91影视

The torque created by the smallest horizontal forceat the edge of a step precisely balances the torque created by the weight of the wheel around the same axis. The typical force no longer exists since the wheel just begins to move at this moment.

Therefore, we have,

$\begin{array}{r}{蟿}_F+{蟿}_{\text{weight}}=0\\ 鈭�{\mathbf{d}}_{\mathrm{F}}鈰�\mathbf{F}+{\mathbf{d}}_{\mathrm{w}}鈰�\mathbf{mg}=\mathbf{0}\end{array}$ (1)

Where${\mathbf{d}}_{\mathbf{F}}$is the moment of the arm of force and${\mathbf{d}}_{\mathbf{w}}$is the moment of arms of weight.

Here we have given the mass of the wheel is$\mathrm{m}$.

R is the radius of the wheel

h is the curb height of the wheel.

(a)

From equation (1)

We have

$\mathrm{F}=\left(\frac{{\mathrm{d}}_{\mathrm{w}}}{{\mathrm{d}}_{\mathrm{F}}}\right)\mathrm{mg}$ (2)

Here we have to find the magnitude of the force is applied to the center of mass.

So, from the above figure,

In this case, ${\mathrm{d}}_{\mathrm{F}}={\mathrm{d}}_{\mathrm{a}}$. So, equation (2) becomes

$\mathrm{F}=\left(\frac{{\mathrm{d}}_{\mathrm{w}}}{{\mathrm{d}}_{\mathrm{a}}}\right)\mathrm{mg}\left(3\right)$

From the figure,${\mathrm{d}}_{\mathrm{a}}=\mathrm{R}-\mathrm{h}$

Now, to findfrom the figure,

Apply Pythagoras theorem:

$\begin{array}{r}{\mathrm{d}}_{\mathrm{w}}=\sqrt{{\mathrm{R}}^2鈭�{\mathrm{d}}_{\mathrm{a}}^2}\\ =\sqrt{{\mathrm{R}}^2鈭�(\mathrm{R}鈭�\mathrm{h}{)}^2}\\ =\sqrt{\mathrm{h}(2\mathrm{R}鈭�\mathrm{h})}\end{array}$

So, from equation (3),

$F=\frac{\sqrt{h(2R鈭�h)}}{R鈭�h}mg$

(b)

Here we have to find the magnitude of the force is applied at the top of the wheel.

So, from the above figure,

In this case, ${\mathrm{d}}_{\mathrm{F}}={\mathrm{d}}_{\mathrm{b}}$So, equation (2) becomes

$\mathrm{F}=\left(\frac{{\mathrm{d}}_{\mathrm{w}}}{{\mathrm{d}}_{\mathrm{b}}}\right)\mathrm{mg}\left(4\right)$

From the figure, ${\mathrm{d}}_{\mathrm{b}}=2\mathrm{R}-\mathrm{h}$

Now, to find ${\mathrm{d}}_{\mathrm{w}}$from the figure,

Apply Pythagoras theorem:

$\begin{array}{r}{\mathrm{d}}_{\mathrm{w}}=\sqrt{{\mathrm{R}}^2鈭�{{\mathrm{d}}_{\mathrm{b}}}^2}\\ =\sqrt{{\mathrm{R}}^2鈭�(\mathrm{R}鈭�\mathrm{h}{)}^2}\\ =\sqrt{\mathrm{h}(2\mathrm{R}鈭�\mathrm{h})}\\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(4\right),\\ \mathrm{F}=\frac{\sqrt{\mathrm{h}(2\mathrm{R}鈭�\mathrm{h})}}{2\mathrm{R}鈭�\mathrm{h}}\mathrm{mg}\end{array}$

(c)

The force applied at the center of the wheel is $\mathrm{F}=\frac{\sqrt{\mathrm{h}(2\mathrm{R}-\mathrm{h})}}{\mathrm{R}-\mathrm{h}}\mathrm{mg}$

The force is appliedat the top of the wheel is $\mathrm{F}=\frac{\sqrt{\mathrm{h}(2\mathrm{R}-\mathrm{h})}}{\mathrm{R}-\mathrm{h}}\mathrm{mg}$

From the above equation, we can say that the force applied at the center of the wheel is larger than the force applied at the top of the wheel.

So, less force is required in the second case which is when the force is applied on top of the wheel.