a uniform bridge span weighs 50000N and is 40m long. An automobile weighing 15000N is parked with its center of gravity located 12m from the right pier. What upward support does the left pier provide?

To determine the upward support provided by the left pier, we first need to understand the equilibrium condition of the bridge.

In a state of equilibrium, the sum of all the forces acting on a body is equal to zero. In this case, the total downward force acting on the bridge should be balanced by the total upward force provided by both piers.

Let's break down the forces acting on the bridge:

1. The weight of the bridge span (uniform load): 50,000 N
2. The weight of the automobile: 15,000 N

Now, let's calculate the moments (or torques) caused by these forces about the right pier:

Moment of the bridge span = Weight x Distance = 50,000 N x (40 m - 12 m) = 50,000 N x 28 m = 1,400,000 Nm

Moment of the automobile = Weight x Distance = 15,000 N x 12 m = 180,000 Nm

To achieve equilibrium, the sum of the moments caused by these forces about the right pier must be zero:

Sum of moments = Moment of the bridge span + Moment of the automobile

0 = 1,400,000 Nm + 180,000 Nm

Now, we can find the upward support provided by the left pier:

Upward support by the left pier = Total downward force on the bridge - Total upward force provided by the right pier

Total downward force on the bridge = Weight of the bridge span + Weight of the automobile

Total downward force on the bridge = 50,000 N + 15,000 N = 65,000 N

Total upward force provided by the right pier = Weight of the automobile = 15,000 N

Upward support by the left pier = 65,000 N - 15,000 N = 50,000 N

Therefore, the left pier provides an upward support of 50,000 Newtons.

To find the upward support provided by the left pier, we need to consider the equilibrium of forces acting on the bridge. We know that the bridge span weighs 50000N and the automobile weighs 15000N.

Let's assume that the left pier provides an upward support force of F1 and the right pier provides an upward support force of F2.

To calculate the upward support provided by the left pier, we can use the principle of moments. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

In this case, we can take moments about the right pier as it will simplify the calculation of the moments. The clockwise moment created by the bridge's weight can be calculated as follows:

Moment(cw) = weight of bridge x distance from right pier
= 50000N x 20m (since the center of gravity of the bridge is at the middle, 40m/2 = 20m)
= 1000000 Nm

The anticlockwise moment created by the automobile can be calculated as:

Moment(acw) = weight of automobile x distance from right pier
= 15000N x 12m
= 180000 Nm

Since the bridge is in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.

Therefore, 1000000 Nm = 180000 Nm + (F1 x 40m)

We can rearrange this equation to solve for F1:

F1 = (1000000 Nm - 180000 Nm) / 40m
= 820000 Nm / 40m
= 20500 N

Therefore, the upward support provided by the left pier is 20500N.

29500

To get the upward force F on the left pier, set the total moment about the right pier equal to zero. The 50,000 N weight of the bridge acts through its CG, which is at the middle of the bridge. The weight of the car acts 28 m from the left pier,in the same direction. The force on the left pier produces a moment in the opposite direction.

50,000*20 + 15,000*28 - F*40 = 0
Solve for F