Problem 2B-14**: A ladder against a wall

Consider a uniform ladder of length 2L and mass M that leans against a wall as shown in Fig. 1. The angle ABO is denoted as θ, and the maximum coefficient of static friction between the ladder and the floor as κS.

(a) Suppose that there is no friction between the ladder and the wall. If a weight of sufficiently large mass m is placed on the ladder, how high can the weight be put until the ladder starts to slip?

(b) If there is also static friction between the ladder and the wall, and the maximum coefficient of static friction is also κS, how high can the weight be put before the ladder starts to slip?

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Solution:

(a) The situation is depicted in Fig. 2. Point C is the center of mass of the ladder, and the weight is placed at point P. The distance between P and B is denoted as x. The wall gives the ladder a horizontal resistance force F1 at point A. The floor gives the ladder a vertical resistance force F3 and a horizontal static friction force F2.

For the ladder to be stationary, all the horizontal forces need to cancel, and so are the vertical forces. For the horizontal forces, we have

F1 + F2 = 0.

Thus

|F1| = |F2|.                                 (1)

The cancellation of the vertical forces leads to

|F3| = mg + Mg = (m + M)·g.

At the critical juncture when the ladder starts to slide, we must have

|F2| = κS·|F3| = κS·(m + M)·g.

Thus Eq. (1) becomes

|F1| = |F2| = κS·(m + M)·g .                                 (2)

The other condition for the ladder to be stable is the cancellation of torques around a reference point. The reference point can be either point A, B or C. We choose point B as the reference point. The torque around B generated by force F1 is denoted as τ1; this torque goes into paper at point B. Torques generated by forces F2 and F3 are zero. The torque due to the weight of the ladder at point C is denoted as τM; this torque comes out of paper at point B. The torque due to the weight at point P is denoted as τP; this torque comes out of paper at point B. We must have

τ1 + τM + τP = 0,

or

|τ1| = |τM| + |τP|.                                 (3)

The magnitude of torque τ1 can be calculated with the help of Eq. (2) as

|τ1| = 2L·|F1|·sinθ = 2κS·Lg·(m + M)·sinθ.                                 (4)

The magnitudes of torques τM and τP are

|τM| = L·MG·sin{(π/2) - θ} = LMg·cosθ ,                                 (5)

and

|τP| = x·mg·sin{(π/2) - θ} = xmg·cosθ,                                 (6)

respectively. Substituting those magnitudes of torques from Eqs. (4), (5) and (6) into Eq. (3), we get

xmg·cosθ + LMg·cosθ = 2κS·Lg·(m + M)·sinθ.

Solving for x, we get

x = {2κSLM·(m + M)·sinθ - LM·cosθ}/(m·cosθ) = L·{2κS·tanθ·(1 + M/m) - (M/m)}.

(b) When there is static friction between the wall and the top of the ladder at point A, a force F4 must be added to point A as shown in Fig. 3. The magnitude of force F4 is

|F4| = κS·|F1|.                                 (7)

The balance of forces at the horizontal direction still yields

F1 + F2 = 0,

and thus

|F1| = |F2|= κS· |F3| .                                 (8)

The balance of force at the vertical direction now becomes

|F3| = (M + m)·g - |F4|.

Upon the substitution of Eq. (7) into the above equation, we get

|F3| = (M + m)·g - κS·|F1|.                                 (9)

Substituting Eq. (9) into Eq. (8), we get

|F1| = κS·g·(m + M) - κS2· |F1|.

Solving the above equation for |F1| we get

|F1| = κS·g·(m + M)/(1 + κS2).                                 (10)

The torque generated by force F4 around point B is denoted as τ4. It comes out of paper at point B, and its magnitude is

|τ4| = 2L·|F4|·cosθ.

Using Eqs. (7) and (10), the above equation becomes

|τ4| = 2κS2·Lg·(m + M)/(1 + κS2)·cosθ.                                 (11)

|τ1| of Eq. (4) must be recalculated by using new |F1| of Eq. (10). It becomes

|τ1| = 2κS·Lg·{(m + M)/(1 + κS2)}·sinθ .                                 (12)

Eq. (3) must be modified to read

|τ1| = |τM| + |τP| - |τ4| .                                 (13)

Substituting Eqs. (12), (5), (6) and (11) into Eq. (13), we get

2κS·Lg·{(m + M)/(1 + κS2)}·sinθ = xmg·cosθ +LMg·cosθ - 2κS2·Lg·(m + M)/(1 + κS2)·cosθ.

Rearranging terms, we get

xmg·cosθ = 2κS·Lg·{(m + M)/(1 + κS2)}·sinθ - LMg·cosθ + 2κS2·Lg·(m + M)/(1 + κS2)·cosθ.

Therefore,

x = [2κS·Lg·{(m + M)/(1 + κS2)}·sinθ - LMg·cosθ + 2κS2·Lg·{(m + M)/(1 + κS2)}·cosθ]/ ( mg·cosθ),

or

x = 2κS·L·{(1 + M/m)/(1 + κS2)} ·tanθ - L·(M/m)· + 2κS2·L·{(1 + M/ m)/(1 + κS2)}.