 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?

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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 F4 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 -F4. Thus the left-right symmetry will be broken. The only consistent situation is for F4 to be zero. The horizontal force that the left leg receive from the right leg at point A is denoted as F1 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 F2 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 F3. The balance of forces at the horizontal direction says that

|F1| = |F2| .                                 (1)

The balance of forces at the vertical direction says that

|F3| = Mg.                                 (2)

For the left leg to be stable, the torques around point B should sum to zero. The torque generated by force F1 around point B is denoted as τ1. This torque comes out of the paper perpendicularly at point B, and has a magnitude of

|τ1| = |F1|·L·sinθ.                                 (3)

The torque generated by the force F2 around point B is denoted as τ2. This torque goes into the paper perpendicularly at point B, and has a magnitude of

|τ2| = |F2|·h.                                 (4)

The torque due to the mass of the left leg at point C is denoted as τM. This torque goes into the paper perpendicularly at point B, and has a magnitude of

|τM| = Mg·(L/2)·cosθ.                                 (5)

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

τ1 + τ2 + τM = 0.

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

|τ1| = |τ2| + |τM|,

or as

|τ1| = |τ2| + |τM| .

Inserting Eq. (3), Eq. (4) and Eq. (5) into the above equation and replacing F1 by F2 according to Eq. (1), we get

|F2|·L·sinθ = |F2|·h + Mg·(L/2)·cosθ.

Solving for |F2|, we get

|F2| = Mg·(L/2)·cosθ/(L·sinθ - h).

This |F2| 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

F1 = F1.                                 (6)

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

F1·L·sinθ = F4·L·cosθ + F2·h + Mg·(L/2)·cosθ.

Using Eq. (6) to eliminate F1 from the equation, we have

F2·(L·sinθ - h) = F4·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

F4·L·cosθ + F1·L·sinθ = F2·h + mg·H + Mg·(L/2)·cosθ .

Again using Eq. (6) to write F1 as F2, we get

F2·(L·sinθ - h) = - F4·L·cosθ + mg·H + (MgL/2)·cosθ .                                 (8)

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

2F2·(L·sinθ - h) = mg·H + Mg·L·cosθ.

Thus, we have the tension as

F2 = (mg·H + Mg·L·cosθ)/{2·(L·sinθ - h)}.