Answer
$F_N = 2.45~N$
Work Step by Step
We can find the force $F_T$ that the cable exerts on the cab:
$F_T~d_2 = W$
$F_T = \frac{W}{d_2}$
$F_T = \frac{92,610~J}{10.5~m}$
$F_T = 8820~N$
Let $M$ be the mass of the cab.
We can find the acceleration of the cab:
$Ma = F_T-Mg$
$a = \frac{F_M}{M}-g$
$a = \frac{8820~N}{900~kg}-9.8~m/s^2$
$a = 0$
Thus the acceleration of the cheese must also be zero.
Let $m$ be the mass of the cheese.
We can find $F_N$:
$F_N-mg = ma$
$F_N-mg = 0$
$F_N = mg$
$F_N = (0.25~kg)(9.8~m/s^2)$
$F_N = 2.45~N$
