A bridge is 40.0 m. A car weighs 1.5 x 10^4 N and it is parked with its center of gravity located 12.0 meters from the right pier. What upward support force must be provided by the left pier?
draw the figure. Put the weight of the bridge in the center, and the car as determined.
Sum moments about the right pier.
clockwise positive.
LeftPierforceup*40-weightbridge*20-weightcar*12=0
so, if you know the weight of the bridge, you have it. Surely you don't want to ignore that, however, your teacher is your teacher.
To determine the upward support force provided by the left pier, we need to analyze the forces acting on the car and the bridge.
First, let's understand the forces acting on the car. The weight of the car is given as 1.5 x 10^4 N, which is the force due to gravity acting downwards. We can assume that this force is acting at the center of gravity of the car. Since the car is parked, there are no other forces acting on it except the weight.
Next, let's consider the forces acting on the bridge. The bridge exerts an upward support force on the car, allowing it to be in equilibrium. This upward force is countered by the weight of the car acting downwards. Additionally, there may be other forces acting on the bridge, but for this question, we are only concerned with the upward support force provided by the left pier.
Since the car is parked with its center of gravity located 12.0 meters from the right pier, we can assume that the bridge is symmetric, and the midpoint of the bridge is at the center of gravity of the car.
To calculate the upward support force provided by the left pier, we need to consider the torque equilibrium. Since the bridge is in equilibrium, the sum of all torques acting on the bridge must be zero.
Torque can be calculated using the formula:
Torque = Force x Distance
For the car, the torque due to its weight acting downwards is given by:
Torque_car = Weight_car x Distance_to_center_of_gravity
Substituting the given values, we get:
Torque_car = (1.5 x 10^4 N) x (12.0 m)
To maintain equilibrium, the left pier must exert an equal and opposite torque on the bridge. Since the bridge is symmetric, the distance between the left pier and the center of gravity of the car is also 12.0 m. Therefore, the torque exerted by the left pier is:
Torque_left_pier = Support_force_left_pier x Distance_to_center_of_gravity
Setting up the equation for torque equilibrium:
Torque_car = Torque_left_pier
(1.5 x 10^4 N) x (12.0 m) = Support_force_left_pier x (12.0 m)
Simplifying:
Support_force_left_pier = (1.5 x 10^4 N) x (12.0 m) / (12.0 m)
Support_force_left_pier = 1.5 x 10^4 N
Therefore, the upward support force provided by the left pier is 1.5 x 10^4 N.