A stepladder of negligible weight is constructed as shown in Figure P12.57 where

Question

A stepladder of negligible weight is constructed as shown in Figure P12.57 where x = 2.30 m. A painter of mass 77.0 kg stands on the ladder 3.00 m from the bottom. 2.00 m 3,00 m 2.00 m Figure P12.57 Assuming the floor is frictionless, find the following. (Suggestion: Treat the ladder as a single object, but also each half of the ladder separately.) (a) the tension in the horizontal bar connecting the two halves of the ladder (b) the normal forces at A and B (at A) (at B) (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half (rightward component) (upward component)

Explanation / Answer

Since there is no friction at the floor, the reactions at A and B are vertical. Furthermore, since the ladder is of "negligible weight," the vertical reactions sum to the painters weight:
Fa + Fb = 77.0kg * 9.8m/s² = 754.6 N

Summing the moments about B, we get
M = 0 = 77.0kg * 9.8m/s² * (5/8)*2.30m - Fa*2.30m
Fa = 471.625 N (b)
so Fb = 754.6N - Fa = 282.975 N (b)

Cut the ladder vertically in half and consider the right side.
Sum the moments about the midpoint of the bar -- any forces in the bar create no moment about that point.
M = 0 = Fb*1.1m - Fch*2.00m*sin
where Fch is the horizontal force at C (any vertical component has no moment action)
and = arccos(1.15/2.00) = 54.9º. So
0 = 282.975N * 1.15m - Fch*2.00m*sin54.9º
Fch = 198.9 N part of (c) -- leftward component of right half on left = rightward component of left half on right (by Newton III)

Since there is no horizontal force at the floor, the tension in the rod must be
Frod = 198.9 N (a)

I'm not sure how to get the vertical force at C except to assume that the bar carries no shear (vertical force). Then analyzing the vertical forces (still on the right side of the ladder) gives
Fcy = -Fb = 282.975 N vertical part of (c), upward

Hope this helps!

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