The magnitude of the friction on the cardboard is f = f1+F2 = μ (2m1+m2) g.
(2) If the acceleration of the weight is a 1 and the acceleration of the cardboard is a2, then:
f 1=m 1a 1
F-f 1-f2=m2a2
The relative motion requirement A2 > A 1.
The solution is f > 2μ (m1+m2) g.
(3) The distance the weight moves before the cardboard is pulled out.
x 1 = 1/2a 1t 1？
The moving distance of cardboard d+x1=1/2a2t1?
The distance that the weight moves after the cardboard is pulled out.
x2= 1/2a3t2？
L=x 1+x2
A 1=a3，a 1t 1=a3t2。
F=22.4N replaces data at the same time.
Answer: (1) The magnitude of the friction on the cardboard is μ (2m1+m2) g;
(2) the required pulling force f > 2μ (m1+m2) g;
(3) The required tension of paperboard is at least 22.4 Newton.