= . N Thus the reactions of the support are about N at K and N at K . ⊳ Example . A 3m long ladder weighing kg leans on a frictionless wall.
Its feet rest on the floor m from the wall as shown in Fig. . . Find the reaction forces of the wall and the floor.
Answer Fig. . The ladder AB is m long, its foot A is at distance AC = m from the wall. From Pythagoras theorem, BC = m.
The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces F and F of the wall and the floor respectively. Force F is perpendicular to the wall, since the wall is frictionless. Force F is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.
For translational equilibrium, taking the forces in the vertical direction, N – W = (i) Taking the forces in the horizontal direction, F – F = (ii) For rotational equilibrium, taking the moments of the forces about A, F − ( / ) W = (iii) Now W = g = × . N = . N From (i) N = . N From (iii) .
/ . N W From (ii) . N . N N The force F makes an angle α with the horizontal, tan , tan ( ) N F ≈ ⊳ .
MOMENT OF INERTIA We have already mentioned that we are developing the study of rotational motion parallel to the study of translational motion with which we are familiar. We have yet to answer one major question in this connection. What is the analogue of mass in rotational motion? We shall attempt to answer this question in the present section.
To keep the discussion simple,