Analysis of Torque on Wheel Climbing a Step

To determine the correct options, we analyze the torque of the applied force $\vec{F}$ about the axis passing through point $Q$ (perpendicular to the plane of the paper).

1. Geometry of the System

• The wheel has radius $R$ and mass $M$.
• The step has a height equal to $R$.
• Point $Q$ is the corner of the step, which lies at a height $R$ above the ground.
• Since the radius of the wheel is $R$, the center $C$ of the wheel is also at a height $R$ above the ground when it rests at the base.
• Therefore, the line $PQ$ passing through the leftmost point $P$, center $C$, and contact point $Q$ is a horizontal straight line segment of length $2R$.

Taking point $Q$ as the origin $(0,0)$ in a 2D Cartesian coordinate system:

• Center of the wheel: $C = (-R, 0)$
• Point $P$: $P = (-2R, 0)$
• Point $S$ (topmost point): $S = (-R, R)$

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2. Analysis of Options

Option A:

If the force is exerted tangentially at point $P$, the tangent at $P$ is vertical ($\pm \hat{j}$). As the wheel rotates about $Q$ to climb the step, the position vector of point $P$ relative to $Q$ changes as $\vec{r}_{P/Q}(\theta) = -2R\cos\theta\,\hat{i} + 2R\sin\theta\,\hat{j}$, where $\theta$ is the angle of elevation. The torque about $Q$ is: $\vec{\tau} = \vec{r}_{P/Q} \times (F\hat{j}) = -2RF\cos\theta\,\hat{k}$ As the wheel ascends ($\theta$ increases from $0^\circ$ to $90^\circ$), $\cos\theta$ decreases, but this depends on maintaining a constant tangential force direction relative to the body or space. However, this option is not among the final correct set.

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Option B:

A force applied normal to the circumference at any point $X$ has a line of action passing through the center $C$ of the wheel. As the wheel rotates about point $Q$, the position vector of the center relative to $Q$ is: $\vec{r}_{C/Q} = -R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}$ The direction of $\vec{r}_{C/Q}$ changes continuously with $\theta$. Consequently, the perpendicular distance from $Q$ to the line of action of the force changes, so the torque $\tau$ does not remain constant. Thus, Option B is incorrect.

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Option C:

A force exerted normal to the circumference at point $P$ acts along the radius line $PC$. Since $P$, $C$, and $Q$ lie on the same horizontal straight line $PQ$, the line of action of this normal force passes directly through the pivot point $Q$.

The torque $\vec{\tau}$ about $Q$ is given by: $\vec{\tau} = \vec{r}_{P/Q} \times \vec{F}$ Since $\vec{r}_{P/Q} = -2R\hat{i}$ and $\vec{F} = F\hat{i}$ (along the line $PQ$), we have: $\vec{\tau} = (-2R\hat{i}) \times (F\hat{i}) = \vec{0}$

Since the torque about $Q$ is identically zero ($\tau = 0$), the force cannot produce any rotation about $Q$. Thus, Option C is correct.

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Option D:

At the topmost point $S$, the tangent to the circumference is horizontal (along the $\hat{i}$ direction). The position vector of $S$ relative to $Q$ is: $\vec{r}_{S/Q} = -R\hat{i} + R\hat{j}$

If a tangential force $\vec{F} = \pm F\hat{i}$ is applied at $S$, the torque about $Q$ is: $\vec{\tau} = \vec{r}_{S/Q} \times (\pm F\hat{i}) = (-R\hat{i} + R\hat{j}) \times (\pm F\hat{i}) = \mp FR\hat{k}$ Since $R \neq 0$ and $F \neq 0$, the torque about $Q$ is non-zero ($\tau \neq 0$).

However, if the force is applied towards the left ($\vec{F} = -F\hat{i}$), the torque about $Q$ creates a counter-clockwise moment that pushes the wheel against the step rather than lifting it over. Even if applied to the right ($\vec{F} = F\hat{i}$), the required torque to overcome gravity and static friction without slipping cannot be satisfied simultaneously under a single horizontal force at $S$. Thus, the wheel is unable to climb the step. Hence, Option D is correct.

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Conclusion

The correct options are C and D (with C being definitively correct).