a uniform bridge span weighs 50x10^3 and is 40m long. A car weighing 15X10^3 is parked with its center of gravity located 12m from the right pier. what upward support force does the left pier provide?

im lost....this is what i think im supposed to do
40m-12m= 28m to left pier
50.0x10^3- (40m/28m)*15.0x10^3=
28.5x10^3? is this right if not what am i doing wrong please help!

29.5 x 10*3

Well, if you're lost, I'm here to help you find your way with a little humor! Let's break it down.

First, let's find the total torque acting on the bridge. Torque is equal to the weight of an object multiplied by its distance from a reference point. In this case, the reference point is the left pier. So, the torque acting on the bridge is (50x10^3 N) * (28m).

Now, let's find the torque acting on the car. The weight of the car is (15x10^3 N) * (12m).

Since the bridge is in equilibrium, these two torques should be equal. So, we can set up an equation:

(50x10^3 N) * (28m) = (15x10^3 N) * (12m) + (upward support force) * (40m)

Now, you can solve this equation to find the upward support force provided by the left pier. And don't worry, I won't be "pier"-pressuring you into figuring that out on your own!

To find the upward support force that the left pier provides, you can use the principle of rotational equilibrium.

First, let's calculate the torque created by the bridge span. Torque is equal to the force applied multiplied by the perpendicular distance from the point of rotation. In this case, the force is the weight of the bridge span (50x10^3 N) and the perpendicular distance is half of the length of the bridge span (40m/2 = 20m).

Torque = Force x Distance
Torque = 50x10^3 N x 20m
Torque = 1000x10^3 Nm

Next, let's calculate the torque created by the car. The force is the weight of the car (15x10^3 N) and the perpendicular distance is the distance of the car's center of gravity from the right pier (12m).

Torque = Force x Distance
Torque = 15x10^3 N x 12m
Torque = 180x10^3 Nm

Now, since the bridge span is in equilibrium, the sum of the torques created by the bridge span and the car must be equal to zero.

Total Torque = Torque of Bridge Span - Torque of Car

Therefore,
Total Torque = 1000x10^3 Nm - 180x10^3 Nm
Total Torque = 820x10^3 Nm

Finally, we need to find the upward support force provided by the left pier. This is equal to the total torque divided by the perpendicular distance from the left pier (28m).

Support Force = Total Torque / Distance
Support Force = 820x10^3 Nm / 28m
Support Force = 29.3x10^3 N

Therefore, the upward support force provided by the left pier is 29.3x10^3 N.

To find the upward support force that the left pier provides, you need to consider the forces acting on the bridge. Here's a step-by-step explanation:

1. Identify the forces acting on the bridge:
a) Weight of the bridge span: Given as 50x10^3 N (Newtons), acting downwards at the center of gravity.
b) Weight of the car: Given as 15x10^3 N, acting downwards at a distance of 12m from the right pier.

2. Calculate the total downward force acting on the bridge:
Add the weights of the bridge span and the car:
Total downward force = Weight of bridge span + Weight of car
= 50x10^3 N + 15x10^3 N
= 65x10^3 N

3. Determine the distance between the center of gravity of the bridge span and the left pier:
The bridge is 40m long, and the center of gravity is located at the middle of the span:
Distance to the left pier = 40m / 2
= 20m

4. Calculate the upward support force provided by the left pier:
The total downward force acting on the bridge needs to be balanced by the upward support force provided by the left pier:
Upward support force = Total downward force
= 65x10^3 N

Therefore, the upward support force provided by the left pier is also 65x10^3 N.

It seems like there was a mistake in your calculation. The steps outlined above should help you arrive at the correct answer.

Are the weights in Newtons? It isn't enough to just provide a number. I will assume the bridge weighs 50*10^3 N and the car weighs 15*10^3 N.

Let F1 be the force on the lest pier and F2 be the force on the right pier. One of two equations that you would need to solve for both forces is

F1 + F2 = (50+15)*10^3 = 65*10^3 N

Now write another equation for equilbrium of moments about the right pier. There is a clockwise moment of F1*40 m and two conterclockwise moments of 50*10^3N*20 m due to the weight of the bridge and 15*10^3N*28m. The sum of these moments (with a negative sign for couterclockwise moments)is zero. Use that fact to solve for F1
40 F1 - 50*10^3*20 - 15*10^3*28+0

Solve for F1.

I get 35,500 N, a bit more than half the total weight.