A uniform bridge span weighs 70.0 × 103 N and is 40.0 m long. An automobile weighing 15.0 × 103 N is parked with its center of gravity located 12.0 m from the right pier. What upward support force does the right pier provide?Select one:a.45.5 × 103 Nb.65.0 × 103 Nc.24.5 × 103 Nd.35.5 × 103 N
Question
A uniform bridge span weighs 70.0 × 103 N and is 40.0 m long. An automobile weighing 15.0 × 103 N is parked with its center of gravity located 12.0 m from the right pier. What upward support force does the right pier provide?Select one:a.45.5 × 103 Nb.65.0 × 103 Nc.24.5 × 103 Nd.35.5 × 103 N
Solution
To solve this problem, we need to understand that the total force acting on the bridge must be zero for it to be in equilibrium. This means that the sum of the forces provided by the left and right piers must equal the total weight of the bridge and the car.
The total weight is the weight of the bridge (70.0 × 10^3 N) plus the weight of the car (15.0 × 10^3 N), which equals 85.0 × 10^3 N.
The force provided by the right pier is the weight of the bridge and the car, minus the force provided by the left pier.
The force provided by the left pier can be calculated by considering the moments about the right pier. The moment due to the car is the weight of the car times the distance from the right pier (15.0 × 10^3 N * 12.0 m). The moment due to the bridge is the weight of the bridge times half the length of the bridge (70.0 × 10^3 N * 20.0 m), because the weight of the bridge is evenly distributed along its length.
Setting these two moments equal gives us the equation:
(15.0 × 10^3 N * 12.0 m) = (70.0 × 10^3 N * 20.0 m) - Force_left_pier * 40.0 m
Solving for Force_left_pier gives us:
Force_left_pier = (15.0 × 10^3 N * 12.0 m + 70.0 × 10^3 N * 20.0 m) / 40.0 m = 39.5 × 10^3 N
Subtracting this from the total weight gives us the force provided by the right pier:
Force_right_pier = 85.0 × 10^3 N - 39.5 × 10^3 N = 45.5 × 10^3 N
So, the answer is a. 45.5 × 10^3 N.
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