A 20 kg step-ladder stands on a horizontal frictionless surface. What is the tension in the crossbar, which has negligible mass, and which is 1.0 m from the bottom of each side of the 4.0 m ladder? Assume the two halves of the ladder are identical. The angle between either side of the ladder and the vertical is 15°.

Consider first the free-body diagram of the entire ladder. There are no horizontal forces because there is no friction. The ladder is uniform, so the weight acts at the center of mass, which is halfway up the ladder and halfway between the two legs.

One approach is to say that the ladder is symmetric, so each normal force is half the weight of the ladder. If you don't like this argument, simply take torques about one of the points where the ladder touches the floor. This will give you an equation saying that one normal force is equal to half the ladder's weight. Either way:

N₁ = N₂ = 0.5 mg = 98 N.

Now consider the free-body diagram of one half of the ladder.

Taking torques around the top of the ladder eliminates the unknown contact force (F) coming from the other half of the ladder, and gives (taking clockwise to be positive):

Στ = 0

(4.0 m)*N₁*sin(15°) - (2.0 m)*(0.5 mg)*sin(15°) - (3.0 m)*(T)*sin(75°) = 0.

Solve for the tension in the crossbar:

T

=

mg 2 sin(34°)

= 12.3 N