Answer
(a) $F = 4010~N$
(b) $t = 2.53~s$
Work Step by Step
(a) When the applied force is at a maximum such that the cylinder does not slip, then the friction force can provide enough torque to make the cylinder rotate to match the acceleration up the ramp without slipping.
$F_f~R = I \alpha$
$F_f~R = (\frac{1}{2}MR^2)(\frac{a}{R})$
$a = \frac{2F_f}{M}$
We can use this expression in the force equation for the cylinder.
$\sum F = Ma$
$F - Mg~sin(\theta) - F_f = Ma$
$F = Mg~sin(\theta) + F_f + Ma$
$F = Mg~sin(\theta) + F_f + M(\frac{2F_f}{M})$
$F = Mg~sin(\theta) + F_f + 2F_f$
$F = Mg~sin(\theta) + 3Mg~cos(\theta)~\mu_s$
$F = (460~kg)(9.80~m/s^2)~sin(37.0^{\circ}) + (3)(460~kg)(9.80~m/s^2)~cos(37.0^{\circ})(0.120)$
$F = 4010~N$
(b) We can find the maximum possible acceleration of the cylinder when it does not slip.
$a = \frac{2F_f}{M}$
$a = \frac{2Mg~cos(\theta)~\mu_s}{M}$
$a = \frac{(2)(460~kg)(9.80~m/s^2)~cos(37.0^{\circ})(0.120)}{460~kg}$
$a = 1.88~m/s^2$
We can find the time it takes to go up the ramp.
$d = \frac{1}{2}at^2$
$t = \sqrt{\frac{2d}{a}}$
$t = \sqrt{\frac{(2)(6.00~m)}{1.88~m/s^2}}$
$t = 2.53~s$
