Problem: A uniform ladder of 7 metres rests against a vertical wall with which it makes an angle of 45* ; the coejicient of friction between the ladder and the wall is I /3 and that between ladder and the floor is I/2. If a man, whose weight is one-half of that of the ladder, ascends it, how high will it be when the ladder slips ? (Appeared - Cambridge University) Solution. Given. Length of the ladder =7 m Angle, at which the ladder stands theta = 45° Coeflicient of friction between the ladder and wall, mu(w) = 1/3 Coefficient of friction between the ladder and floor, mu(f) Let W=L0ad of the ladder Load of theman = W/2 z Let Rf·= Normal reaction at A, Ff = Frictiona1 force at A, Rw = Norn1al reaction at B, and Fw = Frictional force at B. Let M be the position of the man, when the ladder is atthe point of sliding, such that AM = x. Resolving
the forces horizontally Rw = Fy = mu(f) X Rf, because Ff = mu(f) X Rf) = 1/2Rf (mu(f), given) or Rf = 2Rw ...(i) Resolving the forces vertically, Rf + Fw = W+ W/2 = 3/2 W We know that Fw = mu(w) * Rw = 1/3 Rw, because mu(w) = 1/2 is given. Substituting the values of Fw, and Rw from equations (i) in equation (ii), 2Rw + 1/3 Rw = 3/2 W Therefore Rw = 2 * 3/7 W = 9/14W ------(iii) We know that Fw = 1/3, Rw = 1/3 * 9/14 = 3/14 W Holding moments about A, and equating them, W * 3.5 cos 45 degree + W/2.x cos 45 degree = Rw * 7 sin 45 degree + Fw * 7 cos 45 degree Subsitituting the values of Fw and Rw from expresions 3 and 4 we get W * 3.5 * cos 45 degree + W/2 . x cos 45 degree = 9/14 W * 7 sin 45 degree + 3/14W * 7 cos 45 degree. Dividing either sides by (Wsin45) we get 3.5 + x/2 = 9/2 + 3/2 = 6 Therefore the answer is x = 5m.