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 F1 to mg. In turn, the force of pressure exerted by the rope on the log also varies, being proportional at each point to the corresponding 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 Ffr (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: Ffr = 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/T1 = 1/(1 - k) since T1 = mg - kmg.
When we want to lift the load, the ends of the rope as if change places. The friction is now directed against the force T2 and plays a harmful role. The ratio of the maximum tension (which is now equal to T2) to the minimum tension (mg) is obviously the same as in the former case: T2/(mg) = 1/(1 - k)= mg/T1. Hence we obtain
