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 73.0 kg stands on the ladder 3.00 m from the bottom. 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

The normal force at A will be somewhat more thanhalf of the weight of the painter (about 703 N) and the normal forceat B somewhat less than half. For the tension in the tie bar we expect about 100 N.
Categorize: Equilibrium of the whole ladder can tell us supportforces but not the tension in the tie bar, because it is internal to theladder. We must then also consider equilibrium of one side of theladder.
Analyze: If we think of the whole ladder, we can solve part (b).
(b) The painter is 3/4 of the way up the ladder, so the lever arm
of her weight about A is 3/4 (1.00 m) = 0.750 m
Fx = 0: 0 = 0
Fy = 0: nA 715.4 N + nB = 0
tA = 0: nA (0) (715.4 N)(0.750 m) + nB (2.30 m) = 0
Thus, nB = (715.4 N) (0.750 m)/(2.30 m) = 233.28 N
and nA = 715.4 N 233.28 N = 482.12 N
Now consider the left half of the ladder. We know the directionof the bar tension, and we make guesses for the directions of the components of the hinge force. If a guess is wrong, the answer will be negative. The side rails make an angle with the horizontal
= cos1(1/4) = 75.5°
Taking torques about the top of the ladder, we have
tc = 0: (482.12 N)(1.00 m) + T(2.00 m) sin 75.5° + (715.4 N)(0.250 m) = 0
Fx = 0: T – Cx = 0
Fy = 0: 482.12 N 715.4 N + Cy = 0
(a) From the torque equation,
T ={(482.12)(1.00m)-(715.4)(.250m)}/2.00xsin75.5=156.62N up
(c) From the force equations, Cx = T = 156.62 N
Cy = 715.4 N 482.12 N = 233.28 N up
This is the force that the right half exerts on the left half. The force that the left half exerts on the right half has opposite components:
156.62 N to the right, and 233.28 N down.

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