Abelt drive was designed to transmit the power of p=7.5 kw with the velocity v=10m/s. the tensile load of the tight side is twice of that of loose side, f1= 2f2. calculate the f1, fe, and fo.

F₁ = 1500 N

F₂ = 750 N

= 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

= 3 x

where is centrifugal force

  = / 3

                          = 1500 / 3

                         = 500 N

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hope this is what you're talking about

answer:

explanation:

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