Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
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:
Torque equilibrium about the step corner: $$\tau_F = \tau_{Mg}$$
Perpendicular distance for $$F$$: $$r_{\perp, F} = R$$
Perpendicular distance for $$Mg$$: $$r_{\perp, Mg} = \sqrt{R^2 - (R-a)^2} = R\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$$
Equating torques: $$F \cdot R = Mg \cdot R\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$$
$$F = Mg\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$$
Create a FREE account and get:
Educational materials for JEE preparation