Answer:
P = 238475 N
Q = 181125
Explanation:
From the diagram attached below,
For the bridge to be at equilibrium,
Sum of upward force = sum of downward force
P+Q = 400000+19600
P+Q = 419600.................. Equation 1
Taking moment about support A
sum of clockwise moment = sum of anti clockwise moment.
400000(10-3)+19600(8-3) = Q(20-6)
2800000+98000 = 16Q
2898000 = 16Q
Q = 2898000/16
Q = 181125 N.
Substitute the value of Q into equation 1
P+181125 = 419600
P = 419600-181125
P = 238475 N.
Hence, support A exert a force of 238475 N and support B exert a force of 181125 N
Answer:
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
