Prove that the ratio of belt tension is given by the T1/T2 = e^{μθ}
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 T_{1} and T_{2} 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(T_{1}/T_{2}) = µ.θ
or, T_{1}/T_{2} = e^{µ.θ}
Ratio of belt tension = T_{1}/T_{2} = e^{µθ}
Belt ratio is represent as 2.3log(T_{1}/T_{2}) = µ.q
Note that θ is in radian
In this formula the important thing is Angle of contact(θ)