A uniform bridge 20.0 m long and weighing 400000 N is supported by two pillars located 3.00 m from each end. If a 19600 N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert?

F1 = 212,600N and F2 = 207,000 N

Explanation:

let; F1 = upward pillar force near the left end of bridge

F2 = upward pillar force near the right end of bridge

WB = bridge weight

WC = car weight

Now, the weight of the bridge and that of the car are acting downwards.

Thus, at translational equilibrium, if we resolve the forces, we'll get;

Σy = 0

Thus, F1 + F2 - WB - WC = 0

We are given the weights of the car and bridge as;

WB = 400000N

WC = 19600N

Thus,we now have ;

F1 + F2 - 400000N - 19600N = 0

F1 + F2 = 400000N + 19600N

F1 + F2 = 419,600N - - - - - (eq1)

let the point where F1 exerts its upward force be the pivot point for system torques.

Now, at rotational equilibrium about F1, we have;

F2•(14) - WC•(5) - WB•(7) = 0

Plugging in values for WC and WB to obtain;

F2•(14) - 19600•(5) - 400000•(7) = 0

14F2 - 98000 - 2800000 = 0

14F2 = 2800000 + 98000

14F2 = 2898000

F2 = 2898000/14

F2 = 207,000 N

Now, let's put this for F2 in eq 1

F1 + 207,000N = 419,600N

Subtract 207,000N from both sides.

F1 = 419,600N - 207,000

F1 = 212,600N

explanation:

newton's third law, for every action there is an equal and opposite reaction.

and 1and 1