Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 334: 10.64

$a = \frac{F}{2M}$ $F_f = \frac{F}{2}$

We can set up a torque equation for the cylinder. $\tau = I\alpha$ $F_f~R = (MR^2)(\frac{a}{R})$ $F_f = Ma$ We can use this expression in the force equation for the cylinder. $\sum F = Ma$ $F-F_f = Ma$ $F-Ma= Ma$ $F = 2Ma$ $a = \frac{F}{2M}$ We can find the friction force. $F_f = Ma$ $F_f = (M)(\frac{F}{2M})$ $F_f = \frac{F}{2}$