Answer
$F_2 = (1740~N)~\hat{j}$
Work Step by Step
In part (a), we found that the force on the beam due to support 1 is $~~1160~N~~$ and this force is directed upward.
The sum of the forces on the beam due to the two supports is equal in magnitude to the sum of the weight of the beam and the gymnast.
We can find the force on the beam due to support 2:
$F_1+F_2 = (250~kg+46.0~kg)(9.8~m/s^2)$
$F_2 = (250~kg+46.0~kg)(9.8~m/s^2)- F_1$
$F_2 = (250~kg+46.0~kg)(9.8~m/s^2)- (1160~N)$
$F_2 = 1740~N$
The force on the beam due to support 2 is $~~1740~N~~$ and this force is directed upward.
We can express the force on the beam due to support 2 in unit-vector notation:
$F_2 = (1740~N)~\hat{j}$
