A 4 metre long ladder weighing 25 kg rests with its upper end against a smooth wall and lower end on rough ground. What should be the minimum coefficient of friction between the ground and the ladder for it to be inclined at 60? with the horizontal without slipping ? (Take $g = {10 \,m \,s^{-2}}$)
Here, length of ladder, $AB = 4 m$ $W $ = 25 kg at the C.G. (C) of the ladder, as shown in the figure. $? ABO = 60?, ? BAO = 30?$ Let $R_1$ be reaction of the wall, normal of the wall and $R_2$ be the reaction of the ground normal to the ground.
Force of friction $(f )$ between the ladder and the ground acts along $BO$. In equilibrium, $R_2 = W$ ...(i) $f = R_1 $ .....(ii) Taking moments about $B$, at equilibrium, $R_2 \times 0 - W \times BD + R_1 \times AO = 0 $ Using (i) and (ii), we get $ - R_2 \times BD + f \times AO = 0$ or $\frac{f}{R_2} = \frac{BD}{AO} = \frac{BC \, \cos \, 60^\circ}{AB \, \sin \, 60^\circ}$ or $\mu = \frac{2}{4\sqrt{3}} = \frac{1}{2\sqrt{3}} = 0.29$
The laws of motion, which are the keystone of classical mechanics, are three statements that defined the relationships between the forces acting on a body and its motion. They were first disclosed by English physicist and mathematician Isaac Newton.
Newton’s First Law of Motion
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion
Newton's 2nd law of motion deals with the relation between force and acceleration. According to the second law of motion, the acceleration of an object as built by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Newton’s Third Law of Motion
Newton's 3rd law of motion states when a body applies a force on another body that there is an equal and opposite reaction for every action.
