# A belt embraces the shorter pulley by an angle of 165º and runs at a speed of 1700 m/min,

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A belt embraces the shorter pulley by an angle of 165º and runs at a speed of 1700 m/min, Dimensions of the belt are Width = 20cm and thickness = 8mm. Its density is 1gm/cm3. Determine the maximum power that can be transmitted at the above speed, if the maximum permissible stress in the belt is not to exceed 250N/cm2 and µ = 0.25.

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Given date:

Angle of contact(θ) =165° = 165° X Π/180 = 2.88rad

Velocity of belt(V) = 1700m/min = 28.33m/sec

Width of belt(b) = 20cm

Thickness of belt(t) = 8mm 0.8cm

density of belt = 1gm/cm3

Maximum permissible stress(f) = 250 N/cm2

Coefficient of friction(µ) = 0.25 Maximum Power = ?

We know that, Tmax = σ.b.t

= 250 X 20 X 0.8 = 4000N ......(i)

Since Unit mass (m) = ρ.b.t.L

= 1/1000 X 20 X 0.8 X 100 = 1.6Kg .... (ii)

Since velocity(V) is given,

So we don't find the velocity using formula TC =1/3 Tmax = mV2, then Using the formula, TC = mV2, For finding TC.

= 1.6(28.33)2 = 1284 N ...(iii)

Using the formula, Tmax

= T1 + Tc , for finding T1

4000 = T1 + 1284

T1 = 2716N ...(iv)

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

2716/T2 = e(0.25)(2.88) T2 = 1321 N ...(v)

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

P = (2716 – 1321) X 28.33/1000 KW

P = 39.52KW.