Answer
The magnitude of the force on the bridge from the support at the farther end is $~~1500~N$
Work Step by Step
To find the magnitude of the force on the bridge from the support at the farther end, we can consider the torque about a rotation axis at the closer end of the bridge:
$\sum \tau = 0$
$L~F - (\frac{L}{4})(73~kg)(9.8~m/s^2)-(\frac{L}{2})(2700~N) = 0$
$L~F = (\frac{L}{4})(73~kg)(9.8~m/s^2)+(\frac{L}{2})(2700~N)$
$F = (\frac{1}{4})(73~kg)(9.8~m/s^2)+(\frac{1}{2})(2700~N)$
$F = 1500~N$
The magnitude of the force on the bridge from the support at the farther end is $~~1500~N$
