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Setler [38]
3 years ago
14

A belt 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 twi

ce of that of loose side, F1= 2F2. Please calculate the F1, Fe, and Fo.
Engineering
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 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,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

                         = 500 N

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