Answer
$N_B=2428lb$
$N_C=1622lb$
$F=200lb$
Work Step by Step
We can determine the normal reaction force $N_B$ as
$\Sigma M_C=0$
$2N_B(9)-G_T(5)-G_A(12)+T(2.5)=0$
We plug in the known values to obtain:
$2N_B(9)-7500(5)-600(12)+400(2.5)=0$
This simplifies to:
$N_B=2428lb$
Now we equate the forces in the y-direction to determine the normal force under each wheel as follows:
$\Sigma F_y=0$
$\implies 2N_C+2N_B-G_T-G_A=0$
We plug in the known values to obtain:
$2N_C+2(2428)-7500-600=0$
This simplifies to:
$N_C=1622lb$
Now we can find the frictional force as
$\Sigma F_x=0$
$\implies 2F-T=0$
$2F-400=0$
$2F=400$
$\implies F=200lb$
