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# Calculate belt force and wrap angle

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?

## 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 °

$\tag{10} \alpha\,’ = 200°$