Question

A bridge of lenght 79 m and mass 1.1 x10^5 kg is supported at each end. A truck fo mass 20900 kg is located 31 m from the left end. What is the force on the bridge at the left point of support?

Answer 1

let x = support at left end
let y = support at right end
now, to get the reaction force at the left support, we sum moments at the right end, so that y will not be included in our calculation, and we can directly solve for x,,
*recall that moment = (force)*(distance perpendicular to that force) or
M = r x F (in vectors)
we use right hand rule to get the direction of the moment
when moment is out of the page, it's (+) and if into the page, it's (-)
anyway, summing moments at right end:
Given:
w,truck = 20900 * 9.8 N downward
W,bridge = 1.1 x 10^5 * 9.8 downward
direction of x and y is upward (assumed)

M, right end = 0 (because in equilibrium)
(79-31)(20900*9.8) + (79/2)(1.1 x 10^5 * 9.8) - (79)x = 0
note:
*distance from right end to truck is 79-31, and the force acting at this point is W,truck
*the W,bridge is acting on the center/middle of the bridge assuming uniform composition, thus the distance from right end to weight of bridge is 79/2 m
*the distance from right end to left end id 79, or simply the length of the bridge

solving,
x = 663447.6 N
since we got a positive answer, this means that the assumed direction is correct, thus
reaction at left end = 663447.6 N upward

hope this helps~ :)