The force of gravity on the bridge below is 9.60 x 105 N. What upward force must be exerted at plsssss?

Question

The force of gravity on the bridge below is 9.60 x 105 N. What upward force must be exerted at end Q to support the bridge and the truck, if the force of gravity on the truck is 4.80 x 104 N? Hint! Treat the bridge as if it was a lever with an imaginary pivot at P. What torques tends to rotate the "lever" about pivot P?

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Answer 1

The resultant torque at P must be zero since the bridge is in equilibrium state. Since Torque = force x distance,

Torque clockwise = 9.6x10^5x8.5 + 4.8x10^4x12

torque anticlockwise = Fx(12+5), where F is the upward force that must be exerted at Q. Since the torques of different direction cancel out each other, we have

9.6x10^5x8.5 + 4.8x10^4x12= Fx(12+5)

=>8.16x10^6+5.76x10^5=17F

=>F=87.36x10^5/17

=>F=5.14x10^5 N #