**Problem 2B-15**: A shaped ladder**

An A-shaped ladder is shown in Fig. 1. The length of each leg is L, and the mass of each leg is M. Each leg can be considered as a uniform rod, and forms an angle θ with the ground. A massless horizontal bar connects two legs. The bar is a distance h above the ground. We also assume that there is no friction between the legs and the ground.

(a) What is the tension in the horizontal bar?

(b) A weight of mass m is placed on the right-hand leg at a height of H as shown in Fig. 2. What is the tension in the horizontal bar?

Scroll down for
solution

**Solution:**

(a) The left leg and the right leg are in a perfectly symmetric position when no
weight is put on the ladder. The situation of the left leg is shown in Fig. 3.
The right leg cannot give the left leg any vertical force at point A. This is because
if a non-zero vertical force **F**_{4} is given to the left
leg at point A by the right leg, Newton's third law says that the left leg will give
the right leg a force -**F**_{4}. Thus the left-right symmetry
will be broken. The only consistent situation is for **F**_{4}
to be zero. The horizontal force that the left leg receive from the right leg at point A
is denoted as **F**_{1} in Fig. 3. The horizontal bar is attached
to the left leg at point D and the tension of the bar becomes a horizontal force
**F**_{2} acting on point D. The center of mass of the left leg is C.
The ground gives the left leg at point B a vertical force **F**_{3}.
The balance of forces at the horizontal direction says that

|**F**_{1}| =
|**F**_{2}| .
(1)

The balance of forces at the vertical direction says that

|**F**_{3}| = Mg.
(2)

For the left leg to be stable, the torques around point B should sum to zero. The torque
generated by force **F**_{1} around point B is denoted as
**τ**_{1}. This torque comes out of the paper perpendicularly
at point B, and has a magnitude of

|**τ**_{1}| =
|**F**_{1}|·L·sinθ.
(3)

The torque generated by the force **F**_{2} around point B is
denoted as **τ**_{2}. This torque goes into the paper
perpendicularly at point B, and has a magnitude of

|**τ**_{2}| =
|**F**_{2}|·h.
(4)

The torque due to the mass of the left leg at point C is denoted as **τ**_{M}.
This torque goes into the paper perpendicularly at point B, and has a magnitude of

|**τ**_{M}| =
Mg·(L/2)·cosθ.
(5)

The relation that torques around point B should sum to zero is represented by the relation

**τ**_{1} +
**τ**_{2} +
**τ**_{M} = 0.

Considering the orientations of the torques, the above equation can be written as

|**τ**_{1}| =
|**τ**_{2}| + |**τ**_{M}|,

or as

|**τ**_{1}| =
|**τ**_{2}| +
|**τ**_{M}| .

Inserting Eq. (3), Eq. (4) and Eq. (5) into the above equation and replacing
**F**_{1} by **F**_{2} according to Eq. (1), we get

|**F**_{2}|·L·sinθ =
|**F**_{2}|·h + Mg·(L/2)·cosθ.

Solving for |**F**_{2}|, we get

|**F**_{2}| =
Mg·(L/2)·cosθ/(L·sinθ - h).

This |**F**_{2}| is the tension of the horizontal bar.

(b) Once a weight of mass m is placed on the right leg, the symmetry between two legs is broken,
and vertical forces at point A can now appear. The forces acting on the left leg and on the right
leg are depicted separately in Fig. 4. In that figure only the magnitudes of forces are shown with
their directions implied by the arrows.

For the left leg, balance of force at the horizontal direction says

F_{1} = F_{1}.
(6)

The torques around point B should add up to zero, thus

F_{1}·L·sinθ =
F_{4}·L·cosθ +
F_{2}·h + Mg·(L/2)·cosθ.

Using Eq. (6) to eliminate F_{1} from the equation, we have

F_{2}·(L·sinθ - h) =
F_{4}·L·cosθ + (MgL/2)·cosθ .
(7)

For the right leg, we still have Eq. (6). The torques around point D should add up to zero, and this gives us the relation

F_{4}·L·cosθ + F_{1}·L·sinθ =
F_{2}·h + mg·H + Mg·(L/2)·cosθ .

Again using Eq. (6) to write F_{1} as F_{2}, we get

F_{2}·(L·sinθ - h) =
- F_{4}·L·cosθ + mg·H + (MgL/2)·cosθ .
(8)

Summing both side sof Eq. (7) and Eq. (8), weget

2F_{2}·(L·sinθ - h) = mg·H + Mg·L·cosθ.

Thus, we have the tension as

F_{2} = (mg·H + Mg·L·cosθ)/{2·(L·sinθ - h)}.