# A belt 100mm wide and 8.0mm thick are transmitting power at a belt speed of 160m/minute.

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A belt 100mm wide and 8.0mm thick are transmitting power at a belt speed of 160m/minute. The angle of lap for smaller pulley is 165º and coefficient of friction is 0.3. The maximum permissible stress in belt is 2MN/m2 and mass of the belt is 0.9Kg/m. find the power transmitted and the initial tension in the belt.

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Given data:

Width of belt(b) = 100mm

Thickness of belt(t) = 8mm

Velocity of belt(V) = 160m/min = 2.66m/sec

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

Coefficient of friction(µ) = 0.3

Maximum permissible stress(f) = 2 X 106 N/m2 = 2N/mm2

Mass of the belt material(m) = 0.9 Kg/m Power = ?

Initial tension (To) = ? We know that,

Tmax = σ.b.t = 2 X 100 X 8 = 1600N ...(i)

Since m and velocity (V) is given, then

Using the formula, TC = mV2, For finding Tc

= 0.9(2.66)2

= 6.4 N ........(ii)

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

1600 = T1 + 6.4

T1 = 1593.6N ......(iii)

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

1593.6/T2 = e(0.3)(2.88)

T2 = 671.69 N ........(iv)

Now Power Transmitted = (T1–T2).v/1000 Kw

P = (1593.6 – 671.69).2.66/1000 Kw

P = 2.45KW

Let To = initial tension in the belt

To = (T1 + T2)/2 + TC

To = (1593.6 + 671.69)/2 + 6.4

To = 1139.045N.