Answer
The force on the beam from roller B is $~~294~N$
Work Step by Step
In part (a), we found that the force on the beam from roller A is $~~196~N$
The sum of the forces on the beam from the rollers is equal in magnitude to the total weight of the beam and the package.
We can find the force on the beam from roller B:
$F_A+F_B = (40.0~kg+10.0~kg)~(9.8~m/s^2)$
$F_B = (40.0~kg+10.0~kg)~(9.8~m/s^2) - F_A$
$F_B = (40.0~kg+10.0~kg)~(9.8~m/s^2) - (196~N)$
$F_B = 294~N$
The force on the beam from roller B is $~~294~N$
