Question

A tourist, who weighs 805 N, is walking through the woods and crosses a small horizontal bridge. The bridge rests on two concrete supports, one at each end. He stops one-fourth of the way along the bridge. Assume that the board of the bridge has negligible weight. What is the magnitude of the vertical force that a concrete support exerts on the bridge at the far end?

Answer 1

$R_1$ and $R_2$ are the vertical force at near and far end respectively.

Taking moment at point A;

$-R_1×0-w×\frac {L} {4}+R_2×L=0$

$=R_2L=W\frac {L} {4}$

$R_2=\frac {w} {4}=\frac {805}{4}$

$R_2=201.25N$

Thus the vertical force acting on the far end is 201.25N