Prove that the ratio of belt tension is given by the T1/T2 = eμθ
Let T1 = Tension in belt on the tight side T2 = Tension in belt on the slack side θ = Angle of contact
μ = Co efficient of friction existing between belt and pulley.
α = Angle of Lap
Consider driven or follower pulley. Belt remains in contact with the EBF. Let T1 and T2 are the tensions in tight side and slack side.
Angle EBF known as angle of contact = ∏.-2α
Consider driven or follower pulley.
Belt remains in contact with the NPM. Let T1 and T2 are tensions in tight side and slack side. Let T be tension at point M & (T + dT) be the tension at point N.
Consider equilibrium in horizontal Reaction 'R' and vertical reaction µR.
Since the whole system is in equilibrium, that is
ΣV = 0; Tsin (90 - δq/2) + µR - (T + δT)sin(90 - δθ/2) = 0
Tcos (δθ/2) + µR = (T + δT) cos (δθ/2)
Tcos (δθ/2) + µR = Tcos(δq/2) + δTcos(δθ/2)
µR = δTcos(δθ/2)
Since δθ/2 is very small and cos0° = 1, So cos(δθ/2) = 1
µR = δT ...(i)
ΣH = 0;
R-Tcos(90 - δq/2)-(T + δT)cos(90 - δθ/2) = 0
R = Tsin(δθ/2) + (T + δT)sin(δθ/2) As δθ/2 is small So sin(δθ/2) = δθ/2
R = T(δθ/2) + T(δθ/2) + δT(δθ/2) R = T.δθ + δT(δθ/2)
As δT(δq/2) is very small So δT(δq/2) = 0
R = T.δq ...(ii)
Putting value of (ii) in equation (i)
µ.T.δθ = δT
or, δT/T = µ.δθ
Integrating both theside: , Where θ = Total angle of contact
ln(T1/T2) = µ.θ
or, T1/T2 = eµ.θ
Ratio of belt tension = T1/T2 = eµθ
Belt ratio is represent as 2.3log(T1/T2) = µ.q
Note that θ is in radian
In this formula the important thing is Angle of contact(θ)