# A belt of density 1gm/cm^3 has a maximum permissible stress of 250N/cm^2. Determine the maximum power

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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.

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Given date:

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

Maximum permissible stress(f) = 250 N/cm

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 TC =1/3 Tmax = mV2, for maximum power

Using the formula, 1/3 Tmax = mV2

V = (Tmax/3m)1/2

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

V = 28.86 m/sec ...(iii)

Using the formula, TC = mV2, For finding TC

= 2.4(28.86)2

= 1998.96N ...(iv)

Using the formula, Tmax = T1 + Tc , for finding T

6000 = T1 + 1998.96

T1 = 4001N...(v)

Now, For T2, Using relation Ratio of belt tension = T1/T2 = eµθ = 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