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A uniform bridge of weight 5.0 × 10⁴ N is supported on pillars A and B, which are 20 m apart. A lorry of weight 3.0 × 10⁴ N is stationary 4.0 m from pillar A.

Find the forces acting on each pillar.____

User Pleshy
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Final answer:

The force acting on pillar A is 3.0 x 10^4 N and the force acting on pillar B is 5.0 x 10^4 N.

Step-by-step explanation:

To find the forces acting on each pillar, we need to consider the balance of forces in the vertical direction. Since the bridge and the lorry are stationary, the vertical forces on both pillars must add up to zero.

Let's assume the force acting on pillar A is FA and the force acting on pillar B is FB. The weight of the bridge, which is the force acting downward, is 5.0 x 10^4 N. The weight of the lorry, which is also acting downward, is 3.0 x 10^4 N. Therefore, we can write the equation:

FA + FB - 5.0 x 10^4 N - 3.0 x 10^4 N = 0

Simplifying, we get:

FA + FB = 8.0 x 10^4 N

Since the pillars are 20 m apart, we can assume that the distance between the pillars is evenly divided between them. Therefore, the distance between the lorry and pillar A is 2 m. Using the concept of moments, we can find that:

FA x 20 m - 3.0 x 10^4 N x 2 m = 0

Simplifying, we get:

FA = 3.0 x 10^4 N

Substituting this value back into the equation FA + FB = 8.0 x 10^4 N, we find that:

FB = 5.0 x 10^4 N

Therefore, the force acting on pillar A is 3.0 x 10^4 N and the force acting on pillar B is 5.0 x 10^4 N.

User Badda
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