Question

A hiker, who weighs 693 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 4070 N, and rests on two concrete supports, one on each end. He stops 1/4 of the way along the bridge.

What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?

Answer 1

a) F₁ = 2554.75 N, b) F₂ = 2208.25 N

Step-by-step explanation:

This is an equilibrium exercise, where all the forces are vertical, write the rotational equilibrium equation, we must set a reference system.

b) Reference system on the first concrete support

∑ τ = 0

The data given are:

Person's weight is W1 = 693 N

Bridge weight is W = 4070 N

-W₁ L / 4 - W L / 2 + F₂ L =

F₂ is the farthest support force; the signs are positive if the turn is counterclockwise

F₂ = W1 / 4 + w / 2

let's calculate

F₂ = 693/4 + 4070/2

F₂ = 2208.25 N

a) Now let's change the reference system to the farthest support, rewrite the equation and balance

W₁ ¾ L + W ½ L - F₁ L = 0

The signs change because the direction of rotation changes when changing the support point

F₁ = W₁ ¾ + W ½

calculate

F₁ = 693 ¾ + 4070 ½

F₁ = 2554.75 N