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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your answer is, orbit around the nucleus

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s