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 = 1.70 m. A painter of mass 67.0 kg stands on the ladder 3.00 m from the bottom.

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

consider the first half of the ladder.

angle made by the left half with horizontal=arccos(0.5*1.7/4)=77.731 degrees

forces acting on the left half:

1.normal force at A, in vertically upward.

let its magnitude be F1.

2. tension in the horizontal bar , horizontally to to the right

let magnitude of the force be F2.

3. weight of the painter=67*9.8=656.6 N

in vertically downward direction

4. reaction force at C

let horizontal component be F3 and to the left

and vertical component be F4 and in vertically upward direction

(note: directions apart from that of normal force is taken randomly. if the value comes out to be negative, then direction will be opposite to that assumed)

balancing forces in horizontal direction:

F2=F3...(1)

balancing forces in vertical direction:

F1+F4=656.6 N....(2)

balancing torque about point A:

F2*2*sin(77.731)+656.6*3*cos(77.731)=F3*4*sin(77.731)+F4*4*cos(77.731)

==>1.9543*F2+418.58=3.9086*F3+0.85*F4

using F2=F3,

0.85*F4+1.9543*F3=418.58...(3)

now as reaction force at C is perpendicular to the left half of the ladder,

F4/F3=tan(77.731)

==>F4=4.5984*F3...(4)

using equation 4 in equation 3,

we get

0.85*4.5984*F3+1.9543*F3=418.58

==>F3=318.58/(0.85*4.5984+1.9543)=54.338 N

then F4=249.87 N

then F1=656.6-F4=406.73 N

F2=F3=54.338 N

answers are:

tension in the horizontal bar=F2=54.338 N

b)normal force at A=F1=406.73 N

c) component of reaction force at the single C that the left ladder exerts on right half:

rightward component=F3=54.338 N

upward component=F4=249.89 N

now consider the right half:

balancing vertical forces:

normal force magnitude=vertical component of reaction force at C=249.89 N

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