Answer to Question #202466 in Mechanical Engineering for Brian

Question

 A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at 1600 m/min. The mass of the belt is 0.9 kg/m length. The angle of lap in the smaller pulley is 165° and the coefficient of friction between the belt and pulley is 0.3. If the maximum permissible stress in the belt is 2 MN/m2, find : 1. maximum power transmitted ; and 2. initial tension in the belt

Answer 1

$\frac{T1}{ T2}= e^{μθ} = e^{0.3×2.88}= 2.372 … … . a$

$T_c = mv^2 = 0.9 × (26.67)^2 = 640N$

$T = σ × b × t = T_c + T_1 → 2 × 106 × 0.1 × 0.008 = 640 + T_1$

$T_1 = 1600 − 640 → T_1 = 960N … … . b$

sub a in b ∶

$\frac{960}{ T2} = 2.372 → T_2 = 404.72N$

$P = (T_1 − T_2) × v → P = (960 − 404.72) × 26.67 = 14809.3W = 14.8kW$

$Tₒ = \frac{T_1 + T_2 + 2T_c}{ 2} =\frac{ 960 + 404.72 + (2 × 640)}{ 2 }= 1322.36N$

But when we use equation below , we get the same result which given for [Tₒ]

$Tₒ = \frac{T_1 + T_2 + T_c}{ 2} = \frac{960 + 404.72 + 640}{ 2} = 1002.36N$