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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

+15-35=-20

going in opposite directions