2 and plays a harmful role.
The ratio of the maximum tension (which is now