Equilibrium of a rigid body?

Question

There is a symmetrical A-shaped ladder. Both sides or the ladder are equal in length. This ladder is standing on a frictionless horizontal surface and only the crossbar(which has a negilible mass) of the "A" keeps the ladder from collapsing. The ladder is uniform and has a mass of 20kg. The ladder makes a 30 deg angle and the legs are 4m, with the crossbar coming up 1m along the legs. Applying the conditions of equilibrium of a rigid boy, determine the tension in the crossbar of the ladder.

Answer 1

Taking the ladder as a whole, each leg is supporting half of the ladder's weight of mg, pushing up vertically.  Because the mass of the crossbars are negligible, any forces acting on the ends must be entirely horizontal, otherwise they would be accelerating upward or downward.  The center of mass of each leg of the ladder is at its center and pulling downward.  At the apex of the ladder is another point where forces can act, horizontally, of unknown and, for this problem, unimportant magnitude.  By computing the torques around the apex of the ladder, we can null the effects of these forces because they have no lever arm to produce a torque about this point.

Because the crossbars "come up 1 m along the legs," presumably from the floor, they are 3 m from the apex and the point of contact with the floor is 4 m from the top.  Because the ladder makes an angle of 30 deg and is symmetrical, each leg leans 15 deg off vertical.  Three torques therefore apply to each leg.  Let us call the torques positive that tend to bring the legs together and those spreading the legs to be negative.  Then the torque at the point of contact with the floor will be -4 m * 10g N * sin 15 deg, the center of mass will be 2 m * 10g N * sin 15 deg, and the tension on the crossbar 3 m * T * cos 15 deg.  Because the ladder is in equilibrium, these three torques must add to zero.  Therefore

-4 m * 10g N * sin 15 deg + 2 m * 10g N * sin 15 deg + 3 m * T * cos 15 deg = 0 or

T = (2 m * 10 g N * sin 15 deg) / (3 m * cos 15 deg)

= 20/3 g tan 15 deg N*m

Using 9.8 m/sec² for g, this becomes

17.5 N total, or 8.75 N for each crossbar.