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. (asked in MU De…

Answer: \(P=648.5\) N at 30°. Minimum \(P=637.6\) N, acting at 40.54° above the horizontal (perpendicular to the line joining the corner of the step to the centre).

\(W=100\times9.81=981\) N. Let O be the centre and E the top corner of the obstruction.

Geometry: vertical distance of O above E \(=50-12=38\) cm. Horizontal distance between O and E \(=\sqrt{50^2-38^2}=\sqrt{1056}=32.50\) cm. If \(\alpha\) is the angle of OE with the vertical, \(\cos\alpha=38/50=0.76\), \(\alpha=40.54^\circ\).

Condition for just starting: the roller is about to lift off the floor, so the floor reaction is zero. The roller is then acted on by W, P and the reaction at E only. Take moments about E (clockwise = tending to roll over the step).

• W (downward, 32.50 cm to the left of E): anticlockwise moment \(981\times32.50=31\,878\) N cm.
• P at O: horizontal component \(P\cos30^\circ\) (arm 38 cm) and vertical component \(P\sin30^\circ\) (arm 32.50 cm), both clockwise: \(P(0.866\times38+0.5\times32.50)=49.16P\).

\[49.16P=31\,878\Rightarrow P=648.5\ \text{N}\]

Minimum P: the moment of P about E is largest for a given P when P is perpendicular to OE, with arm OE = 50 cm:

\[P_{min}\times50=981\times32.50\Rightarrow P_{min}=637.6\ \text{N}\]

Since OE makes \(\alpha=40.54^\circ\) with the vertical, a line perpendicular to OE makes 40.54° with the horizontal. So \(P_{min}=637.6\) N acting at the centre, up and to the right at 40.54° to the horizontal.