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 1.90 m. A painter of mass 70.0 kg stands on the ladder 3.00 m from the bottom. 200 m 200 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 127.929 N (b) the normal forces at A and B 434,875 N at A) 248 N Your response is within 10% of the comect value. This may be due to roundoff enror, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. (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 127.929 N rightward component 248 N Your response is within 10% of the comect value. This may be due to roundoff enror, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. (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 = 70.0kg * 9.8m/s² = 686 N

Summing the moments about B, we get
M = 0 = 70.0kg * 9.8m/s² * (5/8)*1.90m - Fa*1.90m
Fa = 428.75 N (b)
so Fb = 686N - Fa = 257.25 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(1.1/2.00) = 56.6º. So
0 = 257.25N * 1.1m - Fch*2.00m*sin56.6º
Fch = 169.478 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 = 169.478 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 = 257.25 N vertical part of (c), upward

Hope this helps!

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