Question

A belt of density 1gm/cm³ has a maximum permissible stress of 250N/cm². Determine the maximum power that can be transmitted by a belt of 20cm X 1.2cm if the ratio of the tight side to slack side tension is 2. 

Answer 1

Given date: 

Density of belt = 1gm/cm³ = 1/1000 Kg/cm³ 

Maximum permissible stress(f) = 250 N/cm² 

Width of belt(b) = 20cm 

Thickness of belt(t) = 8mm 0.8cm 

Ratio of tension (T₁/T₂) = 2 

Maximum Power = ? 

We know that, Tmax = σ.b.t 

= 250 X 20 X 1.2 = 6000N ...(i) 

Since Unit mass (m) = σ.b.t.L 

 = 1/1000 X 20 X 1.2 X 100 = 2.4Kg ...(ii) 

Since velocity(V) is not given, 

So we find the velocity using formula T_C =1/3 T_max = mV^2, for maximum power 

Using the formula, 1/3 T_max = mV²

V = (Tmax/3m)^1/2

V = (6000/3 X 2.4)^1/2

V = 28.86 m/sec ...(iii) 

Using the formula, T_C = mV², For finding T_C

= 2.4(28.86)² 

= 1998.96N ...(iv) 

Using the formula, T_max = T₁ + T_c , for finding T₁ 

6000 = T₁ + 1998.96

 T₁ = 4001N...(v)

Now, For T_2, Using relation Ratio of belt tension = T₁/T₂ = e^µθ = 2 

4001/T₂ = 2 

T₂ = 2000.5 N ...(vi) 

Now Maximum Power Transmitted = (T₁–T₂).V/1000 KW 

P = (4001 - 2000.5) X 28.86/1000 KW 

P = 57.73KW