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 wherex-2.30 m. A painter of mass 77.0 kg 2.00 m 3.00 m 2.00m 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 Your response differs from the correct answer by more than 10%. Double check your calculations. (b) the normal forces at A and B Your response differs from the correct answer by more than 10%, Double check your calculations. (at A) Your response differs from the correct answer by more than 10%, Double check your calculations. (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 Your response differs from the correct answer by more than 10%, Double check your calculations. (rightward component) Your response differs from the correct answer by more than 10%. Double check your calculations. (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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