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# Belt friction

This is an exercise to introduce the capstan formula. Here you can find an online calculation tool for belt friction.

A load of m = 100 kg is lowered with a rope wrapped around a pole. The wrap angle can be seen in the sketch below. The coefficient of sliding friction is µ = 0,2. What value for the holding force F is needed to lower the load without acceleration (G = m * g)?

## Solution

A solution video will be published soon on this channel.

The holding force F is calculated using the Euler-Eytelwein formula (rope friction formula), which for this case is:

$\tag{1} G = Fe^{µ\alpha}$

With the given values ​​and the angle of wrap alpha measured from the sketch

$\tag{2} G =mg$

$\tag{3} µ = 0.2$

$\tag{4} \alpha = \frac{3}{2}\pi$

$\tag{5} mg = Fe^{\frac{3}{10}\pi}$
$\tag{6} F = \frac{mg}{e^{\frac{3}{10}\pi}}$
$\tag{7} F = 100 kg \cdot 9.81 \frac{m}{s^2} \cdot e^{-\frac{3}{10}\pi}$
$\tag{8} F = 382.26 N$