Posted by Physics on Monday, December 10, 2012 at 12:17am.
Let F1 and F2 be the forces on the two ends of the bridge.
The sum of the forces in the y direction is 0:
F1 + F2 - 1120 - 3775 = 0
The sum of the torques about each support is zero:
1120*(L/5) + 3775*(L/2) - F2*L = 0
Dividing by L:
1120/5 + 3775/2 - F2 = 0
Solve for F2, then solve for F1
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