In the diagram below, the bridge spans 6 m and is supported by two beams. The weight of the bridge is 1000 N. A car of 3000 N is 4 m away from the support FI. Use the information given to calculate the reaction forces F1 and F2

The reaction force F₁ = 1500 N and the reaction force F₂ = 2500 N

The moment of a force about a point, M is the product of the force F and its perpendicular distance from the point, d.

So, M = Fd

Considering the bridge, since it is in equilibrium, clockwise moments, M' = anticlockwise moments M".

The reaction force F₁ = 1500 N

Taking moments about the F₂ force, we have

F₁ × 6 m = 1000 N × (6 m - 3 m) + 3000 N × (6m - 4 m)

F₁ × 6 m = 1000 N × 3 m+ 3000 N × 2 m

F₁ × 6 m = 3000 Nm + 6000 Nm

F₁ × 6 m = 9000 Nm

F₁ = 9000 Nm/6 m

F₁ = 1500 N

The reaction force F₁ = 1500 N

The reaction force F₂ = 2500 N

Taking moments about the F₁ force, we have

F₂ × 6 m = 1000 N × 3 m + 3000 N × 4 m

F₂ × 6 m = 3000 Nm + 12000 Nm

F₂ × 6 m = 15000 Nm

F₂ = 9000 Nm/6 m

F₂ = 2500 N

The reaction force F₁ = 1500 N and the reaction force F₂ = 2500 N

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correct it is c.) b b

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.