Two cars were at the bridge,given that D1=100m,D2=50m,W2=2000N,W1=1000N and D3=200m. Calculate the forces due to the pillars at A and B.
Can't make sense out of your data. Three Ds for two cars? What pillars?
To calculate the forces due to the pillars at A and B, we need to consider the equilibrium of the system.
The total vertical forces acting on the bridge should sum up to zero for it to be in equilibrium.
Let's start by considering the vertical forces acting on the bridge:
At point A:
- Force due to the weight of car 1: W1 (acting downward)
- Force due to the weight of car 2: W2 (acting downward)
- Force due to the pillar at A: Fa (unknown direction)
At point B:
- Force due to the weight of car 2: W2 (acting downward)
- Force due to the pillar at B: Fb (unknown direction)
Now, let's set up the equilibrium equation:
Total vertical forces at point A + Total vertical forces at point B = 0
(W1 + W2) + (W2 + Fa + Fb) = 0
Substituting the given values:
(1000N + 2000N) + (2000N + Fa + Fb) = 0
3000N + 2000N + Fa + Fb = 0
5000N + Fa + Fb = 0
Now, we need to find the horizontal forces acting on the bridge:
At point A:
- Horizontal force due to the weight of car 1: W1 (acting to the right)
- Horizontal force due to the weight of car 2: W2 (acting to the right)
- Horizontal force due to the pillar at A: Fa (unknown direction)
At point B:
- Horizontal force due to the weight of car 2: W2 (acting to the left)
- Horizontal force due to the pillar at B: Fb (unknown direction)
Since the bridge is in equilibrium, the total horizontal forces at point A should be equal to the total horizontal forces at point B.
W1 + W2 + Fa = W2 + Fb
Substituting the given values:
1000N + 2000N + Fa = 2000N + Fb
3000N + Fa = 2000N + Fb
Fa - Fb = -1000N
Now, we have two equations:
1) 5000N + Fa + Fb = 0
2) Fa - Fb = -1000N
We can solve these two equations simultaneously to find the values of Fa and Fb.