This is a math problem.

According to the characteristics of the triangle with big sides facing big angles

Let AB=BC

The pulley point is o.

AB=BC, AO & gtBO & gt commanding officer.

The angle AOB corresponds to the edge ab; The angle COB corresponds to the side CB; The angle ABO corresponds to the edge AO and the angle BCO corresponds to the edge BO.

And angle ABO= angle BCO+ angle cob = angle BAO+ angle AOB；; AB=BC

Available AO-BO & gt；; BO-CO

Let the left and right angles of the original problem be point d and the pulley point be o.

DO=a DC=b CB=c AB=C, then AC = 2 dB = C+B DA = B+2C.

BO-CO= internal radical number (a2+(c+b)2)- internal radical number (a^2+b ^2).

AO-BO= within the root number (a2+(2c+b)2)- within the root number (A 2+(c+b) 2).

(AO-BO)-(BO-CO)= within the radical number (a2+(2c+b)2)- within the radical number (a2+(c+b)2)-(a2+(c+b)2)+ within the radical number (a. 0

So the displacement of the force of the first part is larger than that of the second part.