Under the action of F, the center of mass of the rod AB tends to translate in the direction of F and rotate around the center of mass of the rod.

The center of mass tends to translate along the direction F, and the balance is: F=2Fs 1, (1).

Trend of rotation around the rod center of mass, balance: Fsin30(L/2)=Fs2. L, (2)

It can be obtained from (1)(2): Fs 1=F/2,? Fs2=F/4，Fs 1=2Fs2，

The total friction (vector) Fs=Fs 1+Fs2, and the size of Fs at the A end is greater than that at the B end. Calculate the size of Fs at terminal a:

Fsx=Fs 1.cos30 degrees =Fs 1√3/2=Fs2. √3,? Fsy=-Fs2-Fs 1.sin30 degrees =-2Fs2,

Fs=√(Fsx)^2+(Fsy)^2=√((Fs2.√3)^2+(2Fs2)^2)=Fs2√7

Let Fs=2fs.p and replace it with Fs2=F/4:

2fs.p=？ f√7/4->； Fmax=8fs.p/√7