A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre ‘O’ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is :

Solution

Physics sir · 3 steps

Take torques about the step edge: the contact forces there drop out, and only F and Mg remain.

1. Pick the pivot. Call the step edge P. When the cylinder just starts to rise, the floor stops pushing on it, so its normal force is zero. The only other contact forces act at P, so they give no torque about P.
2. Lever arms. F acts at O perpendicular to OP, so its lever arm is the full OP = R. Weight Mg acts down through O, so its lever arm is the horizontal distance from O to P. O sits R - a above P, which makes that distance √(R² - (R-a)²).

   F × R = Mg√(R² - (R-a)²)

3. So. Dividing by R gives F = Mg√(1 - ((R-a)/R)²), which is option (a).

The idea to keep. To roll something over an edge, take torques about that edge. The lever arm of the weight is the horizontal distance from the centre to the edge, not the step height.

Answer from the NTA final answer key. Solution by PrepWiser.