# How To Calculate A Slipping Ladder Parameters

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 the

man
= 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.