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Calculate belt force and wrap angle

Task

A belt drives a roller with a force A = 890 N in the load side and a wrap angle α = 160 °. The coefficient of static friction is µ0 = 0.3.

a) How big is the force L in the loose belt?

b) How big must the wrap angle α’ be if the force in the work belt is increased to A’ = 1,100 N while the force L in the loose belt remains the same?

Belt with pulley and tension pulley
Belt with pulley and tension pulley

Solution

The relationship between the forces in the work belt and in the loose belt is described by the belt friction formula (Euler-Eytelwein).

\[ \require{cancel} \]

reg. a), value of the force L

\[ \tag{1} A \leq L \cdot e^{µ_0 \cdot \alpha} \]

The inequalition is simplified in the following to

\[ \tag{2} L = A \cdot e^{-µ_0 \cdot \alpha} \]

\[ \tag{3} L = 890 N \cdot e^{-0.3 \cdot \frac{8}{9} \cdot \pi} \]

\[ \tag{4} L = 385 N \]

reg. b), increased force in the work belt

The calculation of the required wrap angle for the increased tensile force in the load side is carried out using the following relationship

\[ \tag{5} L = A’ \cdot e^{-µ_0 \cdot \alpha\,’} \]

\[ \tag{6} \frac{A’}{L} = e^{µ_0 \cdot \alpha\,’} \]

\[ \tag{7} \alpha\,’ = \frac{\ln \left( \frac{A’}{L} \right)}{µ_0} \]

\[ \tag{8} \alpha\,’ = \frac{\ln \left( \frac{1100 N}{385 N} \right)}{0.3} \]

The necessary wrap angle is therefore in radians

\[ \tag{9} \alpha\,’ = 3,5 rad \]

or converted into °

\[ \tag{10} \alpha\,’ = 200° \]