A wheel of radius R and mass M is placed at t | vedclass

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(A, B) To analyze the torque $	au$ about point $Q$,we consider the forces acting on the wheel. The weight $Mg$ acts at the center of the wheel. The t

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If the force is applied at point $P$ tangentially then $	au$ decreases continuously as the wheel climbs

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(A, B) To analyze the torque $	au$ about point $Q$,we consider the forces acting on the wheel. The weight $Mg$ acts at the center of the wheel. The torque due to gravity about $Q$ is $	au_g = Mg \cdot d_{\perp}$,where $d_{\perp}$ is the perpendicular distance from $Q$ to the line of action of gravity. As the wheel climbs,this distance changes.For option $A$: If a constant force $F$ is applied tangentially at $P$,the lever arm of the force $F$ about $Q$ changes as the wheel rotates. As the angle $	heta$ (the angle of the radius to the contact point with the horizontal) increases,the perpendicular distance of the force $F$ from $Q$ decreases,causing the torque $	au$ to decrease continuously.For option $B$: If the force is applied normal to the circumference at $X$,the force vector always passes through the center of the wheel. The lever arm of this force about $Q$ remains constant as the wheel climbs,thus the torque $	au$ is constant.For option $C$: If the force is applied normal to the circumference at $P$,the force vector passes through the center of the wheel. The torque about $Q$ is not zero because the force vector does not pass through $Q$.For option $D$: If the force is applied tangentially at $S$,the force vector is horizontal. The torque about $Q$ is non-zero,but depending on the magnitude of $F$,it may or may not be sufficient to climb the step. However,the statement that it 'never' climbs is not necessarily true for all $F$.

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