Determine the force P applied at 30° to the horizontal just necessary to start a roller having 50 cm radius over a 12 cm high obstruction, if the roller is of 100 kg mass. Also find the magnitude and direction of P when it is minimum.[Figure: roller…

Given: W = 100 × 9.81 = 981 N, r = 50 cm, h = 12 cm. When the roller is just about to climb, the floor reaction becomes zero; the forces are W, P and the reaction at the corner K of the step.

Geometry: vertical distance of C above K = 50 − 12 = 38 cm; horizontal distance = \(\sqrt{50^2 - 38^2} = \sqrt{1056} = 32.50\) cm. Line CK makes \(\tan^{-1}(38/32.5) = 49.46^\circ\) with the horizontal.

Moments about K (eliminates the corner reaction):

\(P\cos30^\circ \times 38 + P\sin30^\circ \times 32.50 = 981 \times 32.50\)

\(P(32.91 + 16.25) = 31\,879 \Rightarrow P = \dfrac{31879}{49.16} = 648.5\) N

Minimum P: P is minimum when its moment arm about K is maximum, i.e. when P is perpendicular to CK (arm = r = 50 cm):

\(P_{min} \times 50 = 981 \times 32.50 \Rightarrow P_{min} = 637.6\) N

Direction: perpendicular to CK, i.e. at 90° − 49.46° = 40.54° above the horizontal.

Answer: P = 648.5 N (at 30°); \(P_{min}\) = 637.6 N acting at 40.54° to the horizontal (up and to the right).