91影视

When an object starts rotating upon the application of force, the product of this force (F) and the perpendicular distance (r) from the axis of rotation to the line of action of the force is termed as torque, i.e.,

\(\tau = rF\)

Here, r is also known as the lever arm.

In this problem,the wheel rotates about the point of contact of the step and the wheel on the action of the horizontal force acting on it. Thus, the torque due to this force will be the product of this force andthe perpendicular distance from the point of contact to the horizontal line of action of the force.

The mass of the wheel is M.

The radius of the wheel is R.

The horizontal force acting on the wheel is F.

The wheel will roll about point B. Two forces acting on the wheel that can exert torque on it are the weight of the wheel (Mg) and the horizontal force F. This is shown in the figure below:

On applying the Pythagoras theorem in\(\Delta OAB\), you will get:

\(\begin{aligned}{c}O{B^2} = O{A^2} + A{B^2}\\A{B^2} = O{B^2} - O{A^2}\\ = {R^2} - {\left( {R - h} \right)^2}\\AB = \sqrt {{R^2} - {{\left( {R - h} \right)}^2}} \end{aligned}\)

The torque acting on the wheel about point B due to the horizontal force F is:

\({\tau _1} = \left( {R - h} \right)F\)

The direction of this torque will be along the clockwise direction.

The torque acting on the wheel about point B due to its weight is:

\(\begin{aligned}{c}{\tau _2} = \left( {AB} \right)Mg\\ = Mg\sqrt {{R^2} - {{\left( {R - h} \right)}^2}} \end{aligned}\)

The direction of this torque will be along the anti-clockwise direction.

Taking the clockwise direction to be positive, the net torque acting on the wheel is:

\(\begin{aligned}{l}\sum {\tau = {\tau _1}} - {\tau _2}\\\;\;\;\;\;\; = \left( {R - h} \right)F - Mg\sqrt {{R^2} - {{\left( {R - h} \right)}^2}} \end{aligned}\)

The force F will be minimum when the net torque acting on the wheel becomes zero, i.e.,

\(\begin{aligned}{c}\sum {\tau = 0} \\\left( {R - h} \right)F - Mg\sqrt {{R^2} - {{\left( {R - h} \right)}^2}} = 0\\F = \frac{{Mg\sqrt {{R^2} - {{\left( {R - h} \right)}^2}} }}{{\left( {R - h} \right)}}\\ = \frac{{Mg\sqrt {{R^2} - \left( {{R^2} + {h^2} - 2Rh} \right)} }}{{\left( {R - h} \right)}}\\ = \frac{{Mg\sqrt {2Rh - {h^2}} }}{{\left( {R - h} \right)}}\end{aligned}\)

Thisis the minimum force Fthat needs to be acted on the axle of the wheel so that it can climb the step.