Answer
The required force is 1300 N.
Work Step by Step
We can find the tension in the rope pulling up the crate.
$\sum F = ma$
$T - mg = ma$
$T = m(g+a)$
$T = (50~kg)(9.80~m/s^2+1.40~m/s^2)$
$T = 560~N$
We can find the angular acceleration of the cylinder.
$\alpha = \frac{a}{R}$
$\alpha = \frac{1.40~m/s^2}{0.25~m}$
$\alpha = 5.6~rad/s^2$
We can use the angular acceleration to find the required force $F$ on the crank.
$\sum \tau = I \alpha$
$(F)(0.12~m) - T~R = I\alpha$
$F= \frac{I\alpha+T~R}{0.12~m}$
$F= \frac{(2.9~kg~m^2)(5.6~rad/s^2)+(560~N)(0.25~m)}{0.12~m}$
$F = 1300~N$
The required force is 1300 N.
