A bridge is supported by two piers located 14 meters apart. Both the left and right piers provide an upward force on the bridge, labeled FL and FR respectively. (a) If a 1250 kg car comes to rest at a point 5 meters from the left pier, how much force (in N) will the bridge provide to the left and right piers? (Enter the magnitudes.)

4375 N, 7875 N

Explanation:

since the body is equilibrium

total upward force = total downward force

w weight of the bridge = FL + FR

when the car was introduced,

total downward force = total upward force

FL₁ + FR₁ = w + (m × acceleration due to gravity) with w = FL + FR

then FL₁ + FR₁  = FL + FR + Mcg

FL₁ + FR₁  - FL - FR = Mcg

ΔFL + ΔFR = 1250 × 9.8 = 12250 N

taken the left as the pivot point and using the principle of moment

ΔFR × 14 m = 12250 N × 5 m + (ΔFL × 0 m) since the left is the pivot point.

ΔFR = 61250 / 14 = 4375 N

but

ΔFL + ΔFR = 12250 N

ΔFL = 12250 - 4375 = 7875 N

increases it

change tension