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ANEK [815]
3 years ago
5

In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff

ort if it losses are negligible​
Physics
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

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Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

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Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

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Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

      w₂ = 20.94 rad / s

Let's replace

       w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

      w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

      w = 20.94 1.28 / 5.6

      w = 4,786 rad / s

Angular velocity and frequency are related.

      w = 2π f

      f = w / 2π

      f = 4.786 / 2π

      f = 0.76176 Hz

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