91Ó°ÊÓ

Obstacle of height$h=3.00cm$

Massof the wheel$0.800kg$

Wheel’s radius is$r=6.00cm$,

Here,at the moment when the wheel leaves the lower floor, the floor no longer exerts a force on it.As the wheel is raised over the obstacle, the only forces acting are the force Fapplied horizontally at the axle, the force of gravity mg acting vertically at the center of the wheel, and the force of the step corner, shown as the two components$\mathit{f}\mathit{h}$androle="math" localid="1661752228965" $\mathit{f}\mathit{v}$.We have to calculate horizontal and vertical component of the force

Formula:

$\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{[}\mathbf{r}\mathbf{-}\mathbf{h}\mathbf{]}}^{\mathbf{2}}}\mathbf{\text{ }}\mathbf{=}\sqrt{\mathbf{2}\mathbf{r}\mathbf{h}\mathbf{-}{\mathbf{h}}^{\mathbf{2}}}$

There are four forces are acting on the wheel. The applied force$\left(F\right)$as given and the downward force$\left(mg\right)$that is due to its weight at the centre. Now if we consider a point at the surface of the wheel, there are two extra forces that are a horizontal to the left and a vertical force to the top.

As the wheel is continuously so the normal force no longer is valid. So we consider the case of net torque to be zero.

If the minimum force is applied the wheel does not accelerate, so both the total force and the total torque acting on it are zero.

We calculate the torque around the step corner as a pivot point in the second diagram (above right) and the third diagram indicates that the distance from the line of F to the corner is$r−h$, where r is the radius of the wheel and h is the height of the step. The distance from the line of mg to the corner is

$\sqrt{r^2−{[r−h]}^2}\text{ }=\sqrt{2rh−h^2}$

Now, applying the net torque as zero at the pivot point to meet our conditions of balancing the forces, we get that

$F[r-h]-mg\text{ }\sqrt{2rh-h^2}\text{ }=0$

The solution for F is

$\begin{array}{c}F=\frac{\sqrt{2rh-h^2}}{[r-h]}mg\text{ }\mathrm{..................................}\left(a\right)\\ =\frac{\sqrt{2(6.00×{10}^{-2}m)(3.00×{10}^{-2}m)-{(3.00×{10}^{-2}m)}^2}}{(6.00×{10}^{-2}m)-(3.00×{10}^{-2}m)}(0.800kg)(9.80m/s^2)\\ =13.6N\end{array}$

From equation (a), we can say that if the height of the pivot point is increased, then the force that must be applied also goes up. Below is the plot F/mg as a function of the ratio h/ r. The required force increases rapidly as h/ r —>1.

Hence, the magnitude of (constant) force $\stackrel{→}{f}$ applied horizontally at the axle of the wheel is $13.6N$.