With a rolling resistance force of 7848 N, a friction coefficient of 0.9 between rubber wheels and dry tarmac, and assuming a direct connection close to the tarmac:

$m_f = \frac{F_r}{\mu \cdot a_g} = \frac{7848 \text{ N}}{0.9 \cdot 9.81 \text{ m/s}^2} \approx 889 \text{ kg}$

People Required to Pull an Aircraft

When people pull the aircraft, the rope angle introduces a force vector relationship. The minimum vertical force (or weight) depends on the bending angle $\theta$ :

$W = \tan(\theta) \cdot F_r$

For a 45° bending angle: $W = \tan(45^\circ) \cdot 7848 \text{ N} = 7848 \text{ N}$ $m_f = \frac{W}{a_g} = \frac{7848 \text{ N}}{9.81 \text{ m/s}^2} \approx 889 \text{ kg}$ At 80 kg per person: $n = 889 \text{ kg} / 80 \text{ kg/person} \approx 11.1 \rightarrow 12 \text{ people}$ .

For a more realistic 60° bending angle: $W = \tan(60^\circ) \cdot 7848 \text{ N} \approx 1.732 \cdot 7848 \text{ N} \approx 13593 \text{ N}$ $m_f = \frac{W}{a_g} = \frac{13593 \text{ N}}{9.81 \text{ m/s}^2} \approx 1386 \text{ kg}$ At 80 kg per person: $n = 1386 \text{ kg} / 80 \text{ kg/person} \approx 17.3 \rightarrow 18 \text{ people}$ .

References

Original Source

Source: engineeringtoolbox.com