Question

What minimum force applied at 8 inches axle of 2 feet wheel is necessary to raise the wheel over an obstacle of 10 inches height? Take r as radius and W as its weight.

Answer 1

$$r=2\ ft=0.6096\ m, h=10\ in=0.254\ m, m=1.17\ kg$$

$$a^2+(r-h)^2=r^2\to a^2+(0.6096-0.254)^2=0.6096^2$$

$$a=0.495\ m$$

Moments of two forces will be equal:

$$F(r-h)=mga$$

$$F(0.6096-0.254)=(1.17)(9.8)(0.495)$$

$$F=16.0\ N$$