Answer
(a) $0.42KN$
(b) less than
Work Step by Step
(a) We can find the required force as follows:
$\Sigma F_y=0$
$Tsin24^{\circ}+Tsin24^{\circ}-mg+N=0$
$\implies 2Tsin24^{\circ}-mg+N=0$
$\implies N=mg-2Tsin24^{\circ}$
We plug in the known values to obtain:
$N=(67Kg)(9.81m/s^2)-2(290N)sin24^{\circ}$
$N=42N=0.42KN$
(b) We know that the force exerted will be less than that found in part (a) because a force of 290 N upward is more than 290 N to the side. The acceleration would remain zero and the normal force from the floor would decrease in the equation of part (a).
